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Chemistry 1 Which of the following is the energy of a possible excited state of hydrogen
(A) minus68 eV
(B) minus34 eV
(C) +68 eV
(D) +136 eV
Solution (B) En = minus136
n2eV
Where n = 2 rArr E2 = minus340 eV
Topic Atomic Structure
Difficulty Level Easy [Embibe predicted high weightage]
2 In the following sequence of reactions
Toluene KMnO4rarr A
SOCl2rarr B
H2 Pdfrasl BaSO4rarr C
The Product C is
(A) C6H5CH3 (B) C6H5CH2OH
(C) C6H5CHO (D) C6H5COOH
Solution (C)
Topic Aldehyde and Ketone
Difficulty Level Easy [Embibe predicted high weightage]
3 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis
(A)
(B)
(C)
Solution (A)
Topic Hydrocarbons
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
4 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively
(A) 136 171 and 140
(B) 171 140 and 136
(C) 171 136 and 140
(D) 136 140 and 171
Solution (B) As Z
euarr ionic radius decreases for isoelectronic species
N3minus(Z
e)=
7
10
O2minus(Z
e)=
8
10
Fminus(Z
e) =
9
10
N3minus gt O2minusgt Fminus
Topic Periodicity
Difficulty Level Easy [Embibe predicted Low Weightage]
5 The color of KMnO4 is due to
(A) dminus d transition
(B) L rarr M charge transfer transition
(C) σ minus σlowasttransition (D) M rarr L charge transfer transition
Solution (B) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)
Topic Coodination compounds
Difficulty Level Easy [Embibe predicted high weightage]
6 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these
do not react to form oxides of nitrogen Reason The reaction between nitrogen and oxygen requires high temperature
(A) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion
(B) The Assertion is incorrect but the Reason is correct (C) Both the Assertion and Reason are incorrect
(D) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion
Solution (D) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion and
reason are correct and reason is correct explanation
Topic Group 15 ndash 18
Difficulty Level Moderate [Embibe predicted high weightage]
7 Which of the following compounds is not an antacid
(A) Cimetidine
(B) Phenelzine
(C) Rantidine
(D) Aluminium hydroxide
Solution (B) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as
antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a type of
medication that can control the acid levels in stomach Working of antacids Antacids counteract
(neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the symptoms of
heartburn and relieves pain
Topic Chemistry in everyday life
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
8 In the context of the Hall-Heroult process for the extraction of Al which of the following
statements is false
(A) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity
(B) Al3+ is reduced at the cathode to form Al
(C) Na3AlF6 serves as the electrolyte (D) CO and CO2 are produced in this process
Solution (C) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion
temperature and provides good conductivity
Topic Metallurgy
Difficulty Level Moderate [Embibe predicted high weightage]
9 Match the catalysts to the correct process
(A) A rarr ii B rarr i C rarr iv D rarr iii
Catalyst Process
A TiCl3 i Wacker process
B PdCl2 ii Ziegler ndash Natta polymerization C CuCl2 iii Contact process`
D V2O5 iv Deaconrsquos process
(B) A rarr ii B rarr iii C rarr iv D rarr i
(C) A rarr iii B rarr i C rarr ii D rarr iv
(D) A rarr iii B rarr ii C rarr iv D rarr i
Solution (A) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in
water in the presence of tetrachloropalladate (II) as the catalyst
In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with
arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred
In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts
including copper chloride(CuCl2)
In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are usually
used in combination with different organo aluminium co-catalyst
Topic Transition metals
Difficulty Level Easy [Embibe predicted high weightage]
10 In the reaction
The product E is
(A)
(B)
(C)
(D)
Solution (B)
11 Which polymer is used in the manufacture of paints and lacquers
(A) Glyptal (B) Polypropene
(C) Poly vinyl chloride (D) Bakelite
Solution (A) Glyptal is polymer of glycerol and phthalic anhydride
Topic Polymers
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
12 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]
+ is (py = pyridine) (A) 3
(B) 4 (C) 6
(D) 2
Solution (A) Complexes with general formula [Mabcd]+ square planar complex can have three
isomers
Topic Coordination compounds
Difficulty Level Moderate [Embibe predicted high weightage]
13 Higher order (gt 3) reactions are rare due to
(A) Increase in entropy and activation energy as more molecules are involved
(B) Shifting of equilibrium towards reactants due to elastic collisions
(C) Loss of active species on collision
(D) Low probability of simultaneous collision of all the reacting species
Solution (D) Probability of an event involving more than three molecules in a collision are remote
Topic Chemical kinetics
Difficulty Level Easy [Embibe predicted high weightage]
14 Which among the following is the most reactive (A) Br2 (B) I2
(C) ICl
(D) Cl2
Solution (C) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive
Topic Group 15 ndash 18
Difficulty Level Moderate [Embibe predicted high weightage]
15 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper deposited
at the cathode is (Atomic mass of Cu = 635 amu)
(A) 635 g
(B) 2 g
(C) 127 g (D) 0 g
Solution (A) 2F equiv 2 Eqs of Cu
= 2times635
2= 635 g
Topic Electrochemistry
Difficulty Level Easy [Embibe predicted high weightage]
16 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was
filtered and the strength of the filtrate was found to be 0042 N The amount of acetic acid adsorbed (per
gram of charcoal) is
(A) 36 mg
(B) 42 mg
(C) 54 mg
(D) 18 mg
Solution (D) Meqs of CH3COOH (initial) = 50 times006 = 3 Meqs
Meqs CH3COOH (final) = 50times 0042 = 21 Meqs
CH3 COOH adsorbed = 3 minus 21 = 09 Meqs
= 9times 10minus1 times 60 g Eq frasl times 10minus3 g
= 540 times10minus4 = 0054 g
= 54 mg
Per gram =54
3= 18 mg gfrasl of Charcoal
Topic Surface chemistry
Difficulty Level Moderate [Embibe predicted high weightage]
17 The synthesis of alkyl fluorides is best accomplished by
(A) Sandmeyerrsquos reaction
(B) Finkelstein reaction
(C) Swarts reaction
(D) Free radical fluorination
Solution (C) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2 CoF2 SbF3
CH3 Br+ AgF rarr CH3 minusF+ AgBr
The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro
hydrocarbons such as CCl2F2 is known as swarts reaction
Topic Alkyl and Aryl halides
Difficulty Level Easy [Embibe predicted high weightage]
18 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin
(A) 1
206
(B) 2
309
(C) 1
412
(D) 1
103
Solution (C) 2C8H7SO3minusNa++ Ca(aq)
2+ rarr (C8H7SO3)2Ca + 2Na+
2 moles (resin) = 412 g equiv 40 g Ca2+ ions
= 1 moles of Ca2+ ions
Moles of Ca2+ gfrasl of resin =1
412
Topic s-Block elements and Hydrogen
Difficulty Level Moderate (Embibe predicted Low Weightage)
19 Which of the vitamins given below is water soluble
(A) Vitamin D (B) Vitamin E (C) Vitamin K (D) Vitamin C
Solution (D) B complex vitamins and vitamin C are water soluble vitamins that are not
stored in the body and must be replaced each day
Topic Biomolecules
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
20 The intermolecular interaction that is dependent on the inverse cube of distance between the
molecules is
(A) Ion-dipole interaction
(B) London force
(C) Hydrogen bond (D) Ion-ion interaction
Solution (A) For Ion-dipole interaction
F prop μ (dE
dr)
μ is dipole moment of dipole
r is distance between dipole and ion
prop μd
dr(1
r2)prop
μ
r3
Ionminus ion prop1
r2
London force prop1
r6
Hydrogen bond minusDipole minus dipole prop1
r4
Topic Chemical Bonding
Difficulty Level Easy [Embibe predicted high weightage]
21 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)
The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is the
standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)
(A) 86600 +R(298) ln(16 times 1012)
(B) 86600 minusln(16times1012)
R(298)
(C) 05[2 times 86600minus R(298) ln(16 times 1012 )]
(D) R(298) ln(16 times 1012) minus 86600
Solution (C) ΔGreaco = minus2303 RTlog KP
= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac
o = 2ΔGf(NO2)o minus2ΔGf(NO)g
o
= 2ΔGf(NO2)
o minus2 times 866 times 103
2ΔGf(NO2)o = minusR(298)ln(16 times 1012) + 2 times 86600
ΔGf(NO2)o = 86600 minus
R(298)
2ln(16 times 1012)
= 05[2times 86600 minus R(298)ln(16times 1012)]
Topic Thermochemistry and thermodynamics
Difficulty Level Moderate [Embibe predicted high weightage]
22 Which of the following compounds is not colored yellow
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution (D) Cyanides not yellow
BaCrO4 minusYellowK3[Co(NO2)6]minus Yellow
(NH4)3[As(Mo3O10)4] minus Yellow
Zn2[Fe(CN)6] is bluish white
Topic Salt Analysis
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
23 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80) (A) 36
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Difficulty Level Easy [Embibe predicted high weightage]
3 Which compound would give 5-keto-2-methyl hexanal upon ozonolysis
(A)
(B)
(C)
Solution (A)
Topic Hydrocarbons
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
4 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively
(A) 136 171 and 140
(B) 171 140 and 136
(C) 171 136 and 140
(D) 136 140 and 171
Solution (B) As Z
euarr ionic radius decreases for isoelectronic species
N3minus(Z
e)=
7
10
O2minus(Z
e)=
8
10
Fminus(Z
e) =
9
10
N3minus gt O2minusgt Fminus
Topic Periodicity
Difficulty Level Easy [Embibe predicted Low Weightage]
5 The color of KMnO4 is due to
(A) dminus d transition
(B) L rarr M charge transfer transition
(C) σ minus σlowasttransition (D) M rarr L charge transfer transition
Solution (B) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)
Topic Coodination compounds
Difficulty Level Easy [Embibe predicted high weightage]
6 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these
do not react to form oxides of nitrogen Reason The reaction between nitrogen and oxygen requires high temperature
(A) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion
(B) The Assertion is incorrect but the Reason is correct (C) Both the Assertion and Reason are incorrect
(D) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion
Solution (D) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion and
reason are correct and reason is correct explanation
Topic Group 15 ndash 18
Difficulty Level Moderate [Embibe predicted high weightage]
7 Which of the following compounds is not an antacid
(A) Cimetidine
(B) Phenelzine
(C) Rantidine
(D) Aluminium hydroxide
Solution (B) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as
antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a type of
medication that can control the acid levels in stomach Working of antacids Antacids counteract
(neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the symptoms of
heartburn and relieves pain
Topic Chemistry in everyday life
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
8 In the context of the Hall-Heroult process for the extraction of Al which of the following
statements is false
(A) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity
(B) Al3+ is reduced at the cathode to form Al
(C) Na3AlF6 serves as the electrolyte (D) CO and CO2 are produced in this process
Solution (C) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion
temperature and provides good conductivity
Topic Metallurgy
Difficulty Level Moderate [Embibe predicted high weightage]
9 Match the catalysts to the correct process
(A) A rarr ii B rarr i C rarr iv D rarr iii
Catalyst Process
A TiCl3 i Wacker process
B PdCl2 ii Ziegler ndash Natta polymerization C CuCl2 iii Contact process`
D V2O5 iv Deaconrsquos process
(B) A rarr ii B rarr iii C rarr iv D rarr i
(C) A rarr iii B rarr i C rarr ii D rarr iv
(D) A rarr iii B rarr ii C rarr iv D rarr i
Solution (A) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in
water in the presence of tetrachloropalladate (II) as the catalyst
In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with
arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred
In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts
including copper chloride(CuCl2)
In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are usually
used in combination with different organo aluminium co-catalyst
Topic Transition metals
Difficulty Level Easy [Embibe predicted high weightage]
10 In the reaction
The product E is
(A)
(B)
(C)
(D)
Solution (B)
11 Which polymer is used in the manufacture of paints and lacquers
(A) Glyptal (B) Polypropene
(C) Poly vinyl chloride (D) Bakelite
Solution (A) Glyptal is polymer of glycerol and phthalic anhydride
Topic Polymers
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
12 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]
+ is (py = pyridine) (A) 3
(B) 4 (C) 6
(D) 2
Solution (A) Complexes with general formula [Mabcd]+ square planar complex can have three
isomers
Topic Coordination compounds
Difficulty Level Moderate [Embibe predicted high weightage]
13 Higher order (gt 3) reactions are rare due to
(A) Increase in entropy and activation energy as more molecules are involved
(B) Shifting of equilibrium towards reactants due to elastic collisions
(C) Loss of active species on collision
(D) Low probability of simultaneous collision of all the reacting species
Solution (D) Probability of an event involving more than three molecules in a collision are remote
Topic Chemical kinetics
Difficulty Level Easy [Embibe predicted high weightage]
14 Which among the following is the most reactive (A) Br2 (B) I2
(C) ICl
(D) Cl2
Solution (C) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive
Topic Group 15 ndash 18
Difficulty Level Moderate [Embibe predicted high weightage]
15 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper deposited
at the cathode is (Atomic mass of Cu = 635 amu)
(A) 635 g
(B) 2 g
(C) 127 g (D) 0 g
Solution (A) 2F equiv 2 Eqs of Cu
= 2times635
2= 635 g
Topic Electrochemistry
Difficulty Level Easy [Embibe predicted high weightage]
16 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was
filtered and the strength of the filtrate was found to be 0042 N The amount of acetic acid adsorbed (per
gram of charcoal) is
(A) 36 mg
(B) 42 mg
(C) 54 mg
(D) 18 mg
Solution (D) Meqs of CH3COOH (initial) = 50 times006 = 3 Meqs
Meqs CH3COOH (final) = 50times 0042 = 21 Meqs
CH3 COOH adsorbed = 3 minus 21 = 09 Meqs
= 9times 10minus1 times 60 g Eq frasl times 10minus3 g
= 540 times10minus4 = 0054 g
= 54 mg
Per gram =54
3= 18 mg gfrasl of Charcoal
Topic Surface chemistry
Difficulty Level Moderate [Embibe predicted high weightage]
17 The synthesis of alkyl fluorides is best accomplished by
(A) Sandmeyerrsquos reaction
(B) Finkelstein reaction
(C) Swarts reaction
(D) Free radical fluorination
Solution (C) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2 CoF2 SbF3
CH3 Br+ AgF rarr CH3 minusF+ AgBr
The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro
hydrocarbons such as CCl2F2 is known as swarts reaction
Topic Alkyl and Aryl halides
Difficulty Level Easy [Embibe predicted high weightage]
18 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin
(A) 1
206
(B) 2
309
(C) 1
412
(D) 1
103
Solution (C) 2C8H7SO3minusNa++ Ca(aq)
