JEE (Advanced)-2020

06-10-2019 A

TEST No. 3A (Paper-1)(XII Studying Students)

JEE (Advanced)-2020

(Divisions of Aakash Educational Services Limited)

Note : It is compulsory to fill Roll No. and Test Booklet Code on answer sheet, otherwise your answer sheet will not be considered.

[Based on Type-2 Aakash Pattern]

1. Read each question carefully.2. It is mandatory to use blue/black ballpoint pen to darken the appropriate

circle in the answer sheet.3. Mark should be dark and should completely fill the circle.4. Rough work must not be done on the answer sheet.5. Do not use white-fluid or any other rubbing material on answer sheet. 6. Student cannot use log table and calculator or any other material in the

examination hall.7. Before attempting the question paper, student should ensure that the

test paper contains all pages and no page is missing.8. Before handing over the answer sheet to the invigilator, candidate should

check that Roll No., Centre Code and Date of Birth have been filled and marked correctly.

9. Immediately after the prescribed examination time is over, the answer sheet to be returned to the invigilator.

10. Pattern of the questions are as under:(i) The question paper consists of 3 parts (Physics, Chemistry and

Mathematics). Each part has 5 sections.(ii) Section-1: This section contains 5 multiple choice questions which

have correct answers. Each question carries +4 one or more than onemarks for correct answer and no negative marks for wrong answer.

(iii) Section-2: This section contains 2 paragraphs. Based upon each

paragraph, 3 multiple choice questions have to be answered. Each

question has only one correct answer and carries +4 marks for correct

answer and –1 mark for wrong answer.

(iv) Section-3: This section contains 4 multiple choice questions which

have correct answer. Each question carries +3 marks for only one

correct answer and –1 mark for wrong answer.

(v) Section-4: This section contains 2 questions. Each question has two

Columns. Column-I has four entries (A), (B), (C) and (D), Column-II

has five entries (P), (Q), (R), (S) and (T). Each entry in Column-I may

match with one or more entries in Column-II. Each entry in Column-I

carries +2 marks for correct answer and for no negative marks

wrong answer.

(vi) Section-5: This section contains 3 questions. The answers to each of

the question is a single-digit integer, ranging from 0 to 9. If the correct

answer to question numbers X, Y and Z (say) are 6, 0 and 9

respectively, then mark 06, 00 and 09 in OMR respectively. Each

question carries +4 marks for correct answer and –1 mark for wrong

answer.


26-04-2020 A

MOCK TEST No. 4 (Paper-1)(XII Studying/XII Passed Students)

JEE (Advanced)-2020

[Based on JEE (Advanced)-2015 Actual Pattern]

Note : It is compulsory to fill Roll No. and Test Booklet Code on answer sheet, otherwise your answer sheet will not be considered.

1. Read each question carefully.

2. It is mandatory to use blue/black ballpoint pen to darken the appropriate

circle in the answer sheet.

3. Mark should be dark and should completely fill the circle.

4. Rough work must not be done on the answer sheet.

5. Do not use white-fluid or any other rubbing material on answer sheet.

6. Student cannot use log table and calculator or any other material in the

examination hall.

7. Before attempting the question paper, student should ensure that the

test paper contains all pages and no page is missing.

8. Before handing over the answer sheet to the invigilator, candidate should

check that Roll No., Centre Code and Date of Birth have been filled and

marked correctly.

9. Immediately after the prescribed examination time is over, the answer

sheet to be returned to the invigilator.

10. Pattern of the questions are as under:

(i) The question paper consists of 3 parts (Physics, Chemistry and

Mathematics). Each part has 3 sections.

(ii) Section-1: This section contains 8 questions. The answer to each of

the question is a single-digit integer, ranging from 0 to 9. If the correct

answer to question numbers X, Y and Z (say) are 6, 0 and 9

respectively, then mark 6, 0 and 9 in OMR sheet respectively. Each

question carries +4 marks for correct answer and no negative mark

for wrong answer.

(iii) Section-2: This section contains 10 multiple choice questions, each

question has correct answer(s). Each question one or more than one

carries +4 marks for correct answer and –2 marks for wrong answer.

(iv) Section-3: This section contains 2 questions. Each question has two

matching Columns. Column I has four entries (A), (B), (C) and (D),

Column II has five entries (P), (Q), (R), (S) and (T). Each entry in

Column I may match with one or more entries in Column II. Each entry

in Column I carries +2 marks for correct answer and –1 mark for wrong

answer.


Mock Test - 4

Paper-1

Complete Syllabus of JEE(Advanced)

Time : 3 Hrs MM: 264

PART – I : PHYSICS

Aakash Educational Services Limited - Regd. Office : Aakash Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456

1/20

(Space for Rough Work)

Mock Test - 4 (Paper-1) (Code-A) All India Aakash Test Series for JEE (Advanced)-2020

SECTION - 1

Integer Value Type

This section contains 8 questions. The answer to each of the questions is a single-digit integer, ranging from 0 to 9.

The answer will have to be appropriately bubbled in the OMR as per the instructions as follows. Examples- If the correct

answer to question numbers X, Y and Z (say) are 6, 0 and 9 respectively, then mark 6, 0 and 9 in OMR respectively.

1. A small metal ball is being pulled gradually on a fixed frictionless hemisphere as shown in figure. Radii of the ball and that of the pulley are much smaller than that of the hemisphere. The value of force required to do so

as function of (angular position) of ball from center of hemisphere comes out to be cos2

mgF P Q= − .

Find the value of P + Q.

(m = mass of ball, R = radius of hemisphere, 2R = height of point pulley from center of hemisphere.)

2. The acceleration a0 of the plank P required to keep the center C of a uniform solid sphere of mass m in a fixed

position with respect to earth during the motion is sinx

gy

. Find the value of x + y. [Here x and y are co-prime]

[There is no slipping between sphere and plank]

X

Y

0 1 2 3 4 5 6 7 8 9

Z

0 1 2 3 4 5 6 7 8 9

0 1 2 3 4 5 6 7 8 9



All India Aakash Test Series for JEE (Advanced)-2020 Mock Test - 4 (Paper-1) (Code-A)

2/20

3. A spherical ball of radius r and density is dropped from rest in a viscous fluid of density (<) and coefficient

of viscosity . The magnitude of power delivered to ball by the viscous force acting on the ball at any time ‘t’

after it is dropped is ( ) ( ) ( )2

29

2 2 5 21 .27

t

rx

P t g r e

−

= − −

Find the value of x.

4. A hollow sphere of radius 3R is charged to V volts and another smaller sphere of radius R is charged to 3

V

volts. Then somehow, the smaller sphere is placed inside the bigger sphere without changing the net charge

on each spheres. The potential difference between the two spheres now is PV

VQ

= volts. Find the value

Q – P. [Here P and Q are co-prime]

5. In the arrangement shown in figure when the switch S2 is open, the galvanometer shows no deflection for

1 3

L= . When switch S2 is closed, galvanometer shows no deflection for

2 4

L= . Then find the internal

resistance r (in ohm) of the 8 V cell. (Switch S1 is in closed state)

6. There exist a long (infinite) conductor along z-axis carrying 5 A of current in positive z-direction. Torque acts

on the loop (symmetric about x and y axis) as shown in figure. Resistance per unit length of the wire forming

the circular portion of the loops are 2 /m, but connectors 1 – 1, 2 – 2, 3 – 3 and 4 – 4 have zero resistance. Potential difference VAB = 55 volts is applied across AB. Radii of loops are R1 = 0.3 m and R2 = 0.9 m. If torque

on the loop comes out to be = k × 7 × 10–6 Nm, then find the value of k.

[Use, 22

7 = ]



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7. A parallel beam of white light falls on a very thin film having air on both sides whose refractive index is equal

to 3

2n = . The angle of incidence 1 = 30°. It is required that yellow colour ( = 6 × 10–7 m) to be most intensively

reflected. Then the minimum thickness of the thin film is k × 10–7m. Find the value of k. (nearest integer)

8. The half-lives of two longest lived radioactive isotopes of phosphorus P 32 and P 33 are 14 days and 28 days respectively. A sample has been prepared by mixing the isotopes in the ratio of 4 : 1 of their atoms (in numbers). Initial activity of the mixed sample is 180 millicurie. Find the activity (in m curie) of the sample after 84 days.

