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RESONANCE Page # 1
QUESTIONS & SOLUTIONS OFKISHORE VAIGYANIK PROTSAHAN YOJANA (KVPY) 2010
TEST PAPER CLASS - XI
One-Mark QuestionsMATHEMATICS
1. A student notices that the roots of the equation x2 + bx + a = 0 are each 1 less than the roots of the equationx2 + ax + b = 0. Then a + b is :(A) possibly any real number (B) � 2
(C) � 4 (D) � 5
Ans. (C)
Sol. x2 + bx + a = 0
x2 + ax + b = 0
x = � 1 x + 1 = (x + 1)2 + a(x + 1) + b = 0x2 + (a + 2)x + 1 + a + b = 0x2 + bx + a = 0 and 1 + a + b = a b = � 1
a = � 3
_________a + b = � 4
2. If x, y are real numbers such that
1yx
3 �
1yx
3 = 24,
then the value of (x + y) / (x � y) is :
(A) 0 (B) 1 (C) 2 (D) 3
Ans. (D)
Sol. 3x/y = t 3t � 3t
= 24 8t = 3 × 24
t = a x = 2y
y�xyx
yy3
3 Ans. (D)
(TEST PAPER HELD ON 31-10-2010)
RESONANCE Page # 2
3. The number of positive integers n in the set {1, 2, 3, ...., 100} for which the number n......321n.....321 2222
is an integer is :(A) 33 (B) 34 (C) 50 (D) 100
Ans. (B)
Sol. )1n(n3)1n2)(1n(n
2
31n2
= k n = 2
1�k3
31 2
1�k3 100 3 3k 201 1 k
3120
1 k = 67k is odd 34 (B)
4. The three different face diagonals of a cuboid (rectangular parallelopiped) have lengths 39, 40, 41. The lengthof the main diagonal of the cuboid which joins a pair of opposite corners :
(A) 49 (B) 249 (C) 60 (D) 260
Ans. (A)
Sol. 2 + b2 = 392
b2 + h2 = 402
h2 + 2 = 412
adding2(2 + b2 + h2) = 392 + 402 + 412
As = 2
414039 222
= 2
)140(40)1�40( 222
= 2
2)40(3 2
= 2
4802 = 2401
= 49 Ans. (A)
5. The sides of a triangle ABC are positive integers. The smallest side has length . What of the followingstatements is true ?(A) The area of ABC is always a rational number.(B) The area of ABC is always an irrational number.(C) The perimeter of ABC is an even integer.(D) The information provided is not sufficient to conclude any of the statements A, B or C above.
Ans. (B)
Sol. It has to be isosceles triangle
= 21
× 141
�x2 = 1�x441 2
perimeter = 1 + 2x odd always irrational (B)
RESONANCE Page # 3
6. Consider a square ABCD of Side 12 and let M, N be the midpoints of AB, CD respectively. Take a point P onMN and let AP = r, PC = s. Then the area of the triangle whose sides are r, s, 12 is :
(A) 72 (B) 36 (C) 2rs
(D) 4rs
Ans. (B)
Sol. = 21
× 12 × 6
= 36
Ans. (B)
7. A cow is tied to a corner (vertex) of a regular hexagonal fenced area of side a meters by a rope of length5a / 2 meters in a grass field. (The cow cannot graze inside the fenced area.) What is the maximum possiblearea of the grass field to which the cow has access to graze ?
(A) 5 a2 (B) 25a2 (C) 6a2 (D) 3a2
Ans. ( ....)
Sol. Area =
4a3
6�2a5 22
= 4
a36�
4a25 22
= 4a
)36�25(2
Note : Options are not match with solution.