2+ rarr (C8H7SO3)2Ca + 2Na+
2 moles (resin) = 412 g equiv 40 g Ca2+ ions
= 1 moles of Ca2+ ions
Moles of Ca2+ gfrasl of resin =1
412
Topic s-Block elements and Hydrogen
Difficulty Level Moderate (Embibe predicted Low Weightage)
19 Which of the vitamins given below is water soluble
(A) Vitamin D (B) Vitamin E (C) Vitamin K (D) Vitamin C
Solution (D) B complex vitamins and vitamin C are water soluble vitamins that are not
stored in the body and must be replaced each day
Topic Biomolecules
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
20 The intermolecular interaction that is dependent on the inverse cube of distance between the
molecules is
(A) Ion-dipole interaction
(B) London force
(C) Hydrogen bond (D) Ion-ion interaction
Solution (A) For Ion-dipole interaction
F prop μ (dE
dr)
μ is dipole moment of dipole
r is distance between dipole and ion
prop μd
dr(1
r2)prop
μ
r3
Ionminus ion prop1
r2
London force prop1
r6
Hydrogen bond minusDipole minus dipole prop1
r4
Topic Chemical Bonding
Difficulty Level Easy [Embibe predicted high weightage]
21 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)
The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is the
standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)
(A) 86600 +R(298) ln(16 times 1012)
(B) 86600 minusln(16times1012)
R(298)
(C) 05[2 times 86600minus R(298) ln(16 times 1012 )]
(D) R(298) ln(16 times 1012) minus 86600
Solution (C) ΔGreaco = minus2303 RTlog KP
= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac
o = 2ΔGf(NO2)o minus2ΔGf(NO)g
o
= 2ΔGf(NO2)
o minus2 times 866 times 103
2ΔGf(NO2)o = minusR(298)ln(16 times 1012) + 2 times 86600
ΔGf(NO2)o = 86600 minus
R(298)
2ln(16 times 1012)
= 05[2times 86600 minus R(298)ln(16times 1012)]
Topic Thermochemistry and thermodynamics
Difficulty Level Moderate [Embibe predicted high weightage]
22 Which of the following compounds is not colored yellow
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution (D) Cyanides not yellow
BaCrO4 minusYellowK3[Co(NO2)6]minus Yellow
(NH4)3[As(Mo3O10)4] minus Yellow
Zn2[Fe(CN)6] is bluish white
Topic Salt Analysis
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
23 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80) (A) 36
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Hydrocarbons
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
4 The ionic radii (in Å) of N3minus O2minus and Fminus are respectively
(A) 136 171 and 140
(B) 171 140 and 136
(C) 171 136 and 140
(D) 136 140 and 171
Solution (B) As Z
euarr ionic radius decreases for isoelectronic species
N3minus(Z
e)=
7
10
O2minus(Z
e)=
8
10
Fminus(Z
e) =
9
10
N3minus gt O2minusgt Fminus
Topic Periodicity
Difficulty Level Easy [Embibe predicted Low Weightage]
5 The color of KMnO4 is due to
(A) dminus d transition
(B) L rarr M charge transfer transition
(C) σ minus σlowasttransition (D) M rarr L charge transfer transition
Solution (B) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)
Topic Coodination compounds
Difficulty Level Easy [Embibe predicted high weightage]
6 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these
do not react to form oxides of nitrogen Reason The reaction between nitrogen and oxygen requires high temperature
(A) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion
(B) The Assertion is incorrect but the Reason is correct (C) Both the Assertion and Reason are incorrect
(D) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion
Solution (D) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion and
reason are correct and reason is correct explanation
Topic Group 15 ndash 18
Difficulty Level Moderate [Embibe predicted high weightage]
7 Which of the following compounds is not an antacid
(A) Cimetidine
(B) Phenelzine
(C) Rantidine
(D) Aluminium hydroxide
Solution (B) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as
antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a type of
medication that can control the acid levels in stomach Working of antacids Antacids counteract
(neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the symptoms of
heartburn and relieves pain
Topic Chemistry in everyday life
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
8 In the context of the Hall-Heroult process for the extraction of Al which of the following
statements is false
(A) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity
(B) Al3+ is reduced at the cathode to form Al
(C) Na3AlF6 serves as the electrolyte (D) CO and CO2 are produced in this process
Solution (C) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion
temperature and provides good conductivity
Topic Metallurgy
Difficulty Level Moderate [Embibe predicted high weightage]
9 Match the catalysts to the correct process
(A) A rarr ii B rarr i C rarr iv D rarr iii
Catalyst Process
A TiCl3 i Wacker process
B PdCl2 ii Ziegler ndash Natta polymerization C CuCl2 iii Contact process`
D V2O5 iv Deaconrsquos process
(B) A rarr ii B rarr iii C rarr iv D rarr i
(C) A rarr iii B rarr i C rarr ii D rarr iv
(D) A rarr iii B rarr ii C rarr iv D rarr i
Solution (A) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in
water in the presence of tetrachloropalladate (II) as the catalyst
In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with
arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred
In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts
including copper chloride(CuCl2)
In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are usually
used in combination with different organo aluminium co-catalyst
Topic Transition metals
Difficulty Level Easy [Embibe predicted high weightage]
10 In the reaction
The product E is
(A)
(B)
(C)
(D)
Solution (B)
11 Which polymer is used in the manufacture of paints and lacquers
(A) Glyptal (B) Polypropene
(C) Poly vinyl chloride (D) Bakelite
Solution (A) Glyptal is polymer of glycerol and phthalic anhydride
Topic Polymers
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
12 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]
+ is (py = pyridine) (A) 3
(B) 4 (C) 6
(D) 2
Solution (A) Complexes with general formula [Mabcd]+ square planar complex can have three
isomers
Topic Coordination compounds
Difficulty Level Moderate [Embibe predicted high weightage]
13 Higher order (gt 3) reactions are rare due to
(A) Increase in entropy and activation energy as more molecules are involved
(B) Shifting of equilibrium towards reactants due to elastic collisions
(C) Loss of active species on collision
(D) Low probability of simultaneous collision of all the reacting species
Solution (D) Probability of an event involving more than three molecules in a collision are remote
Topic Chemical kinetics
Difficulty Level Easy [Embibe predicted high weightage]
14 Which among the following is the most reactive (A) Br2 (B) I2
(C) ICl
(D) Cl2
Solution (C) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive
Topic Group 15 ndash 18
Difficulty Level Moderate [Embibe predicted high weightage]
15 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper deposited
at the cathode is (Atomic mass of Cu = 635 amu)
(A) 635 g
(B) 2 g
(C) 127 g (D) 0 g
Solution (A) 2F equiv 2 Eqs of Cu
= 2times635
2= 635 g
Topic Electrochemistry
Difficulty Level Easy [Embibe predicted high weightage]
16 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was
filtered and the strength of the filtrate was found to be 0042 N The amount of acetic acid adsorbed (per
gram of charcoal) is
(A) 36 mg
(B) 42 mg
(C) 54 mg
(D) 18 mg
Solution (D) Meqs of CH3COOH (initial) = 50 times006 = 3 Meqs
Meqs CH3COOH (final) = 50times 0042 = 21 Meqs
CH3 COOH adsorbed = 3 minus 21 = 09 Meqs
= 9times 10minus1 times 60 g Eq frasl times 10minus3 g
= 540 times10minus4 = 0054 g
= 54 mg
Per gram =54
3= 18 mg gfrasl of Charcoal
Topic Surface chemistry
Difficulty Level Moderate [Embibe predicted high weightage]
17 The synthesis of alkyl fluorides is best accomplished by
(A) Sandmeyerrsquos reaction
(B) Finkelstein reaction
(C) Swarts reaction
(D) Free radical fluorination
Solution (C) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2 CoF2 SbF3
CH3 Br+ AgF rarr CH3 minusF+ AgBr
The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro
hydrocarbons such as CCl2F2 is known as swarts reaction
Topic Alkyl and Aryl halides
Difficulty Level Easy [Embibe predicted high weightage]
18 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin
(A) 1
206
(B) 2
309
(C) 1
412
(D) 1
103
Solution (C) 2C8H7SO3minusNa++ Ca(aq)
2+ rarr (C8H7SO3)2Ca + 2Na+
2 moles (resin) = 412 g equiv 40 g Ca2+ ions
= 1 moles of Ca2+ ions
Moles of Ca2+ gfrasl of resin =1
412
Topic s-Block elements and Hydrogen
Difficulty Level Moderate (Embibe predicted Low Weightage)
19 Which of the vitamins given below is water soluble
(A) Vitamin D (B) Vitamin E (C) Vitamin K (D) Vitamin C
Solution (D) B complex vitamins and vitamin C are water soluble vitamins that are not
stored in the body and must be replaced each day
Topic Biomolecules
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
20 The intermolecular interaction that is dependent on the inverse cube of distance between the
molecules is
(A) Ion-dipole interaction
(B) London force
(C) Hydrogen bond (D) Ion-ion interaction
Solution (A) For Ion-dipole interaction
F prop μ (dE
dr)
μ is dipole moment of dipole
r is distance between dipole and ion
prop μd
dr(1
r2)prop
μ
r3
Ionminus ion prop1
r2
London force prop1
r6
Hydrogen bond minusDipole minus dipole prop1
r4
Topic Chemical Bonding
Difficulty Level Easy [Embibe predicted high weightage]
21 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)
The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is the
standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)
(A) 86600 +R(298) ln(16 times 1012)
(B) 86600 minusln(16times1012)
R(298)
(C) 05[2 times 86600minus R(298) ln(16 times 1012 )]
(D) R(298) ln(16 times 1012) minus 86600
Solution (C) ΔGreaco = minus2303 RTlog KP
= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac
o = 2ΔGf(NO2)o minus2ΔGf(NO)g
o
= 2ΔGf(NO2)
o minus2 times 866 times 103
2ΔGf(NO2)o = minusR(298)ln(16 times 1012) + 2 times 86600
ΔGf(NO2)o = 86600 minus
R(298)
2ln(16 times 1012)
= 05[2times 86600 minus R(298)ln(16times 1012)]
Topic Thermochemistry and thermodynamics
Difficulty Level Moderate [Embibe predicted high weightage]
22 Which of the following compounds is not colored yellow
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution (D) Cyanides not yellow
BaCrO4 minusYellowK3[Co(NO2)6]minus Yellow
(NH4)3[As(Mo3O10)4] minus Yellow
Zn2[Fe(CN)6] is bluish white
Topic Salt Analysis
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
23 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80) (A) 36
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(A) dminus d transition
(B) L rarr M charge transfer transition
(C) σ minus σlowasttransition (D) M rarr L charge transfer transition
Solution (B) Charge transfer from O2minus to empty d-orbitals of metal ion (Mn+7)
Topic Coodination compounds
Difficulty Level Easy [Embibe predicted high weightage]
6 Assertion Nitrogen and Oxygen are the main components in the atmosphere but these
do not react to form oxides of nitrogen Reason The reaction between nitrogen and oxygen requires high temperature
(A) Both Assertion and Reason are correct but the Reason is not the correct explanation for the Assertion
(B) The Assertion is incorrect but the Reason is correct (C) Both the Assertion and Reason are incorrect
(D) Both Assertion and Reason are correct and the Reason is the correct explanation for the Assertion
Solution (D) N2(g) amp O2(g) react under electric arc at 2000oC to form NO(g) Both assertion and
reason are correct and reason is correct explanation
Topic Group 15 ndash 18
Difficulty Level Moderate [Embibe predicted high weightage]
7 Which of the following compounds is not an antacid
(A) Cimetidine
(B) Phenelzine
(C) Rantidine
(D) Aluminium hydroxide
Solution (B) Ranitidine Cimetidine and metal hydroxides ie Aluminum hydroxide can be used as
antacid but not phenelzine Phenelzine is not an antacid It is an antidepressant Antacids are a type of
medication that can control the acid levels in stomach Working of antacids Antacids counteract
(neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the symptoms of
heartburn and relieves pain
Topic Chemistry in everyday life
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
8 In the context of the Hall-Heroult process for the extraction of Al which of the following
statements is false
(A) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity
(B) Al3+ is reduced at the cathode to form Al
(C) Na3AlF6 serves as the electrolyte (D) CO and CO2 are produced in this process
Solution (C) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion
temperature and provides good conductivity
Topic Metallurgy
Difficulty Level Moderate [Embibe predicted high weightage]
9 Match the catalysts to the correct process
(A) A rarr ii B rarr i C rarr iv D rarr iii
Catalyst Process
A TiCl3 i Wacker process
B PdCl2 ii Ziegler ndash Natta polymerization C CuCl2 iii Contact process`
D V2O5 iv Deaconrsquos process
(B) A rarr ii B rarr iii C rarr iv D rarr i
(C) A rarr iii B rarr i C rarr ii D rarr iv
(D) A rarr iii B rarr ii C rarr iv D rarr i
Solution (A) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in
water in the presence of tetrachloropalladate (II) as the catalyst
In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with
arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred
In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts
including copper chloride(CuCl2)
In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are usually
used in combination with different organo aluminium co-catalyst
Topic Transition metals
Difficulty Level Easy [Embibe predicted high weightage]
10 In the reaction
The product E is
(A)
(B)
(C)
(D)
Solution (B)
11 Which polymer is used in the manufacture of paints and lacquers
(A) Glyptal (B) Polypropene
(C) Poly vinyl chloride (D) Bakelite
Solution (A) Glyptal is polymer of glycerol and phthalic anhydride
Topic Polymers
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
12 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]
+ is (py = pyridine) (A) 3
(B) 4 (C) 6
(D) 2
Solution (A) Complexes with general formula [Mabcd]+ square planar complex can have three
isomers
Topic Coordination compounds
Difficulty Level Moderate [Embibe predicted high weightage]
13 Higher order (gt 3) reactions are rare due to
(A) Increase in entropy and activation energy as more molecules are involved
(B) Shifting of equilibrium towards reactants due to elastic collisions
(C) Loss of active species on collision
(D) Low probability of simultaneous collision of all the reacting species
Solution (D) Probability of an event involving more than three molecules in a collision are remote
Topic Chemical kinetics
Difficulty Level Easy [Embibe predicted high weightage]
14 Which among the following is the most reactive (A) Br2 (B) I2
(C) ICl
(D) Cl2
Solution (C) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive
Topic Group 15 ndash 18
Difficulty Level Moderate [Embibe predicted high weightage]
15 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper deposited
at the cathode is (Atomic mass of Cu = 635 amu)
(A) 635 g
(B) 2 g
(C) 127 g (D) 0 g
Solution (A) 2F equiv 2 Eqs of Cu
= 2times635
2= 635 g
Topic Electrochemistry
Difficulty Level Easy [Embibe predicted high weightage]
