KVPY SA Solutions

GENERAL INSTRUCTIONS

� The Test Booklet consists of 80 questions.

� There are Two parts in the question paper. The distribution of marks subjectwise in each part

is as under for each correct response.

MARKING SCHEME :PART-I :

MATHEMATICSQuestion No. 1 to 15 consist of ONE (1) mark for each correct response.

PHYSICSQuestion No. 16 to 30 consist of ONE (1) mark for each correct response.

CHEMISTRYQuestion No. 31 to 45 consist of ONE (1) mark for each correct response.

BIOLOGYQuestion No. 46 to 60 consist of ONE (1) mark for each correct response.

PART-II :MATHEMATICSQuestion No. 61 to 65 consist of TWO (2) marks for each correct response.

PHYSICSQuestion No. 66 to 70 consist of TWO (2) marks for each correct response.

CHEMISTRYQuestion No. 71 to 75 consist of TWO (2) marks for each correct response.

BIOLOGYQuestion No. 76 to 80 consist of TWO (2) marks for each correct response.

Date : 30-10-2011 Duration : 3 Hours Max. Marks : 100

KISHORE VAIGYANIK PROTSAHAN YOJANA - 2012

STREAM - SA

RESONANCE PAGE - 2

KVPY QUESTION PAPER - STREAM (SA)

PART-IOne Mark Questions

MATHEMATICS

1. Suppose a, b, c are three distinct real numbers. Let

P(x) = )ca)(ba()cx)(bx(

+ )ab)(cb(

)ax)(cx(

+ )bc)(ac(

)bx)(ax(

When simplified, P(x) becomes

(A) 1 (B) x (C) )ac)(cb)(ba(

)cabcab)(cba(x2

(D) 0

Sol. (A)

P(x) = )ca)(ba()cx)(bx(

+ )ab)(cb(

)ax)(cx(

+ )bc)(ac(

)bx)(ax(

Let f(x) = P(x) � 1

f(a) = 1 + 0 + 0 � 1 = 0

f(b) = 0 + 1 + 0 � 1 = 0

f(c) = 0 + 0 + 1 � 1 = 0

f(x) is a polynomial of degree atmost 2, and also attains same value (i.e., 0) for 3 distinct values of x(i.e. a,b,c). f(x) is an identity with only value equal to zero. f(x) = 0 x R P(x) = 1, x R

2. Let a, b, x, y be real numbers such that a2 + b2 = 81, x2 + y2 = 121 and ax + by = 99. Then the set of allpossible values of ay � bx is

(A)

119

,0 (B)

119

,0 (C) {0} (D)

,

119

Sol. (C)Using cauchy schwartz's inequality(a2 + b2) (x2 + y2) (ax + by)2

equality holds at xa

= yb

ay � bx = 0

Aliter :a2 + b2 = 81 ......(i)x2 + y2 = 121 ......(ii)ax + by = 99 ......(iii)(a2 + b2) (x2+ y2) = 81 × 121 ......(iv) [(i) × (ii)]

(ax + by)2 = 992 ......(v) [squaring (iii)](iv) � (v)

(ay � bx)2 = 0 ay � bx = 0

3. If ax1

x , bx

1x

32

, then 2

3

x

1x is

(A) a3 + a2 � 3a � 2 � b (B) a3 � a2 � 3a + 4 � b (C) a3 � a2 + 3a � 6 � b (D) a3 + a2 + 3a � 16 � b

Sol. (A)

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Then, ax1

x and bx

1x

32

22

ax1

x

x2 + 2x

1+2 = a2 .........(i)

3

x1

x

= a3

x3 +

x1

x3x

13 = a3 .........(ii)

add equation (1) and (2)

x1

x32x

1x

x

1x

23

32

= a2 + a3

b +

23

x

1x + 2 + 3a = a2 + a3

23

x

1x = a3 + a2 � 3a � b � 2

4. Let a, b, c, d be real numbers such that |a � b| = 2, |b � c| = 3, |c � d| = 4. Then the sum of all possible values

of |a � d| is

(A) 9 (B) 18 (C) 24 (D) 30Sol. (B)

|a � b| = 2, |b � c| = 3, |c � d| = 4

a � b = ± 2

b � c = ± 3

c � d = ± 4

possible values of (a � d) are ± 9, ± 5, ± 3

|a � d| = 9, 5, 3, 1

Sum of all possible values are 18

5. Below are four equations in x. Assume that 0 < r < 4. Which of the following equations has the largestsolution for x ?

(A) 9r

15x

(B) 9

17r

15x

(C) 5 (1 + 2r)x = 9 (D) 9

r

115

x

Sol. (B)Given 0 < r < 4

in all the obtain

(Base)x = 59

the option having least base will give the largest x.

So, in option B base

17

r1 is minimum for 0 < r < 4.

RESONANCE PAGE - 4

KVPY QUESTION PAPER - STREAM (SA)Aliter :

(1 + x = 54

x log (1 + ) = log (1.8)

x = )1(log)8.1(log

For x to be maximumlog (1 + ) should be minimum

=

ror 2r or

17r

orr1

in 0 < r < 4

17r

is minimum.

6. Let ABC be a triangle with B = 90°. Let AD be the bisector of A with D on BC. Suppose AC = 6 cm and thearea of the triangle ADC is 10 cm2. Then the length of BD in cm is equal to

(A) 53

(B) 103

(C) 35

(D) 3

10

Sol. (D)Angle bisector theorem

6x

= zy

x 6

zy

10cm2 xz = 6y 20 = 6y

y = 3

10cm

ALITER :Let ACB =

BC = 6 cos , AB = 6 sin BD : CD = AB : AC

= 6 sin : 6 = sin : 1

BD =

sin

sin1cos6

.........(i)

CD = 1sin1

cos6

.........(ii)

BD = CD sin Now, area of ADC = 10 cm2

21

× 6(CD) sin = 10

CD sin = 10/3 cm

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KVPY QUESTION PAPER - STREAM (SA)7. A piece of paper in the shape of a sector of a circle (see Fig.1) is rolled up to form a right-circular cone

(see Fig. 2). The value of the angle is

(A) 13

10(B)

139

(C) 135

(D) 136

Sol. (A)

Slant height = 13 by = r

2.5 = 13

13

10

8. In the adjoining figure AB = 12 cm, CD = 8 cm, BD = 20 cm; ABD = AEC = EDC = 90°. IF BE = x, then

(A) x has two possible values whose difference is 4(B) x has two possible values whose sum is 28(C) x has only one value and x 12(D) x cannot be determined with the given information

Sol. (A)

x12

= 8

x20

x2 � 20x + 96 = 0

x = 8, 12ALITER :

B(0,0)

C(20,8)

E(x,0) D(20,0)

A(0,12)

AC = 22 420 = 154 2 = 264

(122 + x2) + (82 + (20 � x)2) = 202 + 