SECTION - 2

One or More Options Correct Type

This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), for its answer, out of which one or more than one is/are correct.

9. A disc of mass m rolling with slipping on a rough horizontal surface with speed v0 and angular speed 0 as

shown. With v0 > 0R

(A) The distance of instantaneous axis of rotation from center O is 0

0

2vr =

(B) The distance of instantaneous axis of rotation from the contact point P is 0 0

0

v Rr

− =

(C) Work done by friction till the disc starts pure rolling is ( )2

fr 0 0

1

6W m v R= − −

(D) Work done by friction till the disc starts pure rolling is ( )2

fr 0 012

mW v R= − −

10. Consider a space where uniform magnetic field ˆ1TB i= and uniform electric field 1 ˆN/C2

E i= exist.

A particle of charge q and mass m (with 2q

m= ) is projected from origin with initial velocity

0ˆ4 m/sv k= . [neglect

gravity]




4/20

Which of the following statement(s) is/are correct?

(A) Displacement of the particle at 3

2t

= s is

29

8

meter

(B) Displacement of the particle at 3

2t

= s is

41024 81

8

+ meter

(C) Total distance travelled by particle till t = 2 s is ( )2 22 4 8ln 4 8ln2s = + + + + −

(D) Total distance travelled by particle till t = 2 s is ( )2 22 4 8ln 4 16ln2s = + + + + −

11. Consider a progressive sinusoidal wave in a string as y(t) = Asin(kx – t). If is the linear mass density. Then

(A) Kinetic energy per unit length is given as

21

KE2l

y

t

=

(B) Potential energy stored per unit length is given as

2 21

PE2l

y

k x

=

(C) Maximum potential energy region and maximum kinetic energy region are separated by 2k

distance.

(D) Maximum potential energy region and maximum kinetic energy region happens to be same

12. For a gas under going an adiabatic process, at certain stage its pressure and temperature are P0 and T0

respectively and the magnitude of slope of P-T curve is m, then

(A) Molar specific heat at constant volume is ( )0 0

02

R mT P

P

−

(B) Molar specific heat at constant volume is ( )0 0

0

R mT P

P

−

(C) Molar specific heat at constant pressure is 0

0

mRT

P

(D) Molar specific heat at constant pressure is 0

02

mRT

P



5/20


13. A certain oscillation results from the addition of coherent oscillations in the same direction

( )( )cos –1 ,k

a t k = + where k is the number of oscillation [k = 1, 2, 3….N], is the phase difference

between kth and (k –1)th oscillation. Then

(A) Amplitude of resultant oscillation is

sin2

sin2

NNa

(B) Amplitude of resultant oscillation is

sin2

sin2

Na

(C) Phase difference of resultant oscillation w.r.t. 1 is ( )– 1

2

N

(D) Phase difference of resultant oscillation w.r.t. 1 is 2

N

14. In a hypothetical system a particle of mass m and charge –3q is moving around a very heavy particle of charge

+2q. Assuming Bohr’s model to be valid for this system. Then

(A) Speed of the mass m when it is in 1st excited state is 2

0

3

2

qv

h =

(B) Time period of revolution in 1st excited state is

3 2

0

4

8

9

hT

mq

=

(C) Radius of revolution in the ground state is

2

0

23

hr

mq

=

(D) Effective magnetic moment of revolving charge particle in ground state is 3

4

qh

m =




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15. Consider a quarter circular conducting ring of large radius R with its center at origin, where a magnetic dipole

of moment m is placed as shown in figure. If the ring rotates at constant angular velocity . Then

(A) If axis of rotation is y-axis with angular velocity vector as ˆ,j then end A will be lower potential and end B

will be at higher potential

(B) If axis of rotation is y-axis then, emf induced between its end AB is 0

4

m

R

(C) If axis of rotation is x-axis, then emf induced between its end AB is 0

8

m

R

(D) If axis of rotation is x-axis, then emf induced between its end AB is 0

8

m

R

16. Circuit shown below consists of two identical resistances R and two identical voltmeters A and B are connected

as shown. Voltmeter A reads 60 volts whereas voltmeter B reads 48 volts. Then

(A) emf of the battery connected is V = 116 volts

(B) emf of the battery connected is V = 87 volts

(C) Value of resistance(s) R = 3 and voltmeter resistance(s) r0 could be = 12

(D) Resistance(s) of voltmeters are r0 = 24 and connected resistance(s) R = 3



7/20


17. Two converging lenses of focal length f1 and f2 are placed with their optical axes coinciding. This lens system

is used to form image of an object. It is found that the size of the image does not depend on the distance

between the first lens and the object. Then

(A) The distance between two lenses are d = 2(f1 + f2)

(B) The distance between two lenses are d = f1 + f2

(C) Magnitude of total magnification is 2

1

f

f

(D) Magnitude of total magnification is 1

2

f

f

18. Figure below (sinusoidal curve) shows the hypothetical speed distribution for a sample of N (number of gas

particles) 0

Given 0 fordN

v vdv

=

(A) Total numbers of gas molecules in the sample is 0 0a v

N =

(B) Total numbers of gas molecules in the sample is 0 02a v

N =

(C) Average speed of the sample of gas molecules is 0avg 2

vv =

(D) Average speed of the sample of gas molecules is 0avg 2

vv =

SECTION - 3

Matching Column Type

This section contains 2 questions. Each question contains two Columns (Column-I and Column-II). Column I has four entries (A), (B), (C) and (D), Column-II has five entries (P), (Q), (R), (S) and (T). Match the entries in Column-I with the entries in Column-II. Each entry in Column-I may match with one or more entries in Column-II. The OMR contains a 4 × 5 matrix whose layout will be similar to the one shown below:




8/20

For each entry in Column-I, darken the bubbles of all the matching entries. For example, if entry (A) in Column-I matches with entries (Q), (R) and (T), then darken these three bubbles in the OMR. Similarly, for entries (B), (C) and (D).

19. Variation of specific (molar) heat capacities and the process equation for one mole of ideal monoatomic gas has been given in Column-I and corresponding variation of state variable has been given in Column-II. Match

these two columns appropriately. [C : specific heat capacity; Cv : specific heat capacity at constant volume; ,

, a are positive constant; P : pressure; V : volume; T : temperature]

Column-I Column-II

(A) C = Cv + T (P) If P is increasing then T may be increasing

or decreasing

(B) C = Cv + V (Q) If P is increasing then V would be decreasing

(C) C = Cv + aP (R) If P is increasing then V may be increasing

or decreasing

(D)

7

5 constantPV = (S) If V increases then T will also increase

(T) If V is increasing then T would be decreasing

20. A point object S is located at a distance of 120 cm from a screen. A lens of focal length 22.5 cm mounted on a movable frictionless stand is kept between the source and the screen. Stand is attached to a spring of natural length 50 cm and spring constant 800 N/m as shown in figure. Mass of the stand (including lens is 2 kg)

(A)

(B)

(C)

(D)

(P)

(P)

(P)

(P)

(Q)

(Q)

(Q)

(Q)

(R)

(R)

(R)

(R)

(S)

(S)

(S)

(S)

(T)

(T)

(T)

(T)



9/20


An impulse P is imparted so that the mounted lens start oscillating. In Column-I magnitude of impulse has been listed and corresponding observation for image on screen is listed in column-II. Match them appropriately.

Column-I Column-II

(A) If 4 kg m/sP = (P) For one complete oscillation there would be 2 instant

when clear image will be observed on the screen

(B) If 8 kg m/sP = (Q) For one complete oscillation there would be 1 instant

only when clear image will be observed on the screen

(C) If 12 kg m/sP = (R) No clear image would be observed on the screen

(D) If 16 kg m/sP = (S) Time interval between appearance of image on

the screen in constant

(T) For one complete oscillation there would be 3 instant when clear image will be formed on the screen

PART – II : CHEMISTRY

SECTION - 1

Integer Value Type


The answer will have to be appropriately bubbled in the OMR as per the instructions as follows. Examples- If the correct

answer to question numbers X, Y and Z (say) are 6, 0 and 9 respectively, then mark 6, 0 and 9 in OMR respectively.