16 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was
filtered and the strength of the filtrate was found to be 0042 N The amount of acetic acid adsorbed (per
gram of charcoal) is
(A) 36 mg
(B) 42 mg
(C) 54 mg
(D) 18 mg
Solution (D) Meqs of CH3COOH (initial) = 50 times006 = 3 Meqs
Meqs CH3COOH (final) = 50times 0042 = 21 Meqs
CH3 COOH adsorbed = 3 minus 21 = 09 Meqs
= 9times 10minus1 times 60 g Eq frasl times 10minus3 g
= 540 times10minus4 = 0054 g
= 54 mg
Per gram =54
3= 18 mg gfrasl of Charcoal
Topic Surface chemistry
Difficulty Level Moderate [Embibe predicted high weightage]
17 The synthesis of alkyl fluorides is best accomplished by
(A) Sandmeyerrsquos reaction
(B) Finkelstein reaction
(C) Swarts reaction
(D) Free radical fluorination
Solution (C) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2 CoF2 SbF3
CH3 Br+ AgF rarr CH3 minusF+ AgBr
The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro
hydrocarbons such as CCl2F2 is known as swarts reaction
Topic Alkyl and Aryl halides
Difficulty Level Easy [Embibe predicted high weightage]
18 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin
(A) 1
206
(B) 2
309
(C) 1
412
(D) 1
103
Solution (C) 2C8H7SO3minusNa++ Ca(aq)
2+ rarr (C8H7SO3)2Ca + 2Na+
2 moles (resin) = 412 g equiv 40 g Ca2+ ions
= 1 moles of Ca2+ ions
Moles of Ca2+ gfrasl of resin =1
412
Topic s-Block elements and Hydrogen
Difficulty Level Moderate (Embibe predicted Low Weightage)
19 Which of the vitamins given below is water soluble
(A) Vitamin D (B) Vitamin E (C) Vitamin K (D) Vitamin C
Solution (D) B complex vitamins and vitamin C are water soluble vitamins that are not
stored in the body and must be replaced each day
Topic Biomolecules
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
20 The intermolecular interaction that is dependent on the inverse cube of distance between the
molecules is
(A) Ion-dipole interaction
(B) London force
(C) Hydrogen bond (D) Ion-ion interaction
Solution (A) For Ion-dipole interaction
F prop μ (dE
dr)
μ is dipole moment of dipole
r is distance between dipole and ion
prop μd
dr(1
r2)prop
μ
r3
Ionminus ion prop1
r2
London force prop1
r6
Hydrogen bond minusDipole minus dipole prop1
r4
Topic Chemical Bonding
Difficulty Level Easy [Embibe predicted high weightage]
21 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)
The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is the
standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)
(A) 86600 +R(298) ln(16 times 1012)
(B) 86600 minusln(16times1012)
R(298)
(C) 05[2 times 86600minus R(298) ln(16 times 1012 )]
(D) R(298) ln(16 times 1012) minus 86600
Solution (C) ΔGreaco = minus2303 RTlog KP
= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac
o = 2ΔGf(NO2)o minus2ΔGf(NO)g
o
= 2ΔGf(NO2)
o minus2 times 866 times 103
2ΔGf(NO2)o = minusR(298)ln(16 times 1012) + 2 times 86600
ΔGf(NO2)o = 86600 minus
R(298)
2ln(16 times 1012)
= 05[2times 86600 minus R(298)ln(16times 1012)]
Topic Thermochemistry and thermodynamics
Difficulty Level Moderate [Embibe predicted high weightage]
22 Which of the following compounds is not colored yellow
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution (D) Cyanides not yellow
BaCrO4 minusYellowK3[Co(NO2)6]minus Yellow
(NH4)3[As(Mo3O10)4] minus Yellow
Zn2[Fe(CN)6] is bluish white
Topic Salt Analysis
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
23 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80) (A) 36
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
medication that can control the acid levels in stomach Working of antacids Antacids counteract
(neutralize) the acid in stomach thatrsquos used to aid digestion This helps reduce the symptoms of
heartburn and relieves pain
Topic Chemistry in everyday life
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
8 In the context of the Hall-Heroult process for the extraction of Al which of the following
statements is false
(A) Al2O3 is mixed with CaF2 which lowers the melting point of the mixture and brings conductivity
(B) Al3+ is reduced at the cathode to form Al
(C) Na3AlF6 serves as the electrolyte (D) CO and CO2 are produced in this process
Solution (C) Hall-Heroult process for extraction of Al Al2O3 is electrolyte Na3AlF6 reduces the fusion
temperature and provides good conductivity
Topic Metallurgy
Difficulty Level Moderate [Embibe predicted high weightage]
9 Match the catalysts to the correct process
(A) A rarr ii B rarr i C rarr iv D rarr iii
Catalyst Process
A TiCl3 i Wacker process
B PdCl2 ii Ziegler ndash Natta polymerization C CuCl2 iii Contact process`
D V2O5 iv Deaconrsquos process
(B) A rarr ii B rarr iii C rarr iv D rarr i
(C) A rarr iii B rarr i C rarr ii D rarr iv
(D) A rarr iii B rarr ii C rarr iv D rarr i
Solution (A) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in
water in the presence of tetrachloropalladate (II) as the catalyst
In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with
arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred
In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts
including copper chloride(CuCl2)
In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are usually
used in combination with different organo aluminium co-catalyst
Topic Transition metals
Difficulty Level Easy [Embibe predicted high weightage]
10 In the reaction
The product E is
(A)
(B)
(C)
(D)
Solution (B)
11 Which polymer is used in the manufacture of paints and lacquers
(A) Glyptal (B) Polypropene
(C) Poly vinyl chloride (D) Bakelite
Solution (A) Glyptal is polymer of glycerol and phthalic anhydride
Topic Polymers
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
12 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]
+ is (py = pyridine) (A) 3
(B) 4 (C) 6
(D) 2
Solution (A) Complexes with general formula [Mabcd]+ square planar complex can have three
isomers
Topic Coordination compounds
Difficulty Level Moderate [Embibe predicted high weightage]
13 Higher order (gt 3) reactions are rare due to
(A) Increase in entropy and activation energy as more molecules are involved
(B) Shifting of equilibrium towards reactants due to elastic collisions
(C) Loss of active species on collision
(D) Low probability of simultaneous collision of all the reacting species
Solution (D) Probability of an event involving more than three molecules in a collision are remote
Topic Chemical kinetics
Difficulty Level Easy [Embibe predicted high weightage]
14 Which among the following is the most reactive (A) Br2 (B) I2
(C) ICl
(D) Cl2
Solution (C) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive
Topic Group 15 ndash 18
Difficulty Level Moderate [Embibe predicted high weightage]
15 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper deposited
at the cathode is (Atomic mass of Cu = 635 amu)
(A) 635 g
(B) 2 g
(C) 127 g (D) 0 g
Solution (A) 2F equiv 2 Eqs of Cu
= 2times635
2= 635 g
Topic Electrochemistry
Difficulty Level Easy [Embibe predicted high weightage]
16 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was
filtered and the strength of the filtrate was found to be 0042 N The amount of acetic acid adsorbed (per
gram of charcoal) is
(A) 36 mg
(B) 42 mg
(C) 54 mg
(D) 18 mg
Solution (D) Meqs of CH3COOH (initial) = 50 times006 = 3 Meqs
Meqs CH3COOH (final) = 50times 0042 = 21 Meqs
CH3 COOH adsorbed = 3 minus 21 = 09 Meqs
= 9times 10minus1 times 60 g Eq frasl times 10minus3 g
= 540 times10minus4 = 0054 g
= 54 mg
Per gram =54
3= 18 mg gfrasl of Charcoal
Topic Surface chemistry
Difficulty Level Moderate [Embibe predicted high weightage]
17 The synthesis of alkyl fluorides is best accomplished by
(A) Sandmeyerrsquos reaction
(B) Finkelstein reaction
(C) Swarts reaction
(D) Free radical fluorination
Solution (C) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2 CoF2 SbF3
CH3 Br+ AgF rarr CH3 minusF+ AgBr
The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro
hydrocarbons such as CCl2F2 is known as swarts reaction
Topic Alkyl and Aryl halides
Difficulty Level Easy [Embibe predicted high weightage]
18 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin
(A) 1
206
(B) 2
309
(C) 1
412
(D) 1
103
Solution (C) 2C8H7SO3minusNa++ Ca(aq)
2+ rarr (C8H7SO3)2Ca + 2Na+
2 moles (resin) = 412 g equiv 40 g Ca2+ ions
= 1 moles of Ca2+ ions
Moles of Ca2+ gfrasl of resin =1
412
Topic s-Block elements and Hydrogen
Difficulty Level Moderate (Embibe predicted Low Weightage)
19 Which of the vitamins given below is water soluble
(A) Vitamin D (B) Vitamin E (C) Vitamin K (D) Vitamin C
Solution (D) B complex vitamins and vitamin C are water soluble vitamins that are not
stored in the body and must be replaced each day
Topic Biomolecules
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
20 The intermolecular interaction that is dependent on the inverse cube of distance between the
molecules is
(A) Ion-dipole interaction
(B) London force
(C) Hydrogen bond (D) Ion-ion interaction
Solution (A) For Ion-dipole interaction
F prop μ (dE
dr)
μ is dipole moment of dipole
r is distance between dipole and ion
prop μd
dr(1
r2)prop
μ
r3
Ionminus ion prop1
r2
London force prop1
r6
Hydrogen bond minusDipole minus dipole prop1
r4
Topic Chemical Bonding
Difficulty Level Easy [Embibe predicted high weightage]
21 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)
The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is the
standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)
(A) 86600 +R(298) ln(16 times 1012)
(B) 86600 minusln(16times1012)
R(298)
(C) 05[2 times 86600minus R(298) ln(16 times 1012 )]
(D) R(298) ln(16 times 1012) minus 86600
Solution (C) ΔGreaco = minus2303 RTlog KP
= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac
o = 2ΔGf(NO2)o minus2ΔGf(NO)g
o
= 2ΔGf(NO2)
o minus2 times 866 times 103
2ΔGf(NO2)o = minusR(298)ln(16 times 1012) + 2 times 86600
ΔGf(NO2)o = 86600 minus
R(298)
2ln(16 times 1012)
= 05[2times 86600 minus R(298)ln(16times 1012)]
Topic Thermochemistry and thermodynamics
Difficulty Level Moderate [Embibe predicted high weightage]
22 Which of the following compounds is not colored yellow
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution (D) Cyanides not yellow
BaCrO4 minusYellowK3[Co(NO2)6]minus Yellow
(NH4)3[As(Mo3O10)4] minus Yellow
Zn2[Fe(CN)6] is bluish white
Topic Salt Analysis
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
23 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80) (A) 36
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(B) A rarr ii B rarr iii C rarr iv D rarr i
(C) A rarr iii B rarr i C rarr ii D rarr iv
(D) A rarr iii B rarr ii C rarr iv D rarr i
Solution (A) The Wacker process originally referred to the oxidation of ethylene to acetaldehyde by oxygen in
water in the presence of tetrachloropalladate (II) as the catalyst
In contact process Platinum used to be the catalyst for this reaction however as it is susceptible to reacting with
arsenic impurities in the sulphur feedstock vanadium (V) oxide (V2O5) is now preferred
In Deacons process The reaction takes place at about 400 to 450oC in the presence of a variety of catalysts
including copper chloride(CuCl2)
In Ziegler-Natta catalyst Homogenous catalysts usually based on complexes of Ti Zr or Hf used They are usually
used in combination with different organo aluminium co-catalyst
Topic Transition metals
Difficulty Level Easy [Embibe predicted high weightage]
10 In the reaction
The product E is
(A)
(B)
(C)
(D)
Solution (B)
11 Which polymer is used in the manufacture of paints and lacquers
(A) Glyptal (B) Polypropene
(C) Poly vinyl chloride (D) Bakelite
Solution (A) Glyptal is polymer of glycerol and phthalic anhydride
Topic Polymers
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
12 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]
+ is (py = pyridine) (A) 3
(B) 4 (C) 6
(D) 2
Solution (A) Complexes with general formula [Mabcd]+ square planar complex can have three
isomers
Topic Coordination compounds
Difficulty Level Moderate [Embibe predicted high weightage]
13 Higher order (gt 3) reactions are rare due to
(A) Increase in entropy and activation energy as more molecules are involved
(B) Shifting of equilibrium towards reactants due to elastic collisions
(C) Loss of active species on collision
(D) Low probability of simultaneous collision of all the reacting species
Solution (D) Probability of an event involving more than three molecules in a collision are remote
Topic Chemical kinetics
Difficulty Level Easy [Embibe predicted high weightage]
14 Which among the following is the most reactive (A) Br2 (B) I2
(C) ICl
(D) Cl2
Solution (C) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive
Topic Group 15 ndash 18
Difficulty Level Moderate [Embibe predicted high weightage]
15 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper deposited
at the cathode is (Atomic mass of Cu = 635 amu)
(A) 635 g
(B) 2 g
(C) 127 g (D) 0 g
Solution (A) 2F equiv 2 Eqs of Cu
= 2times635
2= 635 g
Topic Electrochemistry
Difficulty Level Easy [Embibe predicted high weightage]
16 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was
filtered and the strength of the filtrate was found to be 0042 N The amount of acetic acid adsorbed (per
gram of charcoal) is
(A) 36 mg
(B) 42 mg
(C) 54 mg
(D) 18 mg
Solution (D) Meqs of CH3COOH (initial) = 50 times006 = 3 Meqs
Meqs CH3COOH (final) = 50times 0042 = 21 Meqs
CH3 COOH adsorbed = 3 minus 21 = 09 Meqs
= 9times 10minus1 times 60 g Eq frasl times 10minus3 g
= 540 times10minus4 = 0054 g
= 54 mg
Per gram =54
3= 18 mg gfrasl of Charcoal
Topic Surface chemistry
Difficulty Level Moderate [Embibe predicted high weightage]
17 The synthesis of alkyl fluorides is best accomplished by
(A) Sandmeyerrsquos reaction
(B) Finkelstein reaction
(C) Swarts reaction
(D) Free radical fluorination
Solution (C) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2 CoF2 SbF3
CH3 Br+ AgF rarr CH3 minusF+ AgBr
The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro
hydrocarbons such as CCl2F2 is known as swarts reaction
Topic Alkyl and Aryl halides
Difficulty Level Easy [Embibe predicted high weightage]
18 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin
(A) 1
206
(B) 2
309
(C) 1
412
(D) 1
103
Solution (C) 2C8H7SO3minusNa++ Ca(aq)
2+ rarr (C8H7SO3)2Ca + 2Na+
2 moles (resin) = 412 g equiv 40 g Ca2+ ions
= 1 moles of Ca2+ ions
Moles of Ca2+ gfrasl of resin =1
412
Topic s-Block elements and Hydrogen
Difficulty Level Moderate (Embibe predicted Low Weightage)
19 Which of the vitamins given below is water soluble
(A) Vitamin D (B) Vitamin E (C) Vitamin K (D) Vitamin C
Solution (D) B complex vitamins and vitamin C are water soluble vitamins that are not
stored in the body and must be replaced each day
Topic Biomolecules
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
20 The intermolecular interaction that is dependent on the inverse cube of distance between the
molecules is
(A) Ion-dipole interaction
(B) London force
(C) Hydrogen bond (D) Ion-ion interaction
Solution (A) For Ion-dipole interaction
F prop μ (dE
dr)
μ is dipole moment of dipole
r is distance between dipole and ion
prop μd
dr(1
r2)prop
μ
r3
Ionminus ion prop1
r2
London force prop1
r6
Hydrogen bond minusDipole minus dipole prop1
r4
Topic Chemical Bonding
Difficulty Level Easy [Embibe predicted high weightage]
21 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)
The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is the
standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)
(A) 86600 +R(298) ln(16 times 1012)