8. A closed conical vessel is filled with water fully and is placed with its vertex down. The water is let out at aconstant speed. After 21 minutes, it was found that the height of the water column is half of the originalheight. How much more time in minutes does it empty the vessel ?(A) 21 (B) 14 (C) 7 (D) 3
Ans. (D)
RESONANCE Page # 4
Sol.dtdv
= c.
v = hr31 2 tan =
hr
= 3
h3 tan
dtdv
= h2 tan dtdh
ct + k = 3tanh3
when t = 0, h = H k = 3tanH3
ct =
3
(h3 � H3) tan
when t = 21 , h = 2H
c = 36
H3
87�
h = 0
tan)H(�3
3 = t87�
H6
33
247
= 728
7
t
t = 24more time in minutes does it empty the vessel is 3
9. I carried 1000 kg of watermelon in summer by train. In the beginning, the water content was 99%. By the timeI reached the destination, the water content had dropped to 98%. The reduction in the weight of the water-melon was:(A) 10 kg (B) 50 kg (C) 100 kg (D) 500 kg
Ans. (D)
Sol. Water + solid = 1000
Water is 100010099
= 990
water evaporated is x.
so 100x�1000x�990
= 98
99000 � 100 x = 96000 � 98x
1000 = 2xx = 500
RESONANCE Page # 5
10. A rectangle is divided into 16 sub-rectangles as in the figure; the number in each sub-rectangle representsthe area of that sub-rectangle. What is the area of the rectangle KLMN ?
(A) 20 (B) 30 (C) 40 (D) 50
Ans. (D)
Sol.
xy = 10 and yz = 4 z = 5x2
y
also
x
52
a = 12 a = x
30
and
x
30 (b) = 15 b =
2x
and
2x
(c) = 25 c = x
50
Area = x
x50
= 50
RESONANCE Page # 6
PHYSICS
11. A hollow pendulum bob filled with water has a small hole at the bottom through which water escapes at aconstant rate. Which of the following statements describes the variation of the time period (T) of the pendu-lum as the water flows out ?(A) T decreases first and then increases. (B) T increases first and then decreases.(C) T increases throughout. (D) T does not change.
Ans. (B)
Sol. T = g2
First distance of com from suspension point will increase then decrease. So,
T .
12. A block of mass M rests on a rough horizontal table. A steadily increasing horizontal force is applied suchthat the block starts to slide on the table without toppling. The force is continued even after sliding hasstarted. Assume the coefficients of static and kinetic friction between the table and the block to be equal.The correct representation of the variation of the frictional forces, �, exerted by the table on the block with
time t is given by :
(A) (B) (C) (D)
Ans. (A)
Sol. when sliding has started
till acceleration of block is zero F � fk = ma
F � fs = 0
fs = F
RESONANCE Page # 7
13. A soldier with a machine gun, falling from an airplane gets detached from his parachute. He is able to resistthe downward acceleration if he shoots 40 bullets a second at the speed of 500 m/s. If the weight of a bulletis 49 gm, what is the weight of the man with the gun ? Ignore resistance due to air and assume theacceleration due to gravity g = 9.8 m/s2
(A) 50 kg (B) 75 kg (C) 100 kg (D) 125 kg
Ans. (C)
Sol. Mg = 40 ×
100049
× 1
500
M = 9810
1054940
= 100 kg.
14. A planet of mass m is moving around a star of mass M and radius R in a circular orbit of radius r. The starabruptly shrinks to half its radius without any loss of mass. What change will be there in the orbit of theplanet ?(A) The planet will escape from the star. (B) The radius of the orbit will increase.(C) The radius of the orbit will decrease. (D) The radius of the orbit will not change.Ans. (D)
Sol.
2r
GMm =
rmv2
No change because distance between them will be from centre to centre distance which is unchanged.
15. Figure (a) below shows a Wheatstone bridge in which P, Q, R, S are fixed resistances, G is a galvanometerand B is a battery. For this particular case the galvanometer shows zero deflection. Now, only the positionsof B and G are interchanged,. as shown in figure (b). The new deflection of the galvanometer.
(A) is to the left. (B) is to the right.(C) is zero. (D) depends on the values of P, Q, R, SAns. (C)
RESONANCE Page # 8
Sol.
Since, ig = 0
PR = QSStill it will be a balanced W.S.B.So, again i
g = 0.
16. 12 positive charges of magnitude q are placed on a circle of radius R in a manner that they are equallyspaced. A charge +Q is placed at the centre. If one of the charges q is removed, then the force on Q is :(A) zero
(B) 20R4
away from the position of the removed charge.