21. There is a lattice which contains two type of atoms A and B. A atoms lie at all the corners and at all the face

centers and B atoms lie at all the octahedral voids. If A’ and B’ are the number of atoms of A and B which lie

on the following shaded plane, then calculate the value of A’ : B’ if any atom lying on the plane is to be counted

as one atom.

X

Y

0 1 2 3 4 5 6 7 8 9

Z

0 1 2 3 4 5 6 7 8 9

0 1 2 3 4 5 6 7 8 9




10/20

22. Consider the following structures.

How many of the following is/are the true statement(s) regarding P, Q and R?

(1) P and Q are resonating structures

(2) Q is most stable resonating structure among all possible resonating structures for this carbocation

(3) Q and R are positional isomers

(4) P and R are positional isomers

(5) All P, Q and R are resonating structures with respect to each other

(6) Only 3 resonating structures are possible for Q

(7) Only 4 resonating structures are possible for P

(8) Only 4 resonating structures are possible for R

(9) All three P, Q and R have complete octet

23. A compound with molecular formula C5H10 (which can give only single monobromo derivative in presence of

sunlight) is allowed to react with Br2 in presence of sunlight. All the obtained dibromo derivatives are allowed

to react with alc. KOH.

How many different types of alkadiene can be produced in the above reaction?

24. The energy required to excite 0.1 mole of H atom to second excited state of atomic hydrogen is xy

90 joule. The

ionisation energy of gaseous hydrogen atom is y J/mol. Find the value of x.

25. The solubility of A(OH)3 in a buffer of pH = 4 is 27 × 10–2 mol lit–1 and in pure water it is equal to

p × 10–q mol L–1 (scientific notation). Calculate the value of p + q . [Given pKw = 14]

26. In the electrolysis of brine solution using mercury cathode, 44.6 g of Na — Hg solution is obtained at cathode

in which both Na and Hg are in equimolar proportions. If a current of 50 A is passed through the brine solution,

the time (in minutes) for which the current is passed is 193x

60. Find the value of x. (molar mass Hg = 200 g/mol)

27. When LiNO3 is heated, how many different types of paramagnetic gases are liberated?

28. According to molecular orbital theory the number of electrons in * (anti-bonding ) orbital in – –

2 2 2 2 2N , N , O , O and F+ + + is a, b, c, d and e.

Find the value of a + b + c + d + e.



11/20


SECTION - 2


This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), for its answer, out of which one or more than one is/are correct.

29. Which of the following statement(s) is/are correct about Tetraamminedithiocyanato cobalt (III) trioxalato

cobaltate (III)

(A) The molecular formula of the complex is [Co(SCN)2(NH3)4] [Co(OX)3]

(B) One of the complex ions in the given coordination compound always exists in optically active form

(C) The positively charged complex in the given coordination compound can show six linkage isomers

(D) The negatively charged complex in the given coordination compound can show geometrical isomers.

30. Consider the molecules P, Q, R and S given below. Each of these molecules undergoes nucleophilic

substitution reaction following second order kinetics with the following kind of curve.

Given molecules are

P. CH2 = CH – CH2 – Br

Q. C6H5 – CH2 – Br

R. CH3 – O – CH2 – Br

S.

Which of the following order correctly represents the order for G# for the given molecules?

(A) S > R > Q > P (B) P > Q > S > R

(C) P > Q > R > S (D) Q > P > S > R




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31. Consider the following statements and identify the incorrect one(s).

(A) There are 4 orbitals which participate in 3C – 2e– bond formation in each molecule of diborane

(B) In one molecule of diborane there are only 12 bonding electrons

(C) Al2Cl6 sublimes on heating and forms a polymer of AlCl3 in vapour state

(D) Because of transfer of lone pair from nitrogen to vacant p-orbital of boron in borazine, hybridisation of nitrogen is sp2.

32. The reversible expansion of an ideal gas under the following mentioned conditions is shown in the figure.

Consider the following statements.

S1 : Curve A is for adiabatic process if T1 = T2

S2 : Curve B can be obtained for adiabatic process, if T1 T3

S3 : If T1 = T2, then T2 > T3 and curve B is possible for adiabatic process

S4 : |wA| > |wB|

S5 : If curve A is for adiabatic expansion of diatomic gas and B is for adiabatic expansion of monoatomic gas,

then |UA| > |UB|

Which of the above statement(s) is/are true?

(A) S1 and S2 (B) S2 and S3

(C) S4 and S5 (D) S4

33. Kinetic energy (EK) of a molecule is related to its velocity (v) by the expression 2K

1E mv

2= . EK follows

Boltzmann distribution as schematically given below.

Consider the chemical reaction 2N2O5 → 4NO2 + O2.



13/20


If this reaction follows Boltzmann distribution law. Then which of the following statement(s) is/are true?

(A) To complete the reaction, all N2O5 molecules must possess EK activation energy (Ea) of the reaction

(B) As we can see from the given curve, with the increase in temperature, EK increases that’s why Ea decreases

(C) Rate constant increases with increase in temperature

(D) In the presence of catalyst, rate constant increases as it causes the increase in EK of the N2O5 molecules

34. In the nuclear reaction

1 11 0X H Y n+ → +

The X and Y respectively, can be

(A) 23 2311 12Na, Mg (B)

23 2312 11Mg, Na

(C) 23 2412 13Mg, Al (D)

12 126 7C, N

35. Using crystal field theory, predict the cations having different theoretical and observed (experimental) values

of hydration energy in hexa aqua complex [M(H2O)6]n+ ion.

(A) Mn2+ (B) Zn2+

(C) Cr3+ (D) Ni2+

36. Consider the following reactions.

(1)

(2)

(3)

(4)




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Choose the correct statement(s).

(A) Reagent in reaction (1) is CuSO4 (aq) complexed with sodium potassium tartrate

(B) Reagent in reaction (2) is PCC/THF

(C) Racemisation may occur in reaction (3)

(D) Product z is the major product in reaction (4)

37. Consider the reaction of zinc with dilute and concentrated nitric acid, in both the cases an oxide of nitrogen is

produced. Which of the following statement(s) is/are true for the above reaction?

(A) In both the cases, same oxide is produced

(B) In case of dilute acid, a neutral oxide is produced

(C) In case of concentrated acid, an acidic oxide is produced

(D) In both cases the oxide liberated is different but both are acidic in nature

38. Consider the structure of H2O2 in gas phase and solid phase.

Identify the correct statement(s).

(A) In gas phase O – O bond length is shorter

(B) In gas phase dihedral angle is larger

(C) In solid phase the O – H bond length is greater

(D) In solid phase O – O – H bond angle is smaller

SECTION - 3


This section contains 2 questions. Each question contains two Columns (Column-I and Column-II). Column I has

four entries (A), (B), (C) and (D), Column-II has five entries (P), (Q), (R), (S) and (T). Match the entries in Column-

I with the entries in Column-II. Each entry in Column-I may match with one or more entries in Column-II. The OMR

contains a 4 × 5 matrix whose layout will be similar to the one shown below :

For each entry in Column-I, darken the bubbles of all the matching entries. For example, if entry (A) in Column-I

matches with entries (Q), (R) and (T), then darken these three bubbles in the OMR. Similarly, for entries (B), (C)

and (D).

(A)

(B)

(C)

(D)

(P)

(P)

(P)

(P)

(Q)

(Q)

(Q)

(Q)

(R)

(R)

(R)

(R)

(S)

(S)

(S)

(S)

(T)

(T)

(T)

(T)



15/20


39. Match Column-I with Column-II.

Column-I Column-II

Reactions Intermediates / Reaction type

(A) (P) Addition

(B) (Q) Substitution

(C) (R) Elimination

(D) (S) Carbocation

(T) Carbanion

40. Column-I contains some titrating mixtures with labelled components as (I) and (II). Column-II contains volume of the components required for complete neutralisation.