(B) 86600 minusln(16times1012)
R(298)
(C) 05[2 times 86600minus R(298) ln(16 times 1012 )]
(D) R(298) ln(16 times 1012) minus 86600
Solution (C) ΔGreaco = minus2303 RTlog KP
= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac
o = 2ΔGf(NO2)o minus2ΔGf(NO)g
o
= 2ΔGf(NO2)
o minus2 times 866 times 103
2ΔGf(NO2)o = minusR(298)ln(16 times 1012) + 2 times 86600
ΔGf(NO2)o = 86600 minus
R(298)
2ln(16 times 1012)
= 05[2times 86600 minus R(298)ln(16times 1012)]
Topic Thermochemistry and thermodynamics
Difficulty Level Moderate [Embibe predicted high weightage]
22 Which of the following compounds is not colored yellow
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution (D) Cyanides not yellow
BaCrO4 minusYellowK3[Co(NO2)6]minus Yellow
(NH4)3[As(Mo3O10)4] minus Yellow
Zn2[Fe(CN)6] is bluish white
Topic Salt Analysis
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
23 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80) (A) 36
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(C)
(D)
Solution (B)
11 Which polymer is used in the manufacture of paints and lacquers
(A) Glyptal (B) Polypropene
(C) Poly vinyl chloride (D) Bakelite
Solution (A) Glyptal is polymer of glycerol and phthalic anhydride
Topic Polymers
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
12 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]
+ is (py = pyridine) (A) 3
(B) 4 (C) 6
(D) 2
Solution (A) Complexes with general formula [Mabcd]+ square planar complex can have three
isomers
Topic Coordination compounds
Difficulty Level Moderate [Embibe predicted high weightage]
13 Higher order (gt 3) reactions are rare due to
(A) Increase in entropy and activation energy as more molecules are involved
(B) Shifting of equilibrium towards reactants due to elastic collisions
(C) Loss of active species on collision
(D) Low probability of simultaneous collision of all the reacting species
Solution (D) Probability of an event involving more than three molecules in a collision are remote
Topic Chemical kinetics
Difficulty Level Easy [Embibe predicted high weightage]
14 Which among the following is the most reactive (A) Br2 (B) I2
(C) ICl
(D) Cl2
Solution (C) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive
Topic Group 15 ndash 18
Difficulty Level Moderate [Embibe predicted high weightage]
15 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper deposited
at the cathode is (Atomic mass of Cu = 635 amu)
(A) 635 g
(B) 2 g
(C) 127 g (D) 0 g
Solution (A) 2F equiv 2 Eqs of Cu
= 2times635
2= 635 g
Topic Electrochemistry
Difficulty Level Easy [Embibe predicted high weightage]
16 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was
filtered and the strength of the filtrate was found to be 0042 N The amount of acetic acid adsorbed (per
gram of charcoal) is
(A) 36 mg
(B) 42 mg
(C) 54 mg
(D) 18 mg
Solution (D) Meqs of CH3COOH (initial) = 50 times006 = 3 Meqs
Meqs CH3COOH (final) = 50times 0042 = 21 Meqs
CH3 COOH adsorbed = 3 minus 21 = 09 Meqs
= 9times 10minus1 times 60 g Eq frasl times 10minus3 g
= 540 times10minus4 = 0054 g
= 54 mg
Per gram =54
3= 18 mg gfrasl of Charcoal
Topic Surface chemistry
Difficulty Level Moderate [Embibe predicted high weightage]
17 The synthesis of alkyl fluorides is best accomplished by
(A) Sandmeyerrsquos reaction
(B) Finkelstein reaction
(C) Swarts reaction
(D) Free radical fluorination
Solution (C) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2 CoF2 SbF3
CH3 Br+ AgF rarr CH3 minusF+ AgBr
The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro
hydrocarbons such as CCl2F2 is known as swarts reaction
Topic Alkyl and Aryl halides
Difficulty Level Easy [Embibe predicted high weightage]
18 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin
(A) 1
206
(B) 2
309
(C) 1
412
(D) 1
103
Solution (C) 2C8H7SO3minusNa++ Ca(aq)
2+ rarr (C8H7SO3)2Ca + 2Na+
2 moles (resin) = 412 g equiv 40 g Ca2+ ions
= 1 moles of Ca2+ ions
Moles of Ca2+ gfrasl of resin =1
412
Topic s-Block elements and Hydrogen
Difficulty Level Moderate (Embibe predicted Low Weightage)
19 Which of the vitamins given below is water soluble
(A) Vitamin D (B) Vitamin E (C) Vitamin K (D) Vitamin C
Solution (D) B complex vitamins and vitamin C are water soluble vitamins that are not
stored in the body and must be replaced each day
Topic Biomolecules
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
20 The intermolecular interaction that is dependent on the inverse cube of distance between the
molecules is
(A) Ion-dipole interaction
(B) London force
(C) Hydrogen bond (D) Ion-ion interaction
Solution (A) For Ion-dipole interaction
F prop μ (dE
dr)
μ is dipole moment of dipole
r is distance between dipole and ion
prop μd
dr(1
r2)prop
μ
r3
Ionminus ion prop1
r2
London force prop1
r6
Hydrogen bond minusDipole minus dipole prop1
r4
Topic Chemical Bonding
Difficulty Level Easy [Embibe predicted high weightage]
21 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)
The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is the
standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)
(A) 86600 +R(298) ln(16 times 1012)
(B) 86600 minusln(16times1012)
R(298)
(C) 05[2 times 86600minus R(298) ln(16 times 1012 )]
(D) R(298) ln(16 times 1012) minus 86600
Solution (C) ΔGreaco = minus2303 RTlog KP
= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac
o = 2ΔGf(NO2)o minus2ΔGf(NO)g
o
= 2ΔGf(NO2)
o minus2 times 866 times 103
2ΔGf(NO2)o = minusR(298)ln(16 times 1012) + 2 times 86600
ΔGf(NO2)o = 86600 minus
R(298)
2ln(16 times 1012)
= 05[2times 86600 minus R(298)ln(16times 1012)]
Topic Thermochemistry and thermodynamics
Difficulty Level Moderate [Embibe predicted high weightage]
22 Which of the following compounds is not colored yellow
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution (D) Cyanides not yellow
BaCrO4 minusYellowK3[Co(NO2)6]minus Yellow
(NH4)3[As(Mo3O10)4] minus Yellow
Zn2[Fe(CN)6] is bluish white
Topic Salt Analysis
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
23 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80) (A) 36
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(C) Poly vinyl chloride (D) Bakelite
Solution (A) Glyptal is polymer of glycerol and phthalic anhydride
Topic Polymers
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
12 The number of geometric isomers that can exist for square planar [Pt(Cl)(py)(NH3)(NH2OH)]
+ is (py = pyridine) (A) 3
(B) 4 (C) 6
(D) 2
Solution (A) Complexes with general formula [Mabcd]+ square planar complex can have three
isomers
Topic Coordination compounds
Difficulty Level Moderate [Embibe predicted high weightage]
13 Higher order (gt 3) reactions are rare due to
(A) Increase in entropy and activation energy as more molecules are involved
(B) Shifting of equilibrium towards reactants due to elastic collisions
(C) Loss of active species on collision
(D) Low probability of simultaneous collision of all the reacting species
Solution (D) Probability of an event involving more than three molecules in a collision are remote
Topic Chemical kinetics
Difficulty Level Easy [Embibe predicted high weightage]
14 Which among the following is the most reactive (A) Br2 (B) I2
(C) ICl
(D) Cl2
Solution (C) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive
Topic Group 15 ndash 18
Difficulty Level Moderate [Embibe predicted high weightage]
15 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper deposited
at the cathode is (Atomic mass of Cu = 635 amu)
(A) 635 g
(B) 2 g
(C) 127 g (D) 0 g
Solution (A) 2F equiv 2 Eqs of Cu
= 2times635
2= 635 g
Topic Electrochemistry
Difficulty Level Easy [Embibe predicted high weightage]
16 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was
filtered and the strength of the filtrate was found to be 0042 N The amount of acetic acid adsorbed (per
gram of charcoal) is
(A) 36 mg
(B) 42 mg
(C) 54 mg
(D) 18 mg
Solution (D) Meqs of CH3COOH (initial) = 50 times006 = 3 Meqs
Meqs CH3COOH (final) = 50times 0042 = 21 Meqs
CH3 COOH adsorbed = 3 minus 21 = 09 Meqs
= 9times 10minus1 times 60 g Eq frasl times 10minus3 g
= 540 times10minus4 = 0054 g
= 54 mg
Per gram =54
3= 18 mg gfrasl of Charcoal
Topic Surface chemistry
Difficulty Level Moderate [Embibe predicted high weightage]
17 The synthesis of alkyl fluorides is best accomplished by
(A) Sandmeyerrsquos reaction
(B) Finkelstein reaction
(C) Swarts reaction
(D) Free radical fluorination
Solution (C) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2 CoF2 SbF3
CH3 Br+ AgF rarr CH3 minusF+ AgBr
The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro
hydrocarbons such as CCl2F2 is known as swarts reaction
Topic Alkyl and Aryl halides
Difficulty Level Easy [Embibe predicted high weightage]
18 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin
(A) 1
206
(B) 2
309
(C) 1
412
(D) 1
103
Solution (C) 2C8H7SO3minusNa++ Ca(aq)
2+ rarr (C8H7SO3)2Ca + 2Na+
2 moles (resin) = 412 g equiv 40 g Ca2+ ions
= 1 moles of Ca2+ ions
Moles of Ca2+ gfrasl of resin =1
412
Topic s-Block elements and Hydrogen
Difficulty Level Moderate (Embibe predicted Low Weightage)
19 Which of the vitamins given below is water soluble
(A) Vitamin D (B) Vitamin E (C) Vitamin K (D) Vitamin C
Solution (D) B complex vitamins and vitamin C are water soluble vitamins that are not
stored in the body and must be replaced each day
Topic Biomolecules
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
20 The intermolecular interaction that is dependent on the inverse cube of distance between the
molecules is
(A) Ion-dipole interaction
(B) London force
(C) Hydrogen bond (D) Ion-ion interaction
Solution (A) For Ion-dipole interaction
F prop μ (dE
dr)
μ is dipole moment of dipole
r is distance between dipole and ion
prop μd
dr(1
r2)prop
μ
r3
Ionminus ion prop1
r2
London force prop1
r6
Hydrogen bond minusDipole minus dipole prop1
r4
Topic Chemical Bonding
Difficulty Level Easy [Embibe predicted high weightage]
21 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)
The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is the
standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)
(A) 86600 +R(298) ln(16 times 1012)
(B) 86600 minusln(16times1012)
R(298)
(C) 05[2 times 86600minus R(298) ln(16 times 1012 )]
(D) R(298) ln(16 times 1012) minus 86600
Solution (C) ΔGreaco = minus2303 RTlog KP
= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac
o = 2ΔGf(NO2)o minus2ΔGf(NO)g
o
= 2ΔGf(NO2)
o minus2 times 866 times 103
2ΔGf(NO2)o = minusR(298)ln(16 times 1012) + 2 times 86600
ΔGf(NO2)o = 86600 minus
R(298)
2ln(16 times 1012)
= 05[2times 86600 minus R(298)ln(16times 1012)]
Topic Thermochemistry and thermodynamics
Difficulty Level Moderate [Embibe predicted high weightage]
22 Which of the following compounds is not colored yellow
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution (D) Cyanides not yellow
BaCrO4 minusYellowK3[Co(NO2)6]minus Yellow
(NH4)3[As(Mo3O10)4] minus Yellow
Zn2[Fe(CN)6] is bluish white
Topic Salt Analysis
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
23 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80) (A) 36
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Solution (D) Probability of an event involving more than three molecules in a collision are remote
Topic Chemical kinetics
Difficulty Level Easy [Embibe predicted high weightage]
14 Which among the following is the most reactive (A) Br2 (B) I2
(C) ICl
(D) Cl2
Solution (C) I minus Cl bond strength is weaker than I2 Br2 and Cl2 (Homonuclear covalent) I minus Cl bond is polar due to electronegativity difference between I amp Cl hence more reactive
Topic Group 15 ndash 18
Difficulty Level Moderate [Embibe predicted high weightage]
15 Two Faraday of electricity is passed through a solution of CuSO4 The mass of copper deposited
at the cathode is (Atomic mass of Cu = 635 amu)
(A) 635 g
(B) 2 g
(C) 127 g (D) 0 g
Solution (A) 2F equiv 2 Eqs of Cu
= 2times635
2= 635 g
Topic Electrochemistry
Difficulty Level Easy [Embibe predicted high weightage]
16 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was
filtered and the strength of the filtrate was found to be 0042 N The amount of acetic acid adsorbed (per
gram of charcoal) is
(A) 36 mg
(B) 42 mg
(C) 54 mg
(D) 18 mg
Solution (D) Meqs of CH3COOH (initial) = 50 times006 = 3 Meqs
Meqs CH3COOH (final) = 50times 0042 = 21 Meqs
CH3 COOH adsorbed = 3 minus 21 = 09 Meqs
= 9times 10minus1 times 60 g Eq frasl times 10minus3 g
= 540 times10minus4 = 0054 g
= 54 mg
Per gram =54
3= 18 mg gfrasl of Charcoal
Topic Surface chemistry
Difficulty Level Moderate [Embibe predicted high weightage]
17 The synthesis of alkyl fluorides is best accomplished by
(A) Sandmeyerrsquos reaction
(B) Finkelstein reaction
(C) Swarts reaction
(D) Free radical fluorination
Solution (C) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2 CoF2 SbF3
CH3 Br+ AgF rarr CH3 minusF+ AgBr
The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro
hydrocarbons such as CCl2F2 is known as swarts reaction
Topic Alkyl and Aryl halides
Difficulty Level Easy [Embibe predicted high weightage]
18 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin
(A) 1
206
(B) 2
309
(C) 1
412
(D) 1
103
Solution (C) 2C8H7SO3minusNa++ Ca(aq)
2+ rarr (C8H7SO3)2Ca + 2Na+
2 moles (resin) = 412 g equiv 40 g Ca2+ ions
= 1 moles of Ca2+ ions
Moles of Ca2+ gfrasl of resin =1
412
Topic s-Block elements and Hydrogen
Difficulty Level Moderate (Embibe predicted Low Weightage)
19 Which of the vitamins given below is water soluble
(A) Vitamin D (B) Vitamin E (C) Vitamin K (D) Vitamin C
Solution (D) B complex vitamins and vitamin C are water soluble vitamins that are not
stored in the body and must be replaced each day
Topic Biomolecules
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
20 The intermolecular interaction that is dependent on the inverse cube of distance between the
molecules is
(A) Ion-dipole interaction
(B) London force
(C) Hydrogen bond (D) Ion-ion interaction
Solution (A) For Ion-dipole interaction
F prop μ (dE
dr)
μ is dipole moment of dipole
r is distance between dipole and ion
prop μd
dr(1
r2)prop
μ
r3
Ionminus ion prop1
r2
London force prop1
r6
Hydrogen bond minusDipole minus dipole prop1
r4
Topic Chemical Bonding
Difficulty Level Easy [Embibe predicted high weightage]
21 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)
The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is the
standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)
(A) 86600 +R(298) ln(16 times 1012)
(B) 86600 minusln(16times1012)
R(298)
(C) 05[2 times 86600minus R(298) ln(16 times 1012 )]
(D) R(298) ln(16 times 1012) minus 86600
Solution (C) ΔGreaco = minus2303 RTlog KP
= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac
o = 2ΔGf(NO2)o minus2ΔGf(NO)g
o
= 2ΔGf(NO2)
o minus2 times 866 times 103
2ΔGf(NO2)o = minusR(298)ln(16 times 1012) + 2 times 86600
ΔGf(NO2)o = 86600 minus
R(298)
2ln(16 times 1012)
= 05[2times 86600 minus R(298)ln(16times 1012)]
Topic Thermochemistry and thermodynamics
Difficulty Level Moderate [Embibe predicted high weightage]
22 Which of the following compounds is not colored yellow
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution (D) Cyanides not yellow
BaCrO4 minusYellowK3[Co(NO2)6]minus Yellow
(NH4)3[As(Mo3O10)4] minus Yellow
Zn2[Fe(CN)6] is bluish white
Topic Salt Analysis
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
23 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80) (A) 36
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