(C) 20R4
qQ11
away from the position of the removed charge.
(D) 20R4
towards the position of the removed charge.
Ans. (D)
Sol. 2R
KQq =
041
. 2R
Since initially net force on Q was zero by symmetry
So, 0FF 11 mainingRe1
111 mainingRe FF
So, 20 r
Qq.
41
towards the position of the removed charge.
RESONANCE Page # 9
17. An electric heater consists of a nichrome coil and runs under 220 V, consuming 1 kW power. Part of its coilburned out and it was reconnected after cutting off the burnt portion. The power it will consume now is :(A) more than 1 kW. (B) less that 1 kW, but not zero.(C) 1 kW. (D) 0 kW.
Ans. (A)
Sol. Pi =
i
2i
RV
Since, Rf < R
i keeping V = constant
V = Vf
Pf =
f
2
RV
Since, R P .
18. White light is split into a spectrum by a prism and it is seen on a screen. If we put another identical invertedprism behind it in contact, what will be seen on the screen ?(A) Violet will appear where red was(B) The spectrum will remain the same(C) There will be no spectrum, but only the original light with no deviation.(D) There will be no spectrum, but the original light will be laterally displaced.
Ans. (D)
Sol.
the combination will behave as parallel slab so light get laterally displaced without any spectrum.
19. Two identical blocks of metal are at 20ºC and 80ºC, respectively. The specific heat of the material of the two
blocks increases with temperature. Which of the following is true about the final temperature. Which of thefollowing is true about the final temperature T
� when the two blocks are brought into contact (assuming that
no heat is lost to the surroundings) ?(A) T
� will be 50ºC.
(B) T� will be more than 50ºC.
(C) T� will be less than 50ºC.
(D) T� can be either more than or less than 50ºC depending on the precise variation of the specific heat with
temperature.
Ans. (B)Sol. 20ºC 80ºC
S as T
d = m.s.d d = ms dSince, average S of body which is initially at 80ºC is higher then body initially at temperature 20ºC so
temperature decreases of earlier will be less then temperature increases of letter.So, T
f > 50ºC.
RESONANCE Page # 10
20. A new temperature scale uses X as a unit of temperature, where the numerical value of the temperature tX in
this scale is related to the absolute temperature T by tX = 3T + 300. If the specific heat of a material using this
unit is 1400 J kg�1 X�1 its specific heat in the S.I. system of units is :(A) 4200 J kg�1 K�1 (B) 1400 J kg�1 K�1
(C) 466.7 J kg�1 K�1 (D) impossible to determine from the information provided
Ans. (A)
Sol. tx = 3 T + 300 Q = ms t
x = 3T
S = .m
QS =
.mQ
Since, unit of is joule in both systemX T
m = m0kg m
0kg
Q = Q0J Q
0.J
tx
T
Sx =
x0
0
tmQ
= 1400 ST = Tm
Q
0
0
=
x0 tmQ3
ST = 3 × 1400 = 4200 J-kg�1K�1.
CHEMISTRY
21. The boiling points of 0.01 M aqueous solutions of sucrose NaCl and CaCl2 would be :
(A) the same (B) highest for sucrose solution(C) highest for NaCl solution (D) highest for CaCl
2 solution
Ans. (D)
Sol. Aqueous solution containing more number of particles have more elevation in boiling point.
22. The correct electronic configuration for the ground state of silicon (atomic number 14) is :(A) 1s2 2s2 2p6 3s2 3p2 (B) 1s2 2s2 2p6 3p4 (C) 1s2 2s2 2p4 3s2 3p4 (D) 1s2 2s2 2p6 3s1 3p3
Ans. (A)
Sol.14
Si : 1s2 2s2 2p6 3s2 3p2
RESONANCE Page # 11
23. The molar mass of CaCO3 is 100 g. The maximum amount of carbon dioxide that can be liberated on heating
25 g of CaCO3 is :
(A) 11 g (B) 5.5 g (C) 22 g (D) 2.2 g
Ans. (A)
Sol. CaCO3 (s)
CaO (s) + CO
2 (g)
Number of mole
10025
10025
Amount of CO2 =
10025
× 44 = 11 gram.