Column-I Column-II

(A)

( ) ( )

HCl NaOH

0.5 M 0.05 M

I II

+ (P) With 10 ml of (I) 100 ml of (II) is required

(B)

( ) ( )

2 2 4H C O NaOH

0.5 M0.5 M

III

+ (Q) With 100 ml of (I) 40 ml of (II) is required

(C)

( )

( )

( )

2 2 4 4acidified

H C O KMnO

0.5 M0.5 M

I II

+ (R) With 1000 ml of (I) 400 ml of (II) is required

(D)

( )

( )

( )

2 2 4 4acidified

Na C O KMnO

0.5 M0.5 M

I II

+ (S) With 100 ml of (I) 200 ml of (II) is required

(T) With 100 ml of (I) 1000 ml of (II) is required




16/20

PART - III : MATHEMATICS

SECTION - 1

Integer Value Type


The answer will have to be appropriately bubbled in the OMR as per the instructions as follows. Examples- If the correct answer to question numbers X, Y and Z (say) are 6, 0 and 9 respectively, then mark 6, 0 and 9 in OMR respectively.

41. If 2 (cos x + cos 2x) + (1 + 2cosx) sin 2x = 2 sinx, then number of values of x [– , ] satisfying the equation is

42. A variable chord PQ of the parabola y = 4x2 subtends a right angle at the vertex. The locus of the points of intersection of normals at P and Q is conic, then four times the length of latus rectum of this conic is

43. Three different numbers are selected at random from the set A= {1, 2, 3,…………..,15}. Then the probability

that product of two numbers is equal to the third number is m

n, where m and n are relatively prime then

13

n

m

is

44. If ( ) 3 2–8 24 – 32 17 ,f x x x x x R= + + then the value of ( )( )2

0

f f x dx equals

45. The circle 2 2 6 24 72 0+ + − + =x y x y and hyperbola 2 2– 6 16 – 46 0x y x y+ + = , intersect at four distinct

points. These four points lie on a parabola then maximum distance of a point on circle 2 2 2 10 17 0+ − − + =x y x y from focus of parabola is

46. Let ( )2 1

, ,2 3

nR n n I

+ − +

and f be continuous function. Let ( )( )7 tan

sec

2

2 8−

= −I xf x x dx and

( )( )2

2

7 tan

sec8J f x x dx

−

= − then the value of

I

Jis equal to

47. Let ( ) 3 3 32 4sin sin sin

3 3

= + + + +

f x x x x dx and ( )

90

4=f , then maximum value of 4f(x) is

48. The area bounded by the curve y = f(x) tangent to it at x = 2 and x axis where ( )0

2 8= x

f x t dt is m

nsquare

units where m, n are relatively prime then (2m + n) is equal to

X

Y

0 1 2 3 4 5 6 7 8 9

Z

0 1 2 3 4 5 6 7 8 9

0 1 2 3 4 5 6 7 8 9



17/20


SECTION - 2


This section contains 10 multiple choice questions. Each question has 4 choices (A), (B), (C) and (D), for its answer, out

of which one or more than one is/are correct.

49. f(x) is a differentiable function satisfying the relation ( ) ( )2

0

−= + −x tf x x e f x t dt , then

(A) ( )11

1

1958k

f k=

= (B) Local maximum value of f(x) is 4

3

(C) Local minimum value of f(x) is 4

3

− (D) Number of real solutions of f(x) = 1 is 3

50. If A and B are respectively a symmetric and a skew symmetric matrix such that AB = BA, then

(A) (A – B)–1 (A + B) is orthogonal matrix when (A – B) is non singular

(B) (A + B)–1 (A – B) is orthogonal matrix when (A + B) is non singular

(C) |(A – B)–1 (A + B)| = 1 and |(A + B)–1(A – B)| = 1

(D) |A + B| = |A – B| = 1

51. If f(x) and g(x) are functions such that f(x + y) = f(x).g(y) – g(x).f(y), then

( ) ( ) ( )( ) ( ) ( )( ) ( ) ( )

1 1 1 1

2 2 2 1

3 3 3 1

f x g x f x y

f x g x f x y

f x g x f x y

+

+

+

is

independent of

(A) x1 (B) x2

(C) x3 (D) y1

52. A variable circle is described to pass through the point (0,1) and tangent to the one of the asymptote of

rectangular hyperbola x2 – y2 = 9, passing through IInd and IVth quadrant. The locus of the centre of the circle

is a parabola whose

(A) Equation is x2 – 2xy + y2 – 4y + 2 = 0 (B) Length of latus rectum is 2 units

(C) Vertex has the coordinates 1

,4 4

−

(D) Equation of directrix is x – y = 1




18/20

53. Let ˆ ˆ ˆ ˆ ˆ ˆ ˆ2 2 , 2 3 6a i j k b i j k= + + = + + and ˆ ˆ ˆ4 4 7= + +c i j k . A vector which is equally inclined to these three

vectors is perpendicular to the plane passing through the points , a b and c . If are positive integers,

then the possible values of + + is/are

(A) 74 (B) 78

(C) 195 (D) 185

54. Let P(0, –3, 4) be a point in space and ( )Q r be a point such that ( ) ( )ˆ ˆ ˆ ˆ ˆ ˆ3 3 3 2 3= − − + − +r i j k i j k . A plane

: 2x – 3y + 3z = 2 is given. Then

(A) The vector PQ is parallel to plane = 0 if = 1

(B) Equation of plane passing through point P and parallel to plane = 0 is 2x – 3y + 3z = 21

(C) Distance between plane = 0 and plane passing through point P which is parallel to = 0

is 17

22units

(D) The point (9, 0, 1) lies on the plane passing through point P and parallel to plane = 0

55. The number of positive divisors of (2008)8 that are less than (2008)4 are

(A) 28 (B) 112

(C) 224 (D) 56

56. Let ( )2

2.

4

+ = +

+

x xxe dx f x e C

x, where C is constant of integration then

(A) (1–f(|x|)) is monotonically decreasing for x R+

(B) f(x) is differentiable for all x R – {–4}

(C) f’(–2) = 1

(D) f’(0) is not defined

57. Let ( ) ( ) ( )1 2 1 2sin 6 10 cos 6 10f x x x x x− −= − − + − − be a real valued function, then

(A) Sum of number of elements in domain and range of f(x) is 2

(B) f(x) = Kx has a real solution if 6

=K

(C) ( )2

=f x has infinitely many real solutions

(D) f(x) = 0 has exactly one real solution



19/20


58. If ( ) ( ) ( ) ( )24 2 sin 1 + = − − + −a b c a b b b c and ( ) =c c a c while b and c are non-zero, non collinear

vectors, then

(A) 2 + 2 sin = 3 (B) 3 – cos = 3

(C) 2 sin + 3 cos2 – 2sin2 = 0 (D) . sin + 2 cos = 3

SECTION - 3


This section contains 2 questions. Each question contains two Columns (Column-I and Column-II). Column-I has

four entries (A), (B), (C) and (D), Column-II has five entries (P), (Q), (R), (S) and (T). Match the entries in Column-

I with the entries in Column-II. Each entry in Column-I may match with one or more entries in Column-II. The OMR

contains a 4 × 5 matrix whose layout will be similar to the one shown below :

For each entry in Column-I, darken the bubbles of all the matching entries. For example, if entry (A) in Column-I

matches with entries (Q), (R) and (T), then darken these three bubbles in the OMR. Similarly, for entries (B), (C)

and (D).