16 3 g of activated charcoal was added to 50 mL of acetic acid solution (006N) in a flask After an hour it was
filtered and the strength of the filtrate was found to be 0042 N The amount of acetic acid adsorbed (per
gram of charcoal) is
(A) 36 mg
(B) 42 mg
(C) 54 mg
(D) 18 mg
Solution (D) Meqs of CH3COOH (initial) = 50 times006 = 3 Meqs
Meqs CH3COOH (final) = 50times 0042 = 21 Meqs
CH3 COOH adsorbed = 3 minus 21 = 09 Meqs
= 9times 10minus1 times 60 g Eq frasl times 10minus3 g
= 540 times10minus4 = 0054 g
= 54 mg
Per gram =54
3= 18 mg gfrasl of Charcoal
Topic Surface chemistry
Difficulty Level Moderate [Embibe predicted high weightage]
17 The synthesis of alkyl fluorides is best accomplished by
(A) Sandmeyerrsquos reaction
(B) Finkelstein reaction
(C) Swarts reaction
(D) Free radical fluorination
Solution (C) Alkyl fluorides is best accomplished by swarts reaction ie heating an alkyl chloridebromide in the presence of metallic fluoride such as AgF Hg2F2 CoF2 SbF3
CH3 Br+ AgF rarr CH3 minusF+ AgBr
The reaction of chlorinated hydrocarbons with metallic fluorides to form chlorofluoro
hydrocarbons such as CCl2F2 is known as swarts reaction
Topic Alkyl and Aryl halides
Difficulty Level Easy [Embibe predicted high weightage]
18 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin
(A) 1
206
(B) 2
309
(C) 1
412
(D) 1
103
Solution (C) 2C8H7SO3minusNa++ Ca(aq)
2+ rarr (C8H7SO3)2Ca + 2Na+
2 moles (resin) = 412 g equiv 40 g Ca2+ ions
= 1 moles of Ca2+ ions
Moles of Ca2+ gfrasl of resin =1
412
Topic s-Block elements and Hydrogen
Difficulty Level Moderate (Embibe predicted Low Weightage)
19 Which of the vitamins given below is water soluble
(A) Vitamin D (B) Vitamin E (C) Vitamin K (D) Vitamin C
Solution (D) B complex vitamins and vitamin C are water soluble vitamins that are not
stored in the body and must be replaced each day
Topic Biomolecules
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
20 The intermolecular interaction that is dependent on the inverse cube of distance between the
molecules is
(A) Ion-dipole interaction
(B) London force
(C) Hydrogen bond (D) Ion-ion interaction
Solution (A) For Ion-dipole interaction
F prop μ (dE
dr)
μ is dipole moment of dipole
r is distance between dipole and ion
prop μd
dr(1
r2)prop
μ
r3
Ionminus ion prop1
r2
London force prop1
r6
Hydrogen bond minusDipole minus dipole prop1
r4
Topic Chemical Bonding
Difficulty Level Easy [Embibe predicted high weightage]
21 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)
The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is the
standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)
(A) 86600 +R(298) ln(16 times 1012)
(B) 86600 minusln(16times1012)
R(298)
(C) 05[2 times 86600minus R(298) ln(16 times 1012 )]
(D) R(298) ln(16 times 1012) minus 86600
Solution (C) ΔGreaco = minus2303 RTlog KP
= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac
o = 2ΔGf(NO2)o minus2ΔGf(NO)g
o
= 2ΔGf(NO2)
o minus2 times 866 times 103
2ΔGf(NO2)o = minusR(298)ln(16 times 1012) + 2 times 86600
ΔGf(NO2)o = 86600 minus
R(298)
2ln(16 times 1012)
= 05[2times 86600 minus R(298)ln(16times 1012)]
Topic Thermochemistry and thermodynamics
Difficulty Level Moderate [Embibe predicted high weightage]
22 Which of the following compounds is not colored yellow
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution (D) Cyanides not yellow
BaCrO4 minusYellowK3[Co(NO2)6]minus Yellow
(NH4)3[As(Mo3O10)4] minus Yellow
Zn2[Fe(CN)6] is bluish white
Topic Salt Analysis
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
23 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80) (A) 36
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Difficulty Level Easy [Embibe predicted high weightage]
18 The molecular formula of a commercial resin used for exchanging ions in water softening is C8H7SO3Na (Mol wt 206) What would be the maximum uptake of Ca2+ ions by the resin when expressed in mole per gram resin
(A) 1
206
(B) 2
309
(C) 1
412
(D) 1
103
Solution (C) 2C8H7SO3minusNa++ Ca(aq)
2+ rarr (C8H7SO3)2Ca + 2Na+
2 moles (resin) = 412 g equiv 40 g Ca2+ ions
= 1 moles of Ca2+ ions
Moles of Ca2+ gfrasl of resin =1
412
Topic s-Block elements and Hydrogen
Difficulty Level Moderate (Embibe predicted Low Weightage)
19 Which of the vitamins given below is water soluble
(A) Vitamin D (B) Vitamin E (C) Vitamin K (D) Vitamin C
Solution (D) B complex vitamins and vitamin C are water soluble vitamins that are not
stored in the body and must be replaced each day
Topic Biomolecules
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
20 The intermolecular interaction that is dependent on the inverse cube of distance between the
molecules is
(A) Ion-dipole interaction
(B) London force
(C) Hydrogen bond (D) Ion-ion interaction
Solution (A) For Ion-dipole interaction
F prop μ (dE
dr)
μ is dipole moment of dipole
r is distance between dipole and ion
prop μd
dr(1
r2)prop
μ
r3
Ionminus ion prop1
r2
London force prop1
r6
Hydrogen bond minusDipole minus dipole prop1
r4
Topic Chemical Bonding
Difficulty Level Easy [Embibe predicted high weightage]
21 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)
The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is the
standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)
(A) 86600 +R(298) ln(16 times 1012)
(B) 86600 minusln(16times1012)
R(298)
(C) 05[2 times 86600minus R(298) ln(16 times 1012 )]
(D) R(298) ln(16 times 1012) minus 86600
Solution (C) ΔGreaco = minus2303 RTlog KP
= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac
o = 2ΔGf(NO2)o minus2ΔGf(NO)g
o
= 2ΔGf(NO2)
o minus2 times 866 times 103
2ΔGf(NO2)o = minusR(298)ln(16 times 1012) + 2 times 86600
ΔGf(NO2)o = 86600 minus
R(298)
2ln(16 times 1012)
= 05[2times 86600 minus R(298)ln(16times 1012)]
Topic Thermochemistry and thermodynamics
Difficulty Level Moderate [Embibe predicted high weightage]
22 Which of the following compounds is not colored yellow
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution (D) Cyanides not yellow
BaCrO4 minusYellowK3[Co(NO2)6]minus Yellow
(NH4)3[As(Mo3O10)4] minus Yellow
Zn2[Fe(CN)6] is bluish white
Topic Salt Analysis
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
23 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80) (A) 36
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
20 The intermolecular interaction that is dependent on the inverse cube of distance between the
molecules is
(A) Ion-dipole interaction
(B) London force
(C) Hydrogen bond (D) Ion-ion interaction
Solution (A) For Ion-dipole interaction
F prop μ (dE
dr)
μ is dipole moment of dipole
r is distance between dipole and ion
prop μd
dr(1
r2)prop
μ
r3
Ionminus ion prop1
r2
London force prop1
r6
Hydrogen bond minusDipole minus dipole prop1
r4
Topic Chemical Bonding
Difficulty Level Easy [Embibe predicted high weightage]
21 The following reaction is performed at 298 K 2NO(g) + O2(g) 2NO2(g)
The standard free energy of formation of NO(g) is 866 kJmol at 298 K What is the
standard free energy of formation of NO2(g) at 298 K (KP = 16 times 1012)
(A) 86600 +R(298) ln(16 times 1012)
(B) 86600 minusln(16times1012)
R(298)
(C) 05[2 times 86600minus R(298) ln(16 times 1012 )]
(D) R(298) ln(16 times 1012) minus 86600
Solution (C) ΔGreaco = minus2303 RTlog KP
= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac
o = 2ΔGf(NO2)o minus2ΔGf(NO)g
o
= 2ΔGf(NO2)
o minus2 times 866 times 103
2ΔGf(NO2)o = minusR(298)ln(16 times 1012) + 2 times 86600
ΔGf(NO2)o = 86600 minus
R(298)
2ln(16 times 1012)
= 05[2times 86600 minus R(298)ln(16times 1012)]
Topic Thermochemistry and thermodynamics
Difficulty Level Moderate [Embibe predicted high weightage]
22 Which of the following compounds is not colored yellow
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution (D) Cyanides not yellow
BaCrO4 minusYellowK3[Co(NO2)6]minus Yellow
(NH4)3[As(Mo3O10)4] minus Yellow
Zn2[Fe(CN)6] is bluish white
Topic Salt Analysis
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
23 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80) (A) 36
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Solution (C) ΔGreaco = minus2303 RTlog KP
= minusRT lnKP = minusR(298) ln(16 times 1012) ΔGreac
o = 2ΔGf(NO2)o minus2ΔGf(NO)g
o
= 2ΔGf(NO2)
o minus2 times 866 times 103
2ΔGf(NO2)o = minusR(298)ln(16 times 1012) + 2 times 86600
ΔGf(NO2)o = 86600 minus
R(298)
2ln(16 times 1012)
= 05[2times 86600 minus R(298)ln(16times 1012)]
Topic Thermochemistry and thermodynamics
Difficulty Level Moderate [Embibe predicted high weightage]
22 Which of the following compounds is not colored yellow
(A) K3[Co(NO2)6]
(B) (NH4)3[As(Mo3O10)4]
(C) BaCrO4
(D) Zn2[Fe(CN)6
Solution (D) Cyanides not yellow
BaCrO4 minusYellowK3[Co(NO2)6]minus Yellow
(NH4)3[As(Mo3O10)4] minus Yellow
Zn2[Fe(CN)6] is bluish white
Topic Salt Analysis
Difficulty Level Easy [Embibe predicted easy to score (Must Do)]
23 In Carius method of estimation of halogens 250 mg of an organic compound gave 141 mg of AgBr The percentage of bromine in the compound is (Atomic mass Ag = 108 Br = 80) (A) 36
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(B) 48
(C) 60
(D) 24
Solution (D) Rminus BrCarius methodrarr AgBr
250 mg organic compound is RBr
141 mg AgBr rArr 141 times80
188mg Br
Br in organic compound
= 141 times80
188times
1
250times 100 = 24
Topic Stoichiometry
Difficulty Level Moderate [Embibe predicted high weightage]
24 Sodium metal crystallizes in a body centred cubic lattice with a unit cell edge of 429 Å The
radius of sodium atom is approximately
(A) 322 Å
(B) 572 Å
(C) 093 Å
(D) 186 Å
Solution (D)
For BCC
4r = radic3a
r =radic3
4 a =
1732
4times 429
= 186 Å
Topic Solid State
Difficulty Level Easy [Embibe predicted Low Weightage]
25 Which of the following compounds will exhibit geometrical isomerism
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(A) 3 ndash Phenyl ndash 1- butene
(B) 2 ndash Phenyl ndash 1- butene
(C) 11 ndash Diphenyl ndash 1- propane
(D) 1 ndash Phenyl ndash 2 ndash butene
Solution (D) 1 - Phenyl - 2 - butene
Topic General Organic Chemistry and Isomerism
Difficulty Level Easy [Embibe predicted high weightage]
26 The vapour pressure of acetone at 20oC is 185 torr When 12 g of a non-volatile substance was dissolved in
100 g of acetone at 20oC its vapour pressure was 183 torr The molar mass (g molminus1) of the substance is
(A) 64
(B) 128
(C) 488
(D) 32
Solution (A) ΔP = 185 minus 183 = 2 torr
ΔP
Po=2
185= XB =
12M
(12M) +
10058
M(CH3)2CO = 15 times2 + 16+ 12 = 58 g molefrasl
12
M≪100
58
rArr2
185=12
Mtimes58
100
M =58 times12
100times185
2
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
= 6438 asymp 64 g molefrasl
Topic Liquid solution
Difficulty Level Easy [Embibe predicted high weightage]
27 From the following statements regarding H2O2 choose the incorrect statement
(A) It decomposes on exposure to light
(B) It has to be stored in plastic or wax lined glass bottles in dark
(C) It has to be kept away from dust
(D) It can act only as an oxidizing agent
Solution (D) It can act both as oxidizing agent and reducing agent
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
28 Which one of the following alkaline earth metal sulphates has its hydration enthalpy greater than its lattice enthalpy
(A) BeSO4 (B) BaSO4 (C) SrSO4 (D) CaSO4
Solution (A) ΔHHydration gt ΔHLattice
Salt is soluble BeSO4 is soluble due to high hydration energy of small Be2+ ion Ksp for
BeSO4 is very high
Topic s-Block elements and hydrogen
Difficulty Level Easy [Embibe predicted Low Weightage]
29 The standard Gibbs energy change at 300 K for the reaction 2A B+ C is 24942 J At a given
time the composition of the reaction mixture is [A] =1
2 [B] = 2 and [C] =
1
2 The reaction
proceeds in the [R = 8314 J K minusmol e = 2718]frasl (A) Reverse direction because Q gt KC
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(B) Forward direction because Q lt KC
(C) Reverse direction because Q lt KC
(D) Forward direction because Q gt KC
Solution (A) ∆Go = minusRT In KC
24942 = minus8314times 300 in KC
24942 = minus8314times 300 times 2303log KC
minus24942
2303times300times8314= minus044 = log KC
logKC = minus044 = 1 56
KC = 036
QC =2times
1
2
(12)2 = 4
QC gt KC reverse direction
Topic Chemical Equilibrium
Difficulty Level Moderate [Embibe predicted easy to score (Must Do)]
30 Which one has the highest boiling point
(A) Ne
(B) Kr
(C) Xe (D) He
Solution (C) Due to higher Vander Waalrsquos forces Xe has the highest boiling point
Topic Group 15 ndash 18
Difficulty Level Easy [Embibe predicted high weightage]
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Mathematics 31 The sum of coefficients of integral powers of x in the binomial expansion of (1 minus 2radic119909)
50 is
(A) 1
2(350)
(B) 1
2(350 minus 1)
(C) 1
2(250 +1)
(D) 1
2(350 + 1)
Answer (D)
Solution
Integral powers rArr odd terms
119900119889119889 119905119890119903119898119904 = (1 minus 2radic119909 )
50+ (1 + 2radic119909 )
50
2
119878119906119898 119900119891 119888119900119890119891119891119894119888119894119890119899119905119904 = (1 minus 2)50 + (1+ 2)50
2
= 1 + 350
2
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
32 Let 119891(119909) be a polynomial of degree four having extreme values at 119909 = 1 119886119899119889 119909 = 2 If 119897119894119898119909 rarr 0
[1 +119891(119909)
1199092] = 3 119905ℎ119890119899 119891(2) is equal to
(A) minus4
(B) 0
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(C) 4
(D) minus8
Answer (B)
Solution
Let 119891(119909) = 1198861199094 + 1198871199093 + 1198881199092 + 119889119909 + 119890
119897119894119898119909 ⟶ 0
[1 +119891(119909)
1199092] = 3
119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888 +119889
119909+
119890
1199092] = 3
This limit exists when d = e = 0
So 119897119894119898119909 ⟶ 0
[1 + 1198861199092 +119887119909 + 119888] = 3
rArr 119888 + 1 = 3
119888 = 2
It is given 119909 = 1 119886119899119889 119909 = 2 are solutions of 119891 prime(119909) = 0
119891 prime(119909) = 41198861199093 +31198871199092 + 2119888119909
119909(41198861199092 +3119887119909 + 2119888) = 0
12 are roots of quadratic equation
rArr 119904119906119898 119900119891 119903119900119900119900119905119904 =minus 3119887
4 119886= 1+ 2 = 3
rArr 119887 = minus4119886
Product of roots = 2119888
4119886= 12 = 2
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
rArr 119886 =119888
4
119886 =1
2 119887 = minus2
there4 119891(119909) =1
2 1199094 minus 21199093 + 21199092
119891 (2) = 8minus 16+ 8
= 0
Topic Application of Derivatives
Difficulty level Moderate ( embibe predicted high weightage)
33 The mean of the data set comprising of 16 observations is 16 If one of the observation
valued 16 is deleted and three new observations valued 3 4 and 5 are added to the data then
the mean of the resultant data is
(A) 160
(B) 158
(C) 140
(D) 168
Answer (C)
Solution
Given sum 119909119894+1615119894=1
16= 16
rArr sum119909119894+ 16 = 256
15
119894=1
sum119909119894 = 240
15
119894=1
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Required mean = sum 119909119894+3+4+515119894=1
18=240+3+4+5
18
=252
18= 14
Topic Statistics
Difficulty level Moderate (embibe predicted easy to score (Must Do))
34 The sum of first 9 terms of the series 13
1+13+23
1+3+13+23+33
1+3+5+⋯ 119894119904
(A) 96
(B) 142
(C) 192
(D) 71
Answer (A)
Solution
119879119899 =13+23+⋯ +1198993
1+3+⋯+(2119899minus1)=
1198992 (119899+1)2
4
1198992=(119899+1)2
4
Sum of 9 terms = sum(119899+1)2
4
9119899=1
=1
4times [22 +32 +⋯+102]
=1
4[(12 +22 +⋯+102) minus 12]