24. The atomic radii of the elements across the second period of the periodic table.(A) decrease due to increase in atomic number (B) decrease due to increase in effective nuclear charge
(C) decrease due to increase in atomic weights (D) increase due to increase in the effective nuclear charge
Ans. (B)
Sol. As we move left �to right� in 2nd period atomic radii decreases due to increase in effective nuclear charge.
25. Among NH3, BCl
3, Cl
2 and N
2, the compound that does NOT satisfy the octet rule is :
(A) NH3
(B) BCl3
(C) Cl2
(D) N2
Ans. (B)
Sol. In BCl3 octet rule is not satisfy.
Total number of 6 electrons in outermost shell of B after bonding.
26. The gas produce on heating MnO2 with conc. HCl is
(A) Cl2
(B) H2
(C) O2
(D) O3
Ans. (A)
Sol. MnO2 + 4HCl MnCl
2 + 2H
2O + Cl
2
Cl2 gas produces.
RESONANCE Page # 12
27. The number of covalent bonds in C4H
7Br, is :
(A) 12 (B) 10 (C) 13 (D) 11
Ans. (A)
Sol. C4H
7Br
CH2 = CH � CH
2 � CH
2 � Br
HH||
Br�C�C�CC�H||||HHHH
Number of covalent bond = 12.
28. An aqueous solution of HCl has a pH of 2.0. When water is added to increase the pH to 5.0, the hydrogen ionconcentration :(A) remains the same (B) decreases three-fold(C) increases three-fold (D) decreases thousand-fold
Ans. (D)
Sol. pH = 2 [H+]i = 10�2 M
pH = 5 [H+]f = 10�5 M
i
f
]H[
]H[
= 2
5
10
10
=
10001
So, H+ concentration decrease thousand fold.
29. Consider two sealed jars of equal volume. One contains 2 g of hydrogen at 200 K and the other contains 28g of nitrogen at 400 K. The gases in the two jars will have :(A) the same pressure. (B) the same average kinetic energy.(C) the same number of molecules. (D) the same average molecular speed.
Ans. (C)
Sol. For 1st jar :
Number of moles of H2 (g) =
22
=1 mole.
Number of molecules of H2 (g) = 6.02 × 1023.
For 2nd jar :
Number of moles of N2 (g) =
2828
=1 mole.
Number of molecules of N2 (g) = 6.02 × 1023.
So, both jar have same number of molecules.
RESONANCE Page # 13
30. Identify the stereoisomeric pair from the following choices.(A) CH
3CH
2CH
2OH and CH
3CH
2OCH
3(B) CH
3CH
2CH
2Cl and CH
3CHClCH
3
(C) and (D) and
Ans. (C)
Sol. and
(cis) (trans)
cis and trans are stereoisomeric pair.
BIOLOGY
31. Which of the following is a water-borne disease ?(A) Tuberculosis (B) Malaria (C) Chickenpox (D) Cholera
Ans. (D)
32. In has seminal work on genetics, Gregor Mendel described the physical traits in the pea plant as beingcontrolled by two 'factors'. What term is used to define these factors today ?(A) Chromosomes (B) Genes (C) Alleles (D) Hybrids
Ans. (C)
33. A majority of the tree species of peninsular Indian origin fruit in the months of :(A) April - May (B) August - September(C) December - January (D) All months of the year
Ans. (A)
34. In frogs, body proportions do not change with their growth. A frog that is twice as long as another will beheavier by approximately.(A) Two-fold (B) Four-foldSix (C) -fold (D) Eight-fold
Ans. (A)
35. Which of the following has the widest angle of binocular vision ?(A) Rat (B) Duck (C) Eagle (D) Owl
Ans. (D)
RESONANCE Page # 14
36. The two alleles of a locus which an offspring receives from the male and female gametes are situated on :(A) Two different homologs of the same chromosome.(B) Two different chromosomes.(C) Sex chromosomes.(D) A single chromosome.
Ans. (B)
37. Ants locate sucrose by :(A) Using a strong sense of smell.(B) Using a keen sense of vision.(C) Physical contact with sucrose.(D) Sensing the particular wavelength of light emitted / reflected by sucrose.