59. Match the entries of Column-I with Column-II.

Column-I Column-II

(A) Let ( ) ( )3

1

3

cos sec

= f x dx , then ( )3

3

18f

(P) 1

is equal to

(B) If the range of ( ) ( )2

4| |

1 1tan 2 – 2

x

xf x

− + = is [y1, y2] (Q) 3

then the value(s) which lie in the interval [y1, y2] is /are

(C) If ˆ ˆ ˆ, 1a i j k a b= + + = and ˆ ˆ = −a b j k , then b is (R) 6

equal to

(D) If z be complex number satisfying the equation (S) 2

3

−

22 1 3 0+ − =z i , then the principle argument of

z is/are

(T) 1

2−

(A)

(B)

(C)

(D)

(P)

(P)

(P)

(P)

(Q)

(Q)

(Q)

(Q)

(R)

(R)

(R)

(R)

(S)

(S)

(S)

(S)

(T)

(T)

(T)

(T)




20/20

60. Match the entries of Column-I with Column-II.

Column-I Column-II

(A) A man has 10 one-rupee coins and one of them (P) 11

is known to have two heads. He take one coin

and tosses 5 times and it always fall heads the

chance that the selected coin has two head is p

q

where p and q are relatively prime then p + q is

(B) The number of seven digit integers with sum of (Q) 138

digits equal to 10 and formed by using the digits

1, 2 and 3 only is

(C) There are n A.M.s between 1 and 15. If the ratio (R) 77

of 6th and (n–1)th means is 12 : 19, then n is

equal to

(D) Two numbers x and y are drawn without (S) 72

replacement from the set of the first 15 natural

numbers. The number of ways of drawing them

such that (x4 – y4) is divisible by 5

(T) 73

❑ ❑ ❑

All India Aakash Test Series for JEE (Advanced)-2020

Test Date : 26/04/2020

ANSWERS


MOCK TEST - 4 (Paper-1) - Code-A

Mock Test - 4 (Paper-1) (Code-A)_(Answers) All India Aakash Test Series for JEE (Advanced)-2020

1/14

PHYSICS CHEMISTRY MATHEMATICS

1. (9)

2. (7)

3. (8)

4. (7)

5. (5)

6. (3)

7. (1)

8. (5)

9. (B, C)

10. (B, C)

11. (A, B, D)

12. (B, C)

13. (B, C)

14. (A, B, D)

15. (A, B, D)

16. (B, C)

17. (B, C)

18. (B, C)

19. A → (P, R, S)

B → (P, R, S)

C → (P, R, S)

D → (Q, T)

20. A → (R)

B → (Q, S)

C → (P)

D → (S, T)

21. (2)

22. (3)

23. (1)

24. (8)

25. (9)

26. (2)

27. (2)

28. (8)

29. (B, C)

30. (C)

31. (A, C)

32. (B, C, D)

33. (A, C)

34. (A, D)

35. (C, D)

36. (A, C)

37. (B, C)

38. (B, C)

39. A → (Q, S)

B → (P)

C → (R, T)

D → (R)

40. A → (P, T)

B → (S)

C → (Q, R)

D → (Q, R)

41. (5)

42. (4)

43. (5)

44. (2)

45. (8)

46. (4)

47. (9)

48. (7)

49. (A, B, D)

50. (A, B, C)

51. (A, B, C, D)

52. (A, B, C)

53. (A, D)

54. (A, B, D)

55. (B)

56. (A, B, C)

57. (A, B)

58. (A, B, C)

59. A → (R)

B → (P, Q, R, T)

C → (P)

D → (Q, S)

60. A → (T)

B → (R)

C → (P)

D → (Q)


All India Aakash Test Series for JEE (Advanced)-2020 Mock Test - 4 (Paper-1) (Code-A)_(Hints & Solutions)

2/14

PART - I (PHYSICS)

1. Answer (9)

Hint : mgsin = F2

Solution :

mgsin = Fcos Now sin sin

R r=

sin

sinR

r

= Also 5 4cosr R= −

sin

sin ,5 4cos

=

− ( )

2

= − +

cos = sin( + ) = sincos + cossin

( )

22 sin

1 cos5 4cos

− =

−

2 cos

cos5 4cos

− =

−

So, 2 cos cos sin

sin sin5 4cos 5 4cos

mg F −

= + − −

2

5 4cosmg F

=

−

5 4cos2

mgF = −

2. Answer (7)

Hint : ( ) 2

0

7sin

5

amg ma R mR

R + =

Solution :

For sphere with respect to plank

About point of contact ( ) 2

0

7sin

5

amg ma R mR

R + =

(a : acceleration of C w.r.t. plank)

0

7sin

5g a a + =

But ‘C’ should remain in rest w.r.t. ground

a0 = a

0

2sin

5g a =

0

5sin

2a g=

x + y = 7

3. Answer (8)

Hint : 2

9

2

dvg

dt r

− = −

Solution :

3 3 34 4 46

3 3 3

dvr g r g rv r

dt − − =

( )

2

9

2

dvg v

dt r

− = −

( ) ( )2

92

22

19

t

rr g

v e

−

−

= −

( ) 26P t rv= = ( ) ( )2

29

2 2 5 28

19

t

rg r e

−

− −

4. Answer (7)

Hint : 0

0

4 2

3 4 3

Rvv

R

=

Solution :

3

04 3

QV

R=

Q3 = 40RV.3

Similarly 01

4

3

RVQ

=

Now 0

0

4 2 2

3 4 3 9

RV VV

R

= =

(by comparing)

Q – P = 7

5. Answer (5)

Hint : 2

8 volts;3 4

L Lv

L L

= =


Mock Test - 4 (Paper-1) (Code-A)_(Hints & Solutions) All India Aakash Test Series for JEE (Advanced)-2020

3/14

Solution :

When switch S2 was open then potential drop

across 1 3

Ll = is 8 volt

8 volt3

L

L

=

= 24 volts

Now when switch S2 is closed null point is at

2 3

Ll =

2

246 volt

4 4

LV

L

= = =

So, ( )

815 6 volt

15 r =

+

20 = 15 + r

r = 5

6. Answer (3)

Hint : 6

1

10

2

rd Idr

r

−

=

Solution :

Length of one side of loop

1 2

23 6

R R

+

( )1 22

3R R

+

Resistance = ( )2

2 0.6 0.9 1.53 3

+ = =

current in loop → I = 55

Amp

Torque will be there due to the straight

connectors only

7

05 4 10 5

2 2B

r r

− = =

610

Br

−

= Now 6

1

10

2

rd I dr

r

−

=

6

1 2 1

44 10

2

Id R r−

= = −

6 64 55 67 10 21 10 Nm

2 22 10

− − = =

6 63 7 10 7 10K− − = =

K = 3

7. Answer (1)

Hint : x = 2dsec2 – 2dtan2.sin1 = 2

Solution :

Path difference between interfering wave is

2dsec2 – 2dtan2sin1 = x

2 2 1 1 22 sec tan sin ; sin sinx d = − =

2 2

1

9 1 22 sin 2 8

4 4 2

dx d d = − = − =

For constructive interference 2 22

d

=

7

7 76 101.06 10 1 10

4 2d

−− −

=

8. Answer (5)

Hint : dN

A Ndt

= =

Solution :

For P32 : 1

ln2

14 = ; and for P33 :

2

ln2

28 =

Initially 0 04 ln2 ln2

18014 28

N N+ =

0 0

9 180 28ln2 180 ln2

28 9N N

= =

After 84 days,

ln284

0141 0

44

64

NN N e

−

= =

Similarly 02 8

NN =

Activity (A) = 0 04 ln2 ln2

64 14 8 28

N N+

0

1 1 180 28 1ln2

8 28 8 28 9 4 28A N

= + =

A = 5

9. Answer (B, C)

Hint : 0 0 0v v R

R r r

− =

+



4/14

Solution :

0 0 0v v R

R r r

− =

+

v0r = v0R – 0R2 + v0r – 0R.r

v0R – 0R2 = 0Rr

0 0

0

v Rr

− =

KE(i) =

If for final pure rolling is the angular speed then

2

2

0 0

3

2 2

mRmv R mR+ =

00

3

2 2

RmR v mR R

+ =

( )0 02

3

v R

R

+ =

KE(f) = ( )2 2 2 2

0 0 0 02

1 3 14 4

2 2 9mR v R v R

R + +

loss (of E) =

( )2

2 2 2 2 2

0 0 0 0 0 0

1 14 4

2 2 2 12

mR mmv v R v R+ − + +

KE = ( )2 2 2 2 2 2

0 0 0 0 0 06 3 4 4

12

mv R v R v R+ − − −

KE = 2 2 2

0 0 0 02 4 2

12

mv v R R − +

( )2

0 06

mv R= −

( )2

0 06fr

mW v R

−= −

10. Answer (B, C)

Hint : 22

; , 16 .mv m

R T ds t dtqB qB

= = = +

Solution :

0 42 m

2 1

mvR

qB= = =

Revolution time 0

2 2sec

2

mT

q

= = =

Acceleration in x direction, 20 12 1m/s

2x

qEa

m= = =

So, the particle will move in increasing helical path.