=1
4[101121
6minus 1]
=1
4[384] = 96
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
35 Let O be the vertex and Q be any point on the parabola 1199092 = 8119910 If the point P divides the
line segment OQ internally in the ratio 1 3 then the locus of P is
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(A) 1199102 = 119909
(B) 1199102 = 2119909
(C) 1199092 = 2119910
(D) 1199092 = 119910
Answer (C)
Solution
General point on 1199092 = 8119910 is 119876 (4119905 21199052)
Let P(h k) divide OQ in ratio 13
(h k) = (1 (4119905)+3(0)
1+3 1(21199052 )+3(0)
1+3)
(h k) = (11990521199052
4)
h = t and 119896 =1199052
2
119896 =ℎ2
2
rArr 1199092 = 2119910 is required locus
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Parabola
Difficulty level Moderate (embibe predicted high weightage)
36 Let 120572 119886119899119889 120573 be the roots of equation1199092 minus6119909 minus 2 = 0 If 119886119899 = 120572119899 minus 120573119899 for 119899 ge 1 then the
value of 11988610minus21198868
21198869 is equal to
(A) -6
(B) 3
(C) -3
(D) 6
Answer (B)
Solution
Given 120572 120573 are roots of 1199092 minus6119909 minus 2 = 0
rArr 1205722 minus 6120572 minus 2 = 0 and 1205732 minus6120573 minus 2 = 0
rArr 1205722 minus 6 = 6120572 and 1205732 minus 2 = 6120573 hellip(1)
11988610 minus2119886821198869
= (12057210 minus12057310) minus 2(1205728 minus 1205738)
2(1205729 minus1205739)
= (12057210minus21205728 )minus(12057310minus21205738)
2(1205729minus1205739)
= 1205728 (1205722minus2)minus1205738(1205732minus2)
2(1205729minus1205739)
= 1205728 (6120572)minus1205738(6120573)
2(1205729minus1205739) [∵ 119865119903119900119898 (1)]
= 61205729minus61205739
2(1205729minus1205739)= 3
Topic Quadratic Equation amp Expression
Difficulty level Moderate (embibe predicted high weightage)
37 If 12 identical balls are to be placed in 3 identical boxes then the probability that one of the
boxes contains exactly 3 balls is
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(A) 55(2
3)10
(B) 220(1
3)12
(C) 22(1
3)11
(D) 55
3(2
3)11
Answer (C)
Solution (A)
B1 B2 B3
12 0 0
11 1 0
10 1 1
10 2 0
9 3 0
9 2 1
8 4 0
8 3 1
8 2 2
7 5 0
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
7 4 1
7 3 2
6 6 0
6 5 1
6 4 2
6 3 3
5 5 2
5 4 3
4 4 4
Total number of possibilities = 19
Number of favorable case = 5
Required probability = 5
19
INCORRECT SOLUTION
Required Probability = 121198629 sdot (1
3)3
sdot (2
3)9
= 55 sdot (2
3)11
The mistake is ldquowe canrsquot use 119899119862119903 for identical objectsrdquo
Topic Probability
Difficulty level Difficult (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
38 A complex number z is said to be unimodular if |119911| = 1 suppose 1199111 119886119899119889 1199112 are complex
numbers such that 1199111minus21199112
2minus11991111199112 is unimodular and 1199112 is not unimodular Then the point 1199111lies on a
(A) Straight line parallel to y-axis
(B) Circle of radius 2
(C) Circle of radius radic2
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given 1199111minus 211988522minus11991111199112
is unimodular
rArr | 1199111minus 211988522minus11991111199112
| = 1
|1199111minus 21198852 | = |2 minus 11991111199112|
Squaring on both sides
|1199111minus 21198852 |2= |2 minus 11991111199112|
2
(1199111minus21198852)(1199111 minus 21199112) = (2 minus 11991111199112)(2 minus 11991111199112)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Complex Number
Difficulty level Easy (embibe predicted high weightage)
39 The integral int119889119909
1199092 (1199094+1)34
equals
(A) (1199094+ 1)1
4 + 119888
(B) minus(1199094 +1)1
4 + 119888
(C) minus(1199094+1
1199094)
1
4+ 119888
(D) (1199094+1
1199094)
1
4+ 119888
Answer (A)
Solution
119868 = int119889119909
1199092(1199094+1)34
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
= int119889119909
1199092times 1199093 (1+1
1199094)
34
119875119906119905 1 +1
1199094= 119905
rArr minus4
1199095119889119909 = 119889119905
there4 119868 = intminus119889119905
4
11990534
= minus1
4times int 119905
minus3
4 119889119905
=minus1
4times (
11990514
1
4
)+ 119888
= minus (1+1
1199094)
1
4 + c
Topic Indefinite Integral
Difficulty level Moderate (embibe predicted Low Weightage)
40 The number of points having both co-ordinates as integers that lie in the interior of the
triangle with vertices (0 0) (0 41) and (41 0) is
(A) 861
(B) 820
(C) 780
(D) 901
Answer (C)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Solution
119875(11990911199101) lies inside the triangle rArr 11990911199101 120598 119873
1199091 +1199101 lt 41
there4 2 le 1199091 +1199101 le 40
Number of points inside = Number of solutions of the equation
1199091 +1199101 = 119899
2 le 119899 le 40
1199091 ge 1 1199101 ge 1
(1199091 minus1) + (1199101 minus 1) = (119899 minus 2)
(119873119906119898119887119890119903 119900119891 119899119900119899minus 119899119890119892119886119905119894119907119890 119894119899119905119890119892119903119886119897 119904119900119897119906119905119894119900119899119904 119900119891 1199091 +1199092 + ⋯+119909119899 = 119903 119894119904 119899+119903minus1119862 119903 )
Number of solutions = 2+(119899minus2)minus1119862119899minus2
= 119899minus1 119862119899minus2
= 119899 minus 1
We have 2 le 119899 le 40
there4 Number of solutions = 1 + 2 +hellip+39
= 39 times40
2
= 780
Topic Straight lines
Difficulty level Moderate (embibe predicted high weightage)
41 The distance of the point (1 0 2) from the point of intersection of the line 119909minus2
3=119910+1
4=119911minus2
12
and the plane 119909 minus 119910 + 119911 = 16 is
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(A) 8
(B) 3radic21
(C) 13
(D) 2radic14
Answer (C)
Solution
119909minus2
3=119910+1
4=119911minus1
12= 119905 (119904119886119910)
rArr General point is (2 + 3119905 minus1 + 4119905 2 + 12119905)
It lies on the plane 119909 minus 119910 + 119911 = 16
rArr 2 + 3119905 + 1minus 4119905 + 2+ 12119905 = 16
rArr 11119905 = 11
rArr 119905 = 1
there4 The point of intersection will be
(2 + 3(1)minus1 + 4(1)2 + 12(1))
= (5 3 14)
Distance from (102) = radic(5 minus 1)2 + (3minus 0)2 + (14minus 2)2
= radic42 +32 +122
= 13
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
42 The equation of the plane containing the line 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 and
parallel to the plane 119909 + 3119910 + 6119911 = 1 is
(A) 119909 + 3119910+ 6119911 = minus7
(B) 119909 + 3119910 + 6119911 = 7
(C) 2119909 + 6119910+ 12119911 = minus13
(D) 2119909 + 6119910 + 12119911 = 13
Answer (B)
Solution
Any plane containing the line of intersection of 2119909 minus 5119910 + 119911 = 3 119909 + 119910 + 4119911 = 5 will be of
the form
(2119909 minus 5119910+ 119911 minus 3) + 120582 (119909 + 119910 + 4119911 minus 5) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
(2 + 120582)119909 minus (5 minus 120582)119910 + (1 + 4120582)119911 minus (3 + 5120582) = 0
It is parallel to plane 119909 + 3119910+ 6119911 = 1
rArr 2 + 120582
1= 120582 minus 5
3= 1 + 4120582
6 ne
minus(3 + 5120582)
11
2 + 120582
1= 120582 minus 5
3rArr 6 + 3120582 = 120582 minus 5
2120582 = minus11
120582 = minus11
2
there4 Required plane is
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(2119909 minus 5119910+ 119911 minus 3) minus11
2 (119909 + 119910 + 4119911 minus 5) = 0
2(2119909 minus 5119910 + 119911 minus 3) minus 11 (119909 + 119910 + 4119911 minus 5) = 0
4119909 minus 10119910+ 2119911 minus 6minus 11119909 minus 11119910minus 44119911 + 55 = 0
minus7119909 minus 21119910minus 42119911 + 49 = 0
rArr 119909 + 3119910 + 6119911 minus 7 = 0
Topic 3 - D Geometry
Difficulty level Easy (embibe predicted high weightage)
43 The area (in sq units) of the region described by [(119909 119910)1199102 le 2119909 and 119910 ge 4119909 minus 1] is
(A) 5
64
(B) 15
64
(C) 9
32
(D) 7
32
Answer (C)
Solution
Let us find the points intersections of 1199102 = 2119909 and 119910 = 4119909 minus 1
(4119909 minus 1)2 = 2119909
161199092 minus10119909 + 1 = 0
(8119909 minus 1) (2119909 minus 1) = 0
119909 =1
21
8
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
119909 =1
2 rArr 119910 = 4 (
1
2)minus 1 = 1
119909 = 1
8 rArr 119910 = 4 (
1
8)minus 1 =
minus1
2
Required area = int (1199091minus1
2 2
minus 1199091) 119889119910
= int [1199102
2minus (119910 + 1
4)]
1
minus12
119889119910
=1
6times lfloor1199103rfloorminus1
2
1 minus 1
8 lfloor1199102rfloorminus1
2
1 minus1
4 lfloor119910rfloorminus1
2
1
=1
6 [1 +
1
8] minus
1
8 [1 minus
1
4] minus
1
4 [1 +
1
2]
=1
6times9
8minus1
8times3
4minus1
4times3
2
=3
16minus3
32minus3
8
=6 minus 3 minus 12
32=minus9
32 squnits
Topic Area under Curves
Difficulty level Easy (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
44 If m is the AM of two distinct real numbers 119897 119886119899119889 119899 (119897119899 gt 1)119886119899119889 1198661 1198662 119886119899119889 1198663 are three
geometric means between 119897 119886119899119889 119899 119905ℎ119890119899 11986614 + 21198662
4 + 11986634 equals
(A) 4 1198971198982119899
(B) 4 1198971198981198992
(C) 4 119897211989821198992
(D) 4 1198972119898119899
Answer (A)
Solution
m is the AM of 119897 119899
rArr 119898 =119897+119899
2 hellip(1)
1198661 1198662 1198663 are GM of between l and n
rArr 1198971198661 11986621198663 119899 are in GP
119899 = 119897(119903)4 rArr 1199034 =119899
119897 hellip(2)
1198661 = 119897119903 1198662 = 1198971199032 1198663 = 119897119903
3
11986614 +21198662
4 + 11986634 = 11989741199034 +2 times 11989741199038 + 1198974 times 11990312
= 11989741199034[1 + 21199034 + 1199038]
= 1198974 times119899
119897[1 + 1199034]2 = 1198973119899 times [1 +
119899
119897]2
= 1198973 times 119899 times(119897+119899)2
1198972
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
= 119897119899 times (2119898)2 [∵ 119891119903119900119898 (1)]
= 41198971198982119899
Topic Progression amp Series
Difficulty level Moderate (embibe predicted high weightage)
45 Locus of the image of the point (2 3) in the line (2119909 minus 3119910+ 4) + 119896 (119909 minus 2119910 + 3) = 0119896 isin
119877 is a
(A) Straight line parallel to y-axis
(B) Circle of radius radic2
(C) Circle of radius radic3
(D) Straight line parallel to x-axis
Answer (B)
Solution
Given family of lines 1198711 + 1198712 = 0
Let us take the lines to be
1198712+ 120582(1198711 minus 1198712) = 0
(119909 minus 2119910+ 3) + 120582(119909 minus 119910+ 1) = 0
(1 + 120582)119909 minus (2 + 120582)119910 + (3 + 120582) = 0
Let Image of (2 3) be (h k) ℎ minus 2
1 + 120582=
119896 minus 3
minus(2 + 120582)= minus2(2+ 2120582 minus 6 minus 3120582 + 3+ 120582)
(1 + 120582)2 + (2 + 120582)2
ℎminus2
120582+1=
119896minus3
minus(120582+2)=
minus2(minus1)
(1+120582)2+(2+120582)2 (1)
ℎ minus 2
120582 + 1=
119896 minus 3
minus(120582 + 2)rArr 120582 + 2
120582 + 1= 3 minus 119896
ℎ minus 2
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
1 + 1
120582+1=
3minus119896
ℎminus2
1
120582+1=
5minusℎminus119896
ℎminus2 (2)
From (1) (ℎ minus 2)2 + (119896 minus 3)2 = 4
(120582+1)2+(120582+2)2 (3)
From (1) and (3)
ℎ minus 2
120582 + 1=
2
(120582 + 1)2 + (120582 + 2)2
5 minus ℎ minus 119896 = 1
2 [(ℎ minus 2)2 + (119896 minus 3)2]
10 minus 2ℎminus 2119896 = ℎ2 +1198962 minus4ℎ minus 6119896 + 13
1199092 + 1199102 minus 2119909 minus 4119910+ 3 = 0
Radius = radic1 + 4minus 3
= radic2
Topic Straight lines
Difficulty level Difficult (embibe predicted high weightage)
46 The area (in sq units) of the quadrilateral formed by the tangents at the end points of the
latera recta to the ellipse 1199092
9+1199102
5= 1 is
(A) 18
(B) 27
2
(C) 27
(D) 27
4
Answer (C)
Solution
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
1199092
9+1199102
5= 1
119890 = radic1 minus5
9=2
3
1198862 = 9 1198872 = 5
Focii = (plusmn 1198861198900) = (plusmn 20)
Ends of latus recta = (plusmn 119886119890 plusmn 1198872
119886)
= (plusmn 2plusmn5
3)
Tangent at lsquoLrsquo is T = 0 2 sdot 119909
9+ 5
3 sdot 119910
5= 1
It cut coordinates axes at 119875 (9
2 0) 119886119899119889 119876(03)
Area of quadrilateral PQRS = 4(Area of triangle OPQ)
= 4 (1
2 sdot 9
2 sdot 3) = 27 119904119902119906119886119903119890 119906119899119894119905119904
Topic Ellipse
Difficulty level Easy (embibe predicted high weightage)
47 The number of integers greater than 6000 that can be formed using the digits 3 5 6 7 and
8 without repetition is
(A) 192
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(B) 120
(C) 72
(D) 216
Answer (A)
Solution
4 digit and 5 digit numbers are possible
4 digit numbers
678
uarr uarr uarr3 4 3
uarr2
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 3 ∙ 4 ∙ 3 ∙ 2 = 72
5 digit numbers
uarr uarr uarr5 4 3
uarr uarr2 1
119879119900119905119886119897 119899119906119898119887119890119903119904 119901119900119904119904119894119887119897119890 = 5 ∙ 4 ∙ 3 ∙ 2 ∙ 1 = 120
there4 119879119900119905119886119897 119899119906119898119887119890119903119904 = 72 + 120 = 192
Topic Binomial Theorem
Difficulty level Moderate (embibe predicted high weightage)
48 Let A and B be two sets containing four and two elements respectively Then the number of
subsets of the set 119860 times 119861 each having at least three elements is
(A) 256
(B) 275
(C) 510
(D) 219
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Answer (D)
Solution
119899(119860) = 4
119899(119861) = 2
Number of elements in 119860 times 119861 = 2 sdot 4 = 8
Number of subsets having at least 3 elements
= 28minus81198620 minus 1198628 1 minus 1198628 2
= 256 minus 1minus 8minus 28
= 219
Topic Set and Relations
Difficulty level Moderate(embibe predicted high weightage)
49 Let tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) where |119909| lt
1
radic3 Then a value of y is
(A) 3119909+1199093
1minus31199092
(B) 3119909minus1199092
1+31199092
(C) 3119909+1199093
1+31199092
(D) 3119909minus1199093
1minus31199092
Answer (D)
Solution
tanminus1 119910 = tanminus1 119909 + tanminus1 (2119909
1minus1199092) |119909| lt
1
radic3
= tanminus1 119909 + 2 tanminus1 119909
= 3tanminus1 119909
tanminus1 119910 = tanminus1 (3119909minus1199093
1minus31199092)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
rArr 119910 =3119909minus1199093
1minus31199092
Topic Inverse Trigonometric Function
Difficulty level Moderate (embibe predicted easy to score (Must Do))
50 The integral intlog 1199092
log 1199092+log (36 minus 12119909 + 1199092 )119889119909
4
2 is equal to
(A) 4
(B) 1
(C) 6
(D) 2
Answer (B)
Solution
119868 = intlog 1199092
log1199092 + log(6 minus 119909)2
4
2
119889119909
119868 = intlog(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
[∵ int119891(119909)119889119909 = int119891(119886 + 119887 minus 119909)119889119909
119887
119886
119887
119886
]
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092
4
2
119889119909
2119868 = intlog 1199092 + log(6 minus 119909)2
log(6 minus 119909)2 + log1199092 119889119909
4
2
2119868 = int1 119889119909
4
2
2119868 = [119909]24
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
2119868 = 4 minus 2
119868 = 1
Topic Definite Integral
Difficulty level Easy (embibe predicted high weightage)
51 The negation of ~119904 or (~119903 and 119904) is equivalent to
(A) 119904 and (119903 and sim 119904)
(B) 119904 or (119903 or sim 119904)
(C) 119904 and 119903
(D) 119904 andsim 119903
Answer (C)
Solution
~ ((~119904) or (sim 119903 and 119904)) = 119904 and [sim ((sim 119903) and 119904)] [∵ ~(119901 and 119902) = (sim 119901) or (sim 119902)]