Ans. (A)
38. The interior of a cow-dung pile kept for a few days is quite warm. This is mostly because :(A) Cellulose present in the dung is a good insulator.(B) Bacterial metabolism inside the dung releases heat.(C) Undigested material releases heat due to oxidation by air.(D) Dung is dark and absorbs a lot of heat.
Ans. (B)
39. Which one of these is the correct path for a reflex action ?(A) Receptor-Motor Neuron-Spinal Cord-Sensory Neuron-Effector.(B) Effector-Sensory Neuron-Spinal Cord-Motor Neuron-Receptor.(C) Receptor-Sensory Neuron-Spinal Cord-Motor Neuron-Effector.(D) Sensory Neuron-Receptor-Motor Neuron-Spinal Cord-Effector.
Ans. (C)
40. Insectivorous plants digest insects to get an essential nutrient. Other plants generally get this nutrient fromthe soil. What is this nutrient ?(A) Oxygen (B) Nitrogen (C) Carbon dioxide (D) Phosphates
Ans. (B)
RESONANCE Page # 15
SUBJECTIVE QUESTIONS
DISCLAIMER
The subjective problems are based memory rotations of our students hence they may differ slightly thenthe original questions. The solutions presented here are in accordance with the given problemsstatement.
MATHEMATICS
1. Leela and madan collectea their CD's and sold them such that each C.D. was sold at the same amount asnumber of C.D's. From this amount colluted leela juist borrowed Rs. 10, then madan borrowed Rs. 10, theyborrowed money alternately until amount less than Rs. 10 was left for Madan to borrow. Find has much wasleft for Madan to borrowed at the end.
Sol. Let total amount is n2
total borrowed amount = (2u + 1) 10n2 � (2u + 1) 10 < 10
True for. n = 6 u = 1so the left amount = 6
2. In ABC, DE is drawn parallel to BC, D on A and B on E. Such that ADE has area 3. Now BE and DC arejoined to get intersection point P. DPE = 1 sq. unit. Find area of ABC ?
Sol.
Let ADE is equilateral and D is mid point of AB and E is mid point of AC[given condition is true for above assumption] Area of quad. ADPE = Area of quad. DPFB
= Area of quad. EPFC Area of ABC = 12 sq. units
RESONANCE Page # 16
3. (i) If there is natural number n relative prime with 10. Then show that there exist another natural numberm such that all digits are 1's and m is div. by 'n'.
(ii) Show that every natural number can represented as )110(10
acb
, where a, b, c N
Sol. from the question if m = 1111 or(i) m = 111.11is always divisible by n = 11 which is coprime with 10(ii) by choosing a = k 10b (10c � 1) when k is any natural number we can option any natural number k.
The problem seen to have an error which may be due to memory retersion constraints.
PHYSICSDISCLAIMER
The subjective problems are based on memory relation of our students hence they may differ slightly thenthe original questions. The solutions presented here are in accordance with the given problems state-ment.
4. There is a smooth fixed concave surface. A particle is released from p. Find :
(i) PE as function of (ii) KE as a function (iii) time taken from P to (iv) the reaction force at
Sol.
Assume to be spherical concave.
cos0 =
R)HR(
(i) P.E. = mg(R � R cos) = mgR(1 � cos)(ii) mgH � mgR (1 � cos) = kinetic energy
(iii) m(g sin)R = (mR2) 2
2
dt
d
RESONANCE Page # 17
tPQ
= 4
161
g2
4T
20
=
161
g2
20
(iv)
N � mg = R
mV2
N = mg + R
mV2
where V = gh2 .
5.
Given: RA << R << R
V
Restimated =
V
R = |R � R
est|
(i) Calculate RP
(ii) Calculate RQ
(iii) Find the value of R for which RP = R
Q
RESONANCE Page # 18
Sol.