At the end of 3

sec2

t

= (means 1

12

rotation)

2 21 9 9

1 ; 4 m2 4 8

X y

= = =

Displacement = 4

2 2 8116

64x y

+ = +

Displacement = 411024 81

8+

Speed at any time ( ) 216v t t= +

distance 216ds t dt= +

2

2 2016 8ln 16

2

tS ds t t t

= = + + + +

( ) ( )2 2216 4 8ln 2 16 4 8ln 4

2S

= + + + + −

( )2 22 4 8ln2 8ln 4 16ln2S = + + + + + −

( )2 22 4 8ln 4 8ln2S = + + + + −

11. Answer (A, B, D)

Hint : ( )2

1

2

yP E I T

x

=

Solution :

Kinetic energy is 21

2mv

2 2

( )

1 1KE

2 2I

y yl

t t

=

Also, ( ) ( )22

dl dx dy dx= + −

21

2

ydx

dx

=

2

( )

1PE

2I

yT

x

=

We see that for KE ( )cosy

A kx tt

= −

is

maximum for particles at mean position.

And for PE ( )cosy

Ak kx tt

= −

is maximum at mean position.

So, K.E and P.E are maximum at same region.

12. Answer (B, C)

Hint : ( )1

dp p

dT T

=

−

Solution :

1. constantT p − =

2 22 2 2 2

0 0 0 0

1 1 1 1

2 2 2 2 2 2

mR mRmv mv+ = +



5/14

( ) 1 11 0T p dp p T dT − − −− + =

( )1 0T dp pdT− + =

( )1

dp p

dT T

=

−

( )0

01

pm

T

=

−

p0 = mT0 – mT0 mT0 = (mT0 – p0)

0 0

0 0 0 0

; 1mT p

mT p mT p = − =

− −

( )0 00

1v

R RC mT p

p= = −

−

And ( )( )

00 0

0 0 01p

mTR RC mT p

p mT p

= = −

− −

0

0p

mRTC

p=

13. Answer (B, C)

Hint :

aR = a + aei + aei2 + aei3 ….. aei(N – 1)

Solution :

( )( )cos –1k

a t k = +

1 2 3 4

.....N

= + + +

( )2 3 –1.....i i i i Na ae ae ae ae = + + +

( )( )2 –11 .....i i i Na e e e = + +

( )

( )

( )( )

–2 2 2

–

2 2 2

– 1 –

– 1–

iN iN iNiN

i i ii

a e ae e e

ee e e

= =

( )–1

2

2 sin2

2 sin2

i N

Ni

ae

i

=

( )sin

– 12cos

2sin

2

Na

Nt

= +


Hint : 2 2

2

0

6;

24

mv q hmvr n

r r= =

Solution :

2

0 2

0

6

4

qF

r=

2 2

2

0

6

4

mv q

r r=

Also 2

hmvr n=

( )2

0

6

4

qmvr v =

2

0

3 2

2

qv

nh

=

2

0

3n

qv

nh=

For n = 2 (1st excited state)

2

20

3

2

qv

h=

0

22 3n

nh n hr

m q

=

2 2

0

26n

n hr

mq

=

And

2 3 3

0

4

2

9

n hrT

v mq

= = ( )

2 3

0

2 4

8

9

hT

mq

=

Magnetic moment in ground state is

3

2

q

L m

=

3

2 2

q nh

m =

3

4

qh

m =


Hint : ( ) 0

3

2 cos.

4

mB r

R

=

Solution :

When y is the axis of rotation

( ) 0

3

2 cos

4

mB r

R

=

In figure instantaneous velocity

( )ˆsinv R k= −

So, end B will positive end and end A will be

negative end

2

0

30

2 cossin

4

md R Rd

R

=

( )

2 2 20 0

30

2 2 sinsin cos

4 4 2y

m R md

RR

= =

( )

0

4y

m

R

=



6/14

When x-axis is the rotational axis then

v = Rcos; ( ) 0

3

2 cos

4

mB r

R

=

0

3

2 coscos

4

md R Rd

R

=

( )

2 220 0

30

22 cos2 1cos

4 4 2x

mm Rd d

RR

+ = =

( )

20

0

2 1 sin2

4 2 2x

m

R

= +

( )0 2 1

4 2 2x

m

R

=

( )0

8x

m

R

=

16. Answer (B, C)

Hint : ( )

( )

20 0 00 0

0 0

48; 602 2

R r r ir i

R r R r

+= =

+ +

Solution :

Let i0 is

then current that battery supplies

then ( )

( )( )

0 0 02 00

0 0

48; 602 2

R r r iir

R r R r

+= =

+ +

00 0

0

45 4 4

5

rr R r

R r= = +

+

r0 = 4R, Now 04 4

489

R Ri

R

=

Ri0 = 27 Also 0

9

29i

R

=

= 87 volts

17. Answer (B, C)

Hint : ( )

1 21 2

1 2 1 1 2

f fm m

f f df x d f f=

− + − −

Solution :

For first lens 1 1

1 1 1

v x f+ =

1

1 1

1 x f

v xf

−=

11

1

xfv

x f=

−

1 11

1

v fm

x x f= =

−

Now object distance for 2nd lens is d – v1

2 1 2

1 1 1

v d v f+ =

−

( )

1 2

2 2 1

1 d v f

v f d v

− −=

−

( )2 1

21 2

f d vv

d v f

−=

− −

( )

2 22

1 1 2

v fm

d v d v f= =

− − −

( )( )

1 2 1 21 2

1 1 1 2 1 21 2

1

f f f fm m

xf dx df xf xf f fx f d f

x f

= =− − − +

− − − −

( )( )

1 21 2

1 2 1 1 2

f fm m m

f f df x d f f= =

− + − −

As m is independent of x

d – f1 – f2 = 0

d = f1 + f2

and find ( )

21 21 2

1 2 1

ff fm m m

f f d f= = =

−

18. Answer (B, C)

Hint : 0

0

sinv

N a dvv

=

Solution :

0

0

sinv

dN a dvv

=



7/14

0

00

0 0

cos

vv v

N dN av

= = −

0 02a v

N =

0 0

0

Avg

0

sinv

va v dv

v dN vv

dN N

= =

0

000

sin

vv

a v dvv

0 00

0 0

cos cosvv vv v

a dvv v

= − +

02

0 00 2

0 0 0

cos sin

vv v vv v

av v

= − +

( )2

020

0 0Avg

0 0 0 0

0 0

2 2

va

a vv

a v a v

+ −

= =

0Avg 2

vv =

19. Answer A(P, R, S); B(P, R, S); C(P, R, S); D(Q, T)

Hint : Finding ; ,dP dV dP

dT dT dV

Solution :

For C = Cv + T

Process equation is

T

RVe

−

= constant

ln constantT

vR

− = ,

10dv dT

v R

− =

( )vedv v

dT R

= +

So dv

dT is positive

Also, in (PT) variable –

constant

T

RT

ep

=

ln ln constantT

T PT

− − =

1dP

PdT T R

= −

[could be positive or negative]

Also in (PV) variable2

PV

Rve

−

= constant

( )

2

dP R VR

dV V

− =

(Could be positive or negative)

For next option C = Cv + V; process equation

constant

R

VTe =

2dV V

dT TR

= (positive only)

In (PV) variable :

R

VPVe = constant

( )

2

P R VdP

dV V

− =

(could be positive or negative)

In (PT) variable : constant

RP

TTe =

dP RP T

dT RT

− =

(could be positive or negative)

For next option : C = CV + aP; the process equation

is V – aT = constant

dV

dT= (positive)

In (PT) variable : RT

aTP

− = constant

( )dP R aP

dT TR

−= (could be positive or negative)

In (PT) variable aPV

VR

− = constant

dP R aP

dV aV

−= (could be positive or negative)

For last option

7

5PV = constant

7

5

dP P

dV V

−= (negative)

5

2

dV V

dT T

−= − (negative);

2 3

5 57

5

dPk T P

dT= (positive)

20. Answer A(R); B(Q, S); C(P); D(S, T)

Hint : 0

1 1 22 ;

120 45

mT

k x x= + =

−

Solution :

Time period of oscillation 0

2m

Tk

= and

distance from screen (of lens) when it will form

the image are



8/14

1 1 2

120 45x x+ =

−

x2 – 120x + 2700 = 0

x = 30 cm and 90 cm

Initial position from screen was 50 cm. So for impulse 8 kg m/s or velocity 4 m/s the lens will shift 20 cm right ward so only one time image will be formed. And for impulse 16 kg m/s or velocity 8 m/s the particle will just be able to cross through both of the corresponding position where the image will be formed on the screen.