= 119904 and (119903 or (sim 119904)) [sim (119901 or 119902) = (sim 119901) and (sim 119902)]
= (119904 and 119903) or (119904 and (sim 119904))
= (119904 and 119903) or 119865
= 119904 and 119903 (119865 rarr 119865119886119897119897119886119888119910)
Topic Mathematical Reasoning
Difficulty level Moderate (embibe predicted easy to score (Must Do))
52 If the angles of elevation of the top of a tower from three collinear points A B and C on a
line leading to the foot of the tower are 30deg 45deg 119886119899119889 60deg respectively then the ratio 119860119861 ∶
119861119862 is
(A) radic3 ∶ radic2
(B) 1 ∶ radic3
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(C) 2 ∶ 3
(D) radic3 ∶ 1
Answer (D)
Solution
tan 30119900 = ℎ
119909 + 119910+ 119911tan 450 =
ℎ
119910 + 119911 tan600 =
ℎ
119911
rArr 119909 + 119910 + 119911 = radic3 ℎ
119910 + 119911 = ℎ
119911 =ℎ
radic3
119910 = ℎ (1 minus 1
radic3)
119909 = (radic3minus 1)ℎ
119909
119910= (radic3 minus 1)ℎ
ℎ(radic3minus 1
radic3)
= radic3
1
Topic Height amp Distance
Difficulty level Moderate (embibe predicted easy to score (Must Do))
53 119897119894119898119909 rarr 0
(1minuscos 2119909) (3+cos 119909)
119909 tan 4119909 is equal to
(A) 3
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(B) 2
(C) 1
2
(D) 4
Answer (B)
Solutions
lim119909rarr119900
(1 minus cos2119909) (3+ cos119909)
119909 tan 4119909
= lim119909rarr119900
2 sin2 119909 sdot (3 + cos119909)
119909 sdot tan 4119909
= lim119909rarr119900
2 sdot (sin 119909119909)2
sdot (3 + cos119909)
tan 41199094119909
sdot 4
= 2 sdot (1)2 sdot (3 + 1)
1 sdot 4
= 2
Topic LimitContinuity amp Differentiability
Difficulty level Easy (embibe predicted high weightage)
54 Let 119888 be three non-zero vectors such that no two of them are collinear and ( times ) times 119888
=1
3 || |119888| || If 120579 is the angle between vectors 119886119899119889 119888 then a value of sin 120579 is
(A) minusradic2
3
(B) 2
3
(C) minus2radic3
3
(D) 2radic2
3
Answer (D)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Solution
( times ) times 119888 = 1
3 || |119888|
( sdot 119888) minus ( sdot 119888) = 1
3 || |119888|
rArr sdot 119888 = 0 119886119899119889 sdot 119888 = minus1
3 || |119888|
|| |119888| 119888119900119904120579 = minus1
3 || |119888|
119888119900119904120579 = minus1
3
119904119894119899120579 = radic1 minus1
9= radic
8
9 =
2radic2
3
Topic Vector Algebra
Difficulty level Moderate (embibe predicted high weightage)
55 If 119860 = [1 2 22 1 minus2119886 2 119887
] is a matrix satisfying the equation 119860119860119879 = 9119868 where 119868 119894119904 3 times 3 identity
matrix then the ordered pairs (a b) is equal to
(A) (minus21)
(B) (21)
(C) (minus2minus1)
(D) (2 minus1)
Answer (C)
Solution
119860119860119879 = 9119868
[1 2 22 1 minus2119886 2 119887
] [1 2 1198862 1 22 minus2 119887
] = [9 0 00 9 00 0 9
]
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
[
1(1) + 2(2)
2(1) + 1(2)119886(1) + 2(2)
+ 2(2)
minus 2(2)+ 119887(2)
1(2) + 2(1)
2(2) + 1(1)119886(2) + 2(1)
+ 2(minus2)
minus 2(minus2)+ 119887(minus2)
1(119886) +
2(119886) +119886(119886) +
2(2) + 2(119887)
1(2) minus 2(119887)2(2) + 119887(119887)
]
= [9 0 00 9 00 0 9
]
[9 0 119886 + 2119887 + 40 9 2119886minus 2119887 + 2
119886 + 2119887 + 4 2119886minus 2119887 + 2 1198862 + 1198872 +4
] = [9 0 00 9 00 0 9
]
119862119900119898119901119886119903119894119899119892 119905ℎ119890 119888119900119903119903119890119904119901119900119899119889119894119899119892 119890119897119890119898119890119899119905119904
119886 + 2119887 +4 = 02119886 minus 2119887 +2 = 0
and 1198862 + 1198872 +4 = 9
____________________
3119886 + 6 = 0
119886 = minus2 rArr minus2 + 2119887 + 4 = 0
119887 = minus1
The third equation is useful to verify whether this multiplication is possible
Topic Matrices
Difficulty level Easy (embibe predicted high weightage)
56 If the function 119892(119909) = 119896 radic119909 + 1 0 le 119909 le 3119898119909 + 2 3 lt 119909 le 5
is differentiable then the value of k + m is
(A) 16
5
(B) 10
3
(C) 4
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(D) 2
Answer (D)
Solution
119892(119909) is differentiable at x = 3
rArr 119892 (119909) is continuous at x = 3
there4 LHL = RHL = 119892(3)
lim119909rarr3minus
119892(119909) = lim119909rarr3+
119892(119909) = 119892(3)
119896radic3+ 1 = 119898(3) + 2
2119896 = 3119898 + 2
119896 = 3
2 119898 + 1 helliphellip(1)
119892(119909) is differentiable
rArr 119871119867119863 = 119877119867119863
119892prime(119909) = 119896
2radic119909 + 1119898
0 lt 119909 lt 33 lt 119909 lt 5
lim119909rarr3minus
119892prime(119909) = lim119909rarr3+
119892prime(119909)
119896
2radic3+ 1= 119898
119896 = 4119898
From (1) 4119898 = 3
2119898+ 1
5119898
2= 1
rArr 119898 = 2
5 119896 =
8
5
119896 +119898 = 2
5+
8
5= 2
Topic LimitContinuity amp Differentiability
Difficulty level Moderate (embibe predicted high weightage)
57 The set of all values of 120582 for which the system of linear equations
21199091 minus21199092 + 1199093 = 1205821199091
21199091 minus31199092 + 21199093 = 1205821199092
minus1199091 +21199092 = 1205821199093 has a non-trivial solution
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(A) Is a singleton
(B) Contains two elements
(C) Contains more than two elements
(D) Is an empty set
Answer (B)
Solution
(2 minus 120582)1199091 = 21199092 +1199093 = 0
21199091 minus (120582 + 3)1199092 + 21199093 = 0
minus 1199091 + 21199092 minus 1205821199093 = 0
The systems of linear equations will have a non-trivial solution
rArr |2 minus 1205822minus1
minus2
minus120582 minus 32
12minus120582| = 0
rArr (2 minus 120582)[1205822 +3120582 minus 4] + 2[minus2120582+ 2] + 1 4 minus 120582 minus 3] = 0
21205822 +6120582 minus 8minus 1205823 minus 31205822 +4120582 minus 4120582 + 4+ 1minus 120582 = 0
minus1205823 minus 1205822 + 5120582 minus 3 = 0
1205823 + 1205822 minus 5120582 + 3 = 0
(120582 minus 1) (1205822 +2120582 minus 3) = 0
(120582 minus 1) (120582+ 3) (120582minus 1) = 0
rArr 120582 = 3minus1minus1
Topic Determinants
Difficulty level Easy (embibe predicted high weightage)
58 The normal to the curve 1199092 + 2119909119910minus 31199102 = 0 at (1 1)
(A) Meets the curve again in the second quadrant
(B) Meets the curve again in the third quadrant
(C) Meets the curve again in the fourth quadrant
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(D) Does not meet the curve again
Answer (C)
Solution
1199092 +2119909119910minus 31199102 = 0
(119909 + 3119910) (119909minus 119910) = 0
Pair of straight lines passing through origin
119909 + 3119910 = 0 or 119909 minus 119910 = 0
Normal exists at (1 1) which is on 119909 minus 119910 = 0
rArr Slope of normal at (1 1) = minus1
there4 Equation of normal will be
119910 minus 1 = minus1 (119909 minus 1)
119910 minus 1 = minus119909 + 1
119909 + 119910 = 2
Now find the point of intersection with 119909 + 3119910 = 0
119909 + 119910 = 2
119909 + 3119910 = 0 ____________ minus 2119910 = 2 rArr 119910 = minus1 119909 = 3
(3 1) lies in fourth quadrant
Topic Straight Lines
Difficulty level Easy (embibe predicted high weightage)
59 The number of common tangents to the circles 1199092 +1199102 minus4119909 minus 6119910 minus 12 = 0 and 1199092 +
1199102 +6119909 + 18119910+ 26 = 0 is
(A) 2
(B) 3
(C) 4
(D) 1
Answer (B)
Solution
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
1199092 + 1199102 minus 4119909 minus 6119910minus 12 = 0
1198621(23) 1199031 = radic22 + 32 + 12 = radic25 = 5
1199092 + 1199102 + 6119909 + 18119910+ 26 = 0
1198622(minus3minus9) 1199032 = radic32 +92 minus 26
= radic90 minus 26 = 8
11986211198622 = radic52 + 122 = 13
11986211198622 = 1199031 + 1199032
rArr Externally touching circles
rArr 3 common tangents
Topic Circle
Difficulty level Easy (embibe predicted high weightage)
60 Let 119910(119909) be the solution of the differential equation(119909 log119909)119889119910
119889119909+119910 = 2119909 log119909 (119909 ge 1)
Then 119910 (119890) is equal to
(A) 0
(B) 2
(C) 2e
(D) e
Answer (A)
Solution 119889119910
119889119909+ (
1
119909 log 119909) 119910 = 2
Integrating factor = 119890int119875119889119909
= 119890int
1
119909 log119909 119889119909
= 119890 log (log 119909)
= log 119909
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
The solution will be
119910 sdot 119890int119875119889119909 = int119876 sdot 119890int 119875119889119909 + 119888
119910 sdot log119909 = int2 sdot log 119909 + 119888
119910 sdot log 119909 = 2(119909 log119909 minus 119909) + 119888
Let 119875(11199101) be any point on the curve
1199101(0) = 2(0 minus 1) + 119888
119888 = 2
When 119909 = 119890 119910 (log119890) = 2 (119890 log119890 minus 119890) + 2
119910 = 2
Topic Differential Equation
Difficulty level Easy (embibe predicted easy to score (Must Do))
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Physics
61 As an electron makes transition from an excited state to the ground state of a hydrogen like
atomion
(A) Kinetic energy potential energy and total energy decrease
(B) Kinetic energy decreases potential energy increases but total energy remains same
(C) Kinetic energy and total energy decreases but potential energy increases
(D) Its kinetic energy increases but potential energy and total energy decrease
Answer (D)
Solution 119880 = minus1198902
41205871205760119903
U = potential energy
119896 =1198902
81205871205760119903
K = kinetic energy
119864 = 119880+ 119896 = minus1198902
81205871205760119903
119864 = Total energy
there4 as electron de-excites from excited state to ground state k increases U amp E decreases
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
62 The period of oscillation of a simple pendulum is T = 2πradicL
g Measured value of L is 200 cm
known to 1 mm accuracy and time for 100 oscillations of the pendulum is found to be 90s
using a wrist watch of 1s resolution The accuracy in the determination of g is
(A) 3
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(B) 1
(C) 5 (D) 2
Answer (A)
Solution there4 119879 = 2120587radic119871
119892rArr 119892 = 41205872
119871
1198792
there4 Error in g can be calculated as
Δ119892
119892=ΔL
119871+2ΔT
119879
there4 Total time for n oscillation is 119905 = 119899119879 where T= time for oscillation
rArrΔt
119905=Δ119879
119879
rArrΔ119892
119892=Δ119871
119871+2Δt
119905
Given that Δ119871 = 1 119898119898 = 10minus3 119898119871 = 20times 10minus2119898
Δ119905 = 1119904 119905 = 90119904
error in g
Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
=Δ119892
119892times 100 = (
Δ119871
119871+2Δ119905
119905) times100
= (10minus3
20times 10minus2+2 times1
90) times 100
=1
2+20
9
= 05 + 222
= 272
cong 3
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Unit amp Dimensions
Difficulty Easy (embibe predicted easy to score)
63 A long cylindrical shell carries positive surface charge 120590 in the upper half and negative
surface charge ndash 120590 in the lower half The electric field lines around the cylinder will look like
figure given in (Figures are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Answer (D)
Solution
Consider cross section of cylinders which is circle the half part of circle which has positive charge can be
assume that total positive charge is at centre of mass of semicircle In the same way we can assume that
negative charge is at centre of mass of that semicircle
Now it acts as a dipole now by the properties of dipole and lows of electric field line where two lines should
not intersect the graph would be
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
64 A signal of 5 kHz frequency is amplitude modulated on a carrier wave of frequency 2 MHz
The frequency of the resultant signal isare
(A) 2005 kHz and 1995 kHz
(B) 2005 kHz 2000 kHz and 1995 kHz
(C) 2000 kHz and 1995 kHz (D) 2 MHz only
Answer (B)
Solution
Topic Communication Systems
Difficulty Easy (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
65 Consider a spherical shell of radius R at temperature T the black body radiation inside it can
be considered as an ideal gas of photons with internal energy per unit volume 119906 =119880
119881prop
1198794and pressure 119901 =1
3(119880
119881) If the shell now undergoes an adiabatic expansion the relation
between T and R is
(A) 119879 prop 119890minus3119877
(B) 119879 prop1
119877
(C) 119879 prop1
1198773
(D) 119879 prop 119890minus119877
Answer (B)
Solution ∵ in an adiabatic process
119889119876 = 0
So by first law of thermodynamics
119889119876 = 119889119880+119889119882
rArr 0 = 119889119880+119889119882
rArr 119889119882 = minus119889119880
there4 119889119882 = 119875119889119881
rArr 119875119889119881 = minus119889119880 hellip(i)
Given that 119880
119881prop 1198794rArr 119880 = 1198961198811198794
rArr 119889119880= 119896119889(1198811198794) = 119870(1198794119889119881+ 41198793119881 119889119879)
Also 119875 =1
3
119880
119881=1
3
1198961198811198794
119881=1198701198794
3
Putting these values in equation
rArr1198701198794
3119889119881 = minus119896(1198794119889119881+ 41198793119881 119889119879)
rArr119879119889119881
3= minus119879119889119881minus4119881119889119879
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
rArr4119879
3119889119881 = minus4119881119889119879
rArr
13119889119881
119881= minus
119889119879
119879
rArr1
3ln 119881 = minusln 119879 rArr ln119881 = ln 119879minus3
rArr 1198811198793 = 119888119900119899119904119905119886119899119905
4
31205871198773 1198793 = 119888119900119899119904119905119886119899119905
119877119879 = 119888119900119899119904119905119886119899119905
rArr 119879 prop1
119877
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
66 An inductor (119871 = 003119867) and a resistor (119877 = 015 119896Ω) are connected in series to a battery
of 15V EMF in a circuit shown below The key 1198701 has been kept closed for a long time Then
at 119905 = 0 1198701 is opened and key 1198702 is closed simultaneously At 119905 = 1119898119904 the current in the
circuit will be (1198905 cong 150)
(A) 67 mA
(B) 67 mA
(C) 067 mA (D) 100 mA
Answer (C)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Solution
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
67 A pendulum made of a uniform wire of cross sectional area A has time period T When and
additional mass M is added to its bob the time period changes to 119879119872 If the Youngrsquos
modulus of the material of the wire is Y then 1
119884is equal to
(g = gravitational acceleration)
(A) [(119879119872
119879)2
minus1]119872119892
119860
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(B) [1minus (119879119872
119879)2]119860
119872119892
(C) [1minus (119879
119879119872)2]119860
119872119892
(D) [(119879119872
119879)2
minus1]119860
119872119892
Answer (D)
Solution
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Simple Harmonic Motion
Difficulty Difficult (embibe predicted high weightage)
68 A red LED emits light at 01 watt uniformly around it The amplitude of the electric field of the light at a
distance of 1 m from the diode is
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(A) 245 Vm
(B) 548 Vm
(C) 775 Vm
(D) 173 Vm
Answer (A)
Solution
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Optics
Difficulty Easy (embibe predicted high weightage)
69 Two coaxial solenoids of different radii carry current I in the same direction Let 1198651 be the
magnetic force on the inner solenoid due to the outer one and 1198652 be the magnetic force on