(i) PR R =
V
V =
A
V
V RRR
RR'
V
VAAVR RR
RRRRRRR
P
RP =
V =
V
V
RRRR
+ RA
(ii) PR = A
V
V RRR
R.RR
Since, RA < < R < < R
V
Rest
= R�~RRR
RRA
V
V
PR = 0
(iii) RQ
V = VA
VA
RRRR)RR(
Restimated
= VA
VA
RRRR)RR(
RQ
= R � VA
VA
RRRR)RR(
=
VA
VAVVA2
RRRRRRRRRRRR
= VA
VAA2
RRR
RRRRR
= AV
A
V
2
RR
RRRR
RQ
�~ RA .
(iii)VV
VAA2
RRR
RRRRR
= 0
After solving, R = VARR .
RESONANCE Page # 19
6. An object is placed. 20 cm from a concave lens of focal length 10 cm :(i) find the position of the image(ii) find the position of another concave mirror of focal length 5 cm. where it should be placed right side ofabove concave lens such that final image coincides with the object.(iii) If a plane mirror is placed in place of concave mirror at the same position, then find the position of finalimage.
Sol.
(i) 101
201
V1
201
101
V1
= 20
12
V = 320
(ii) fLM
= �5 cm
320
+ d = 10
d = 10 � 3
20 =
310
cm
(iii)
(A) Ist reference with lens
V = 302
(B) Then mirror,X
im = �X
0 M
(C) Again by lens,
403
V1
=
101
403
101
V1
= 40
34
V = 740
cm
It means right of lens at a distance 740
cm.
RESONANCE Page # 20
CHEMISTRYDISCLAIMER
The subjective problems are based memory rotations of our students hence they may differ slightly thenthe original questions. The solutions presented here are in accordance with the given problemsstatement.
7. There are four bottles 1,2,3,4 containing following compounds.
Bottle-1 : ; Bottle-2 :
Bottle-3 : ; Bottle-4 :
Identify the bottle :(I) (A) whose content does not react with 1N HCl or 1N NaOH.
(B) whose content reacts with 1N NaOH only.(C) whose content reacts with 1N NaOH and 1N HCl both.(D) whose content reacts with 1N HCl only.
(II) The bottle whose content is highly soluble in distilled water.
Sol. (I) (A) Bottle-3 does not react with HCl or NaOH.
(B) Bottle-2 reacts only with NaOH.(C) Bottle-4 reacts both NaOH or HCl both.(D) Bottle-1 reacts with HCl only.
(II) Bottle-4 is highly soluble in distilled water due to zwitter ion formation.
8. A copper ore was treated with HNO3 to form Cu(NO
3)
2, Which on further reaction with K, forms Cu
2, which
decomposes to form Cu2
2 and
2,
2 was titrated with Na
2S
2O
3.
(I) Balance the reactions involved (find out values of a to )(a) a Cu + b HNO
3 c Cu(NO
3)
2 + d NO + e H
2O
(b) f Cu2 g Cu
2
2 + h
2
(c) i Na2S
2O
3 + j
2 k Na
2S
4O
6 + Na
(II) 2.54 g of 2 was evolved : Find the percentage purity of copper in 2 g of ore.
Sol. (i) Balanced equation are :(a) 3 Cu + 8 HNO
3 3 Cu(NO
3)
2 + 4 NO + 3 H
2O
(b) 2 Cu2 Cu
2
2 +
2
(c) 2 Na2S
2O
3 +
2 Na
2S
4O
6 + 2 Na
RESONANCE Page # 21
(ii) Mole of 2 =
25454.2
= 0.01
Mole of Cu2 = 2 × mole of
2 = 0.02
Mole of Cu2 = mole of Cu = 0.02
wt. = 0.02 × 6.25 = 1.27 g
% purity = 227.1
× 100 = 63.5%
9. Human being 2500 K cal of energy per dayC
12H
22O
11 + 12 O
2 12 CO
2 + 11 H
2O
H = � 5.6 × 106 J / mol(i) Human being required ..................... kJ of energy per day.(ii) Amount of sucrose required per day and volume of CO
2 evolved during the process.
Sol. (i)18.4
2500kJ = 598 kJ
(ii) Mole of sucrose required = 36 10106.5
598
= 0.106
wt. of sucrose required = 0.101 × 342 = 36.52 gm