PART - II (CHEMISTRY)

21. Answer (2)

Hint : 4 A atoms from the corner and 2A atoms

from the faces lie on the plane. Two B atoms at

the edge centres and one at body centre lie on

the plane.

Solution : Value of A = 6

Value of B = 3

22. Answer (3)

Hint : Possible resonating structures for P are as

follows

Solution : Q and R are identical structures

P, Q and R have incomplete octet

Q and R are not the resonating structures that’s

why statement 5 is incorrect. Statements (1), (7)

and (8) are correct.

23. Answer (1)

Hint :

Solution :

24. Answer (8)

Hint : 1.12 L of H2 gas at STP 1

20= mole of H2

1

20 moles of H2 =

1

10 mole of H atoms.

Solution : Energy required to excite 1 mole of

H-atoms from ground state to 2nd excited state

2 2

1 1E y

1 3

= = −

8

y J/mol9

=

For 0.1 mole of H-atom 8

y J90

=

25. Answer (9)

Hint : Ksp remain same under all conditions at a

constant temperature

Solution : ( ) 33

2 10

A OH A 3OH

27 10 10

+ −

− −

+

Ksp = (27 × 10–2) (10–10)3

= 27 × 10–32

Solubility in pure water

( ) 3

3A OH A 3OH

s 3s

+ −+

Ksp = 27s4 = 27 × 10–32 s = 1 × 10–8



9/14

26. Answer (2)

Hint : 44.6 g of NaHg in which both are mixed in equal molar proportions

x × 200 + x × 27 = 44.6

x = 1

5mole

Na amount which is reduced = 23

g5

.

Solution : w = Z × I × t

23 23

50 t5 1 96500

=

t = 386 sec

27. Answer (2)

Hint : 3 2 2 2LiNO Li O NO O

⎯⎯→ + +

Solution : Both NO2 and O2 are paramagnetic.

28. Answer (8)

Hint : Number of electrons in * orbital in

( )2N 0 a+ =

( )2N 1 b− =

( )2O 1 c+ =

( )2O 3 d− =

( )2F 3 e+ =

Solution : a + b + c + d + e = 8

29. Answer (B, C)

Hint : The correct molar formula of the given coordination, compound is [Co(SCN)2 (NH3)4]3 [Co (OX)3]

Solution : The positively charged radical is [Co(SCN)2 (NH3)4]+1

It can show six linkage isomers

The negatively charged anion is [Co(OX)3]–3. It is

optically active and it cannot show geometrical

isomerism.

30. Answer (C)

Hint : Greater the ease of the molecule towards

SN2, lesser will be the G#.

Solution : Given plot is for SN2 nucleophilic

substitution reaction. The correct order for SN2

among the given molecules is S > R > Q > P.

The order for G# is P > Q > R > S.

31. Answer (A, C)

Hint : There are 6 orbitals which participate in

3C – 2e– bond formation in diborane

Solution :

Each boron is contributing 2 orbitals and one

orbital by each hydrogen. AlCl3 exist in the form

of dimer in the vapour state.

32. Answer (B, C, D)

Hint : During adiabatic change, temperature gets changed that’s why S1 is false.

Solution : |wA| > |wB| as work done is area under curve.

Wad = U = Cv T

For monoatomic gas v

3C R

2=

For diatomic gas v

5C R

2=

33. Answer (A, C)

Hint : To complete the reaction reacting

molecules must have EK Ea then only they can cross the barrier of the reaction.

Solution : Catalyst lowers the activation energy by changing the path of reaction not by increasing the velocity of particles.



10/14

34. Answer (A, D)

Hint : Y must have same mass number as that of X and atomic number one more than X.

Solution : 23 1 23 111 1 12 0Na H Mg n+ → +

12 1 12 16 1 7 0C H N n+ → +

35. Answer (C, D)

Hint : d5 and d10 system has zero CFSE value

Solution : Both Mn2+ and Zn2+ have zero CFSE value which does not contribute to hydration energy

36. Answer (A, C)

Hint : aq. CuSO4 complexed with sodium potassium tartarate is Fehling solution with can oxidise aldehyde group.

Solution : PCC can oxidise only alcohols

Product x in reaction (d) is the major product.

37. Answer (B, C)

Hint : ( )

( )3 3 2dil.

4Zn 10HNO 4Zn NO+ → +

2 25H O N O+

( )

( )3 3 2 22conc.

Zn 4HNO Zn NO 2H O 2NO+ → + +

Solution : N2O is neutral

NO2 is acidic.

38. Answer (B, C)

Hint :

Solution : We can compare the bond parameters

of H2O2 in gas phase and in solid phase by

observing the structure.

39. Answer A(Q, S); B(P); C(R, T); D(R)

Hint :

Solution :

40. Answer A(P, T); B(S); C(Q, R); D(Q, R)

Hint : Apply the concept of chemical equivalence

N1V1 = N2V2

Solution : ( ) ( )2 2 4 4H C O KMnO

1 1 2 2N V N V=

1 22 0.5 V 5 0.5 V =

2

1

V 2

V 5=

PART - III (MATHEMATICS)

41. Answer (5)

Hint : (1 + sinx) (cosx + cos2x) = 0

Solution :

2cos x + 2cos2x + sin2x + 2cosx.sinx – 2sinx = 0

2cosx + 2cos2x + 2sinx.cosx + 2sinx – 4sin3x = 0

2cosx + 2cos2x +2sinx.cosx + 2cos2x.sinx = 0

(1 + sinx) (cosx + cos2x) = 0

( )3

2cos .cos 1 sin 02 2

x xx + =

, , , and3 3 3

x

= − − − −

42. Answer (4)

Hint :

Equation of normal at (t) is 3

1 1 18 32x t y t t+ = +

Solution :

A point on parabola y = 4x2 is (t, 4t2).

Let P = (t1, 4t12) and Q = (t2, 4t22)

If A is vertex of parabola, then



11/14

Slope of AP × slope of AQ = –1

1 21

16t t = −

Equation of normal at P and Q are

31 1 18 32x t y t t+ = + …(i)

and 32 2 28 32x t y t t+ = + …(ii)

Let normal (i) and (ii) intersect at (x1, y1), then

( )1 1 22x t t= + and ( )2

1 1 23

48

y t t= + +

On eliminating t1 and t2 we get

21 1

3

8y x= +

Length of latus rectum = 1

4

43. Answer (5)

Hint :

Total number of selection = 15

C3

Solution :

Total number of ways to select 3 different

numbers = 15C3 = 455

The favourable pair of numbers are (2, 3, 6), (2,

4, 8), (2, 5, 10), (2, 6, 12), (2, 7, 14), (3, 4, 12) and

(3, 5, 15)

Required probability 7 1

455 65= =

44. Answer (2)

Hint :

f(x) + f(2 – x) = 2

Solution :

f(x) + f(2 – x) = 2

So, ( )( )2 2

0 0

f f x dx dx=

= 2

45. Answer (8)

Hint :

Equation of conic is 5 + S = 0.

Solution :

The equation of required parabola is

(x2 + y2 + 6x – 24y + 72) + (x2 – y2 + 6x + 16y –46) = 0

2x2 + 12x – 8y + 26 = 0

Focus of parabola = (–3, 2)

The centre and radius of circle x2 + y2 – 2x – 10y

+ 17 = 0 are (1, 5) and 3 units respectively

Maximum distance

( ) ( )2 2

3 1 2 5 3 8= − − + − + =

46. Answer (4)

Hint :

( ) ( )–

b b

a a

f x dx f a b x dx= + .

Solution :

(( )2

2

7 tan

sec. 8I x f x c dx

−

= −

( ) ( )( )2

2

7 tan

sec8 8 .I x f x x dx

−

= − −

( )( ) ( )2 2

2 2

7 tan 7 tan

sec sec8 8I f x x dx xf x

− −

= − − −

I = 8J – I

4 =I

J

47. Answer (9)

Hint :

3 3 3 32 4 4

sin sin sin – sin3 3 3

+ + + + =

.