the outer solenoid due to the inner one Then
(A) 1198651 is radially inwards and 1198652 is radially outwards
(B) 1198651 is radially inwards and 1198652 = 0
(C) 1198651 is radially outwards and 1198652 = 0
(D) 1198651 = 1198652 = 0
Answer (D)
Solution
1198782 is solenoid with more radius than 1198781 field because of 1198781 on 1198782 is o
there4 force on 1198782 by 1198781 = 0
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
In the uniform field of 1198782 1198781 behaves as a magnetic dipole
there4 force on 1198781 by 1198782 is zero because force on both poles are equal in magnitude and opposite indirection
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
70 Consider an ideal gas confined in an isolated closed chamber As the gas undergoes an adiabatic
expansion the average time of collision between molecules increase as Vq where V is the volume of the
gas The value of q is
(γ =CPCv)
(A) 3γminus5
6
(B) γ+1
2
(C) γminus1
2
(D) 3γ+5
6
Answer (B)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Heat ampThermodynamics
Difficulty Difficult (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
71 An LCR circuit is equivalent to a damped pendulum In an LCR circuit the capacitor is charged
to 1198760 and then connected to the L and R as shown below
If a student plots graphs of the square of maximum charge (119876119872119886119909
2 ) on the capacitor with time (t) for two
different values 1198711 and 1198712(1198711 gt 1198712) of L then which of the following represents this graph correctly
(plots are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (D)
Solution
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Magnetism
Difficulty Difficult (embibe predicted high weightage)
72 From a solid sphere of mass M and radius R a spherical portion of radius 119877
2 is removed as
shown in the figure Taking gravitational potential 119881 = 0 and 119903 = infin the potential at the
center of the cavity thus formed is
(G = gravitational constant)
(A) minus119866119872
119877
(B) minus2119866119872
3119877
(C) minus2119866119872
119877
(D) minus119866119872
2119877
Answer (A)
Solution
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Gravitation
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Difficulty Moderate (embibe predicted high weightage)
73 A train is moving on a straight track with speed 20 119898119904minus1 It is blowing its whistle at the
frequency of 1000 Hz The percentage change in the frequency heard by a person standing
near the track as the train passes him is (speed of sound = 320 119898119904minus1) close to
(A) 12
(B) 18
(C) 24
(D) 6
Answer (A)
Solution
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Wave amp Sound
Difficulty Moderate (embibe predicted high weightage)
74
Given in the figure are two blocks A and B of weight 20N and 100N respectively These are being pressed
against a wall by a force F as shown If the coefficient of friction between the blocks is 01 and between
block B and the wall is 015 the frictional force applied by the wall on block B is
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(A) 80 N
(B) 120 N
(C) 150 N
(D) 100 N
Answer (B)
Solution
For complete state equilibrium of the system The state friction on the block B by wall will balance the total
weight 120 N of the system
Topic Laws of Motion
Difficulty Moderate (embibe predicted Low Weightage)
75 Distance of the centre of mass of a solid uniform cone its vertex is z0 If the radius of its base is R and its
height is h then z0 is equal to
(A) 3ℎ
4
(B) 5ℎ
8
(C) 31198672
8119877
(D) ℎ2
4119877
Answer (A)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Solution
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Centre of Mass
Difficulty Easy (embibe predicted Low Weightage)
76 A rectangular loop of sides 10 cm and 5cm carrying a current I of 12 A is placed in different
orientations as shown in the figure below
If there is a uniform magnetic field of 03 T in the positive z direction in which orientations the
loop would be in (i) stable equilibrium and (ii) unstable equilibrium
(A) (a) and (c) respectively
(B) (b) and (d) respectively
(C) (b) and (c) respectively (D) (a) and (b) respectively
Answer (B)
Solution For a magnetic dipole placed in a uniform magnetic field the torque is given by 120591 =
times and potential energy U is given as
119880 = minus = minus119872 119861 119888119900119904120579
When is in the same direction as then 120591 = 0 and U is min = - MB as 120579 = 0119900
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
rArr Stable equilibrium is (b) and when then 120591 = 0 and U is max = + MB
As 120579 = 180119900
Unstable equilibrium in (d)
Topic Magnetism
Difficulty Easy (embibe predicted high weightage)
77
In the circuit shown the current in the 1Ω resistor is
(A) 0A
(B) 013 A from Q to P
(C) 013 A from P to Q (D) 13 A from P to Q
Answer (B)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
78 A uniformly charged solid sphere of radius R has potential 1198810 (measured with respect to infin)
on its surface For this sphere the equipotential surfaces with potential 31198810
251198810
431198810
4 and
1198810
4
have radius 1198771 1198772 1198773 and 1198774 respectively Then
(A) 1198771 ne 0 and (1198772minus1198771) gt (1198774minus1198773)
(B) 1198771 = 0 and 1198772 lt (1198774minus1198773)
(C) 2119877 lt 1198774 (D) 1198771 = 0 and 1198772 gt (1198774minus1198773)
Answer (B)
Solution
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
79 In the given circuit charge Q2 on the 2μF capacitor changes as C is varied from 1μF to 3μF Q2 as a
function of lsquoCrsquo is given properly by (figure are drawn schematically and are not to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(C)
(D)
Answer (A)
Solution
∵ 1 amp 2 120583119891 are in parallel
there4 Equivalent capacitance of the series combination is
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
119862119890119902 is=3119888
119862+3
So total charge supplied by battery is 119876 = 119862119890119902 =3119862119864
119862+3
there4 Potential difference across parallel combination of 1120583119891 ad 2120583119891 is
∆119881 =119876
3=
119862119864
119862+3
So charge on 2120583119891 capacitor is
1198762 = 1198622∆119881=2119862119864
119862+3
rArr1198762
2119864=
119862
119862+3rArr1198762
2119864=119862+3minus3
119862+3
rArr1198762
2119864= 1 minus
3
119862+3rArr (
1198762
2119864minus 1) = minus
3
119862+3
rArr (1198762minus 2119864)(119862 + 3) = minus6119864
Which is of the form (119910 minus 120572)(119909 + 120573) lt 0
So the graph in hyperbola With down ward curve line ie
Topic Electrostatics
Difficulty Moderate (embibe predicted high weightage)
80 A particle of mass m moving in the x direction with speed 2v is hit by another particle of mass 2m moving
in the y direction with speed v If the collision is perfectly inelastic the percentage loss in the energy
during the collision is close to
(A) 50
(B) 56
(C) 62 (D) 44
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Answer (B)
Solution
The initial momentum of system is 119875119894 = 119898(2119881)119894 + (2119898)119907
According to question as
On perfectly inelastic collision the particles stick to each other so
119875119891 = 3119898 119881119891
By conservation of linear momentum principle
119875119891 = 119875119894 rArr 3119898 119881119891 = 1198982119881 119894 + 2119898119881
rArr 119881119891 =2119881
3(119894 + ) rArr 119881119891 =
2radic2
3119881
there4 loss in KE of system = 119870119894119899119894119905119894119886119897minus119870119891119894119899119886119897
=1
2119898(2119881)2+
1
2(2119898)1198812 minus
1
2(3119898)(
2radic2119881
3)
2
= 21198981198812 +1198981198812 minus4
31198981198812 = 31198981198812 minus
41198981198812
3
=5
31198981198812
change in KE = 100 times∆119870
119870119894=
5
31198981199072
31198981198812=5
9times 100
=500
9= 56
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Conservation of Momentum
Difficulty Moderate (embibe predicted easy to score (Must Do))
81 Monochromatic light is incident on a glass prism of angle A If the refractive index of the
material of the prism is 120583 a ray incident at an angle 120579 on the face AB would get
transmitted through the face AC of the prism provided
(A) 120579 lt sinminus1 [120583 sin (119860minus sinminus1 (1
120583))]
(B) 120579 gt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(C) 120579 lt cosminus1[120583 sin (119860 + sinminus1 (1
120583))]
(D) 120579 gt sinminus1 [120583 119904119894119899 (119860 minus sinminus1 (1
120583))]
Answer (D)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Difficulty Moderate (embibe predicted high weightage)
82 From a solid sphere of mass M and radius R a cube of maximum possible volume is cut
Moment of inertia of cube about an axis passing through its centre and perpendicular to one
of its faces is
(A) 1198721198772
16radic2120587
(B) 41198721198772
9radic3120587
(C) 41198721198772
3radic3120587
(D) 1198721198772
32radic2120587
Answer (B)
Solution Let a be length of cube for cube with maximum possible volume diagonal length = 2R
rArr radic3119886 = 2119877 rArr 119886 =2119877
radic3
As densities of sphere and cube are equal Let Mrsquo be mass of cube
119872431205871198773 =
119872prime
431198863
119872prime =31198721198863
41205871198773
Moment of inertial of cube about an axis passing through centre
=119872prime(21198862)
12
=31198721198863
41205871198773times21198862
12
=1198721198865
81205871198773
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
119886 =2
radic3119877
=119872 times32 1198775
8120587 times9radic31198773
=41198721198772
9radic3120587
Topic Rotational Mechanics
Difficulty Moderate (embibe predicted Low Weightage)
83 Match list-I (Fundamental Experiment) with List-II (its conclusion) and select the correct option from the
choices given below the list
List-I List-II A Franck-Hertz Experiment (i) Particle nature of light
B Photo-electric experiment (ii) Discrete energy levels of atom
C Davison-Germer Experiment (iii) Wave nature of electron D (iv) Structure of atom
(A) Aminus (ii) B minus (iv)Cminus (iii)
(B) Aminus (ii) B minus (i) C minus (iii)
(C) Aminus (iv) B minus (iii) Cminus (ii)
(D) Aminus (i) B minus (iv)Cminus (iii)
Answer (B)
Solution Frank-Hertz experiment demonstrated the existence of excited states in mercury atoms helping to
confirm the quantum theory which predicted that electrons occupied only discrete quantized energy states
Phot-electric experiment = Demonstrate that photon is the field particle of light which can transfer
momentum and energy due to collision
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Davisson-Germer experment = this experiment shows the wave nature of electron
Topic Modern Physics
Difficulty Easy (embibe predicted high weightage)
84 When 5V potential difference is applied across a wire of length 01 m the drift speed of
electrons is 25 times10minus4 119898119904minus1 If the electron density in the wire is 8times 1028 119898minus3 the
resistivity of the material is close to
(A) 16 times10minus7Ω119898
(B) 16 times10minus6 Ω119898
(C) 16 times10minus5 Ω119898 (D) 16 times10minus8Ω119898
Answer (C)
Solution
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Electrostatics
Difficulty Easy (embibe predicted high weightage)
85 For a simple pendulum a graph is plotted between its kinetic energy (KE) and potential
energy (PE) against its displacement d which one of the following represents these
correctly (Graphs are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(B)
(C)
(D)
Answer (A)
Solution
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Topic Simple Harmonic Motion
Difficulty Easy (embibe predicted high weightage)
86 Two stones are thrown up simultaneously from the edge of a cliff 240 m high with initial
speed of 10 ms and 40 ms respectively Which of the following graph best represents the
time variation of relative position of the second stone with respect to the first (Assume stones do not rebound after hitting the ground and neglect air resistance take 119892 =
10 119898119904minus2)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(the figure are schematic and not drawn to scale)
(A)
(B)
(C)
(D)
Answer (B)
Solution 1198781 = 10119905 minus1
21198921199052
When 1198781 = minus240
rArr minus240 = 10119905 minus 51199052
rArr 119905 = 8119904
So at 119905 = 8 seconds first stone will reach ground
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
1198782 = 20119905 minus1
21198921199052
Till 119905 = 8 119904119890119888119900119899119889119904
1198782minus1198781 = 30119905
But after 8 second 1198781 is constant = minus240
Relative to stone 1199051 gt 8 119904119890119888119900119899119889119904 displacements of stone 2 1198782 +240
rArr 1198782+240 = 20119905 minus1
21198921199052
And at t = 12s seconds stone will reach ground
The corresponding graph of relative position of second stone wrt first is
Topic Kinematics
Difficulty Moderate (Embibe predicted high weightage)
87 A solid body of constant heat capacity 1119869 frasl is being heated by keeping it in contact with
reservoirs in two ways
(i) Sequentially keeping in contact with 2 reservoirs such that each reservoir supplies same
amount of heat
(ii) Sequentially keeping in contact with 8 reservoirs such that each reservoir supplies same amount of heat
In both the cases body is brought from initial temperature 100119900119862 to final temperature 200119900119862
Entropy change of the body in the two cases respectively is
(A) 1198971198992 1198971198992 (B) 119897119899221198971198992
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
(C) 2119897119899281198971198992 (D) 119897119899241198971198992
Answer (A)
Solution
Topic Heat ampThermodynamics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
88 Assuming human pupil to have a radius of 025 cm and a comfortable viewing distance of 25
cm the minimum separation between two objects that human eye can resolve at 500 nm
wavelength is
(A) 30 120583119898
(B) 100 120583119898 (C) 300 120583119898
(D) 1 120583119898
Answer (A)
Solution
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Difficulty Moderate (embibe predicted high weightage)
89
Two long current carrying thin wires both with current I are held by insulating threads of length L and
are in equilibrium as shown in the figure with threads making an angle lsquoθrsquo with the vertical If wires have
mass λ per unit length then the value of I is
(g = gravitational acceleration)
(A) 2sinθradicπλgL
μ0 cosθ
(B) 2radicπgL
μ0tanθ
(C) radicπλgL
μ0tanθ
(D) 2radicπgL
μ0tanθ
Answer (A)
Solution
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
By equilibrium of mix
Topic Magnetism
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)
Difficulty Moderate (embibe predicted high weightage)
90 On a hot summer night the refractive index of air is smallest near the ground and increases with height
from the ground When a light beam is directed horizontally the Huygensrsquo principle leads us to conclude
that as it travels the light beam
(A) Goes horizontally without any deflection
(B) Bends downwards
(C) Bends upwards (D) Becomes narrower
Answer (C)
Solution Consider air layers with increasing refractive index
At critical angle it will bend upwards at interface This process continues at each layer and light ray bends
upwards continuously
Topic Optics
Difficulty Moderate (embibe predicted high weightage)