Solution :

3 3 32 4sin sin sin

3 3x x x

+ + + +

3 2 4

sin sin sin4 3 3

x x x

= + + + + −

( ) ( )( )1

sin3 sin 3 2 sin 3 44

x x x+ + + +

3

0 sin34

= − x

( )3

sin34

f x x dx= −

1

cos34

= +x C

When ( )9

0, 0 24

x f C= = =

4 f(x) = cos3x + 8

Maximum value of 4 + (x) is 9



12/14

48. Answer (7)

Hint :

Differentiate both sides.

Solution :

( ) ( )0

2 8 2. 2 8x

f x t dt f x x= =

f’(x) = 2x

f(x) = x2 {f(0) = 0}

Equation of tangent at x = 2 is, y = 4x – 4.

Required area ( )2 2

02= − x dx sq. units

2

3= sq. units


Hint :

( ) ( )

0 0

–

a a

f x dx f a x dx=

Solution :

( ) ( )2

0

x tf x x e f x t dt−= + − …(i)

( ) ( )2

0

xx tf x x e e f t dt−= + …(ii)

On differentiating both sides w.r.t. x we get

( ) ( ) ( )0

' 2 .xx t x xf x x e e f t dt e e f x− −= − +

( ) ( )( ) ( )2' 2f x x f x x f x= − − + [from (ii)]

f’(x) – 2x + x2 …(iii)

( )3

2

3

xf x x C= + +

but f(0) = 0 C = 0

( ) 3 21

3f x x x= +

For maxima and minima f(x) = 0 x = 0, –2

Local maximum value ( )8 4

2 43 3

= − = − =f

Local minimum value = f (0) = 0

Number of real solutions of f(x) = 1 is 3.

( ) ( )3 3 311

1

11 2 ..... 11

3k

f k

=

= + + +

( )2 2 21 2 .... 11 1958+ + + + =

50. Answer (A, B, C)

Hint :

Use A.AT = I.

Solution :

(A + B)–1 (A – B) (A – B)T ((A+ B)T)–1

= (A + B)–1 (A – B) (A + B) (A – B)–1

= (A + B)–1 (A + B) (A – B) (A – B)–1

= I

51. Answer (A, B, C, D)

Hint :

Use properties of determinant.

Solution : Here C3 → C3 – g(y1)C1 + f(y1) C2

gives the value of determinant is equal to zero.

It is independent of x1, x2, x3 and y1.


Hint :

Focus and directrix of parabola is (0, 1) and x + y

= 0 respectively.

Solution :

From required conditions it is obvious that the

focus of parabola is (0,1) and directrix x + y = 0

Equation of parabola is

( ) ( )( )

22 2

0 12

+− + − =

x yx y

x2 – 2xy + y2 – 4y + 2 = 0

Length of latus rectum 0 1

2 22

+= =

Coordinates of foot of perpendicular form (0,1) to

x + y = 0 is 1 1

,2 2

−

Coordinates of vertices are1 3

,4 4

−

53. Answer (A, D)

Hint :

( ) ( ) ( )cos , cos , cos ,r a r b r c= =

Solution :

Let r be the required vector

( ) ( ) ( )cos , cos , cos ,= =r a r b r c



13/14

2 2 2 3 6

3 7

4 4 7(Say)

9

x y z x y z

x y zk

+ + + + =

+ += =

x + 2y + 2z = 3 k …(i)

2x + 3y + 6z = 7k … (ii)

4x + 4y + 7z = 9k … (iii)

From equations (i), (ii) and (iii)

x : y : z = 3 : 5 : 7

ˆ ˆ ˆ3 5 7r i j k= + +

r is perpendicular to the plane passing

through points , a b and c

( ) ( ). 0, . 0r a b r b c − = − =

( ) ( )r a r b = and ( ) ( ) = r b r c

. 27 = . 63 = . 81

21 9 7

= =

Possible values of ( + + )

= (21 + 9 + 7) , where I–1

= 37 , I–1

37 × 2 = 74 and 37 × 5 = 185


Hint :

The vector

( ) ( ) ( )ˆ ˆ ˆ3 3 2 2 3 7= + + − + + −PQ i j k

Solution :

The vector

( ) ( ) ( )ˆ ˆ ˆ3 3 2 2 3 7= + + − + + −PQ i j k

PQ is parallel to plane = 0

2(3 + 3) – 3 (– 2 + 2) + 3(3 – 7) = 0

6 + 6 + 6 – 6 + 9 – 21 = 0

= 0.1

Equation of plane parallel to plane = 0 is

2x – 3y + 3z = k …(i)

Above plane is contains vector PQ.

k = 21

Equation of plane is 2x – 3y + 3z = 21

55. Answer (B)

Hint :

Concept of factors.

Solution :

1……….. n ………..n2

2008 = 23 × 251

(2008)8 = 224 × 2518

Total number of factors = 225

Hence required answer is 225 – 3

12

+

= 112


Hint :

22

4

x xe dx

x

+

+

( )

2

2

4 4

4

+ + = +

x x x

e dxx

Solution :

22

4

x xe dx

x

+

+

( )

2

2

4 4

4

+ + = +

x x x

e dxx

( )

2

4

4 4

= + + +

x x

e dxx x

4

= ++

x xe C

x

( )4

xf x

x=

+

( )4

14

f xx

− =+

which decreasing for x R+

andf’ (x) is not differentiable at x = –4

( )( )

( )2

42 1

4f x f

x = − =

+

57. Answer (A, B)

Hint :

–1 –1sin cos2

x x

+ =

Solution :

( ) ( ) ( )1 2 1 2sin 6 10 cos 6 10− −= − − + − −f x x x x x

( ) ( )( ) ( )( )2 21 1sin 3 1 cos 3 1− −= − − + + − − +f x x x

( )( )( ) ( )( )2 21 1sin 3 1 cos 3 1− −= − − + + − +x x



14/14

When x = 3, then ( )2

=f x

Domain of f(x) = {3}

And range of ( )2

=

f x

( )2

=f x has only one solution as x = 3


Hint :

( ) ( )

( ) ( )24 2 sin 1

+ =

− − + −

a b c a b b

b c

Solution :

( ) ( )

( ) ( )24 2 sin 1

+ =

− − + −

a b c a b b

b c

( ) ( )

( ) ( )24 2 sin 1

a c a b b a b c

b c

+ − =

− − + −

4 2 sina c a b + = − − and

21 = − a b

21 1 4 2 sin+ − = − − 1 =a c

2 2 2 sin 0 − + − =

1and sin 1 = =

cos = 0

59. Answer A(R); B(P, Q, R, T); C(P); D(Q, S)

Hint :

Use Leibnitz's Rule

Solution :

(A) ( ) ( )3

3

1cos secf x dx

=

( ) ( )3

3

1cos secf x dx

=

( ) ( )

( ) ( )

32 3

1

3 2 3

' 3 sin sec

cos 3 sec

f x dx

= −

+

( )3

2 3

13 1 sin sec x dx

= −

( ) 2/33' 3f =

( )3

1 3

18 6f

=

(B) The range of 2

4

1+

x

xis [0,2]

Range of ( ) 1,tan 24

− = −

f x

(C) Clearly ˆ=b i

1b =

(D)

22 3

1 3

2

iiz e

− +

= =

3 3i i

z e e

= = and

2

3

−

i

e

arg (z) are 3

and

2

3

−

60. Answer A(T); B(R); C(P); D(Q)

Hint :

If x4 N then remainder when x4 is divisible by 5

is either 1 or 0.

Solution :

(A) By Bay’s theorem the required

Probability 5

11

10

1 9 11 .

10 10 2

−

=

+

33

41=

p

q

p + q = 32 + 41 = 73

(B) Total number of numbers

= Coefficient of x10 in (x + x2 + x3)7

= Coefficient of x3 in (1 + x + x2)7

= 9C3 – 7 = 77

(C) 1, a1, a2,…………., an, 15 are in A.P.

The common difference is given by

15 = 1 +(n + 1) d

14

1d

n=

+

6

–1

12 1 6

19 15 – 2n

a d

a d

+= =

7 14

6 1d

n= =

+

n = 11

(D) x4 – y4 = (x2 + y2) (x – y) (x +y) is divisible by 5

This is possible by 15 + 54 = 69 ways

❑ ❑ ❑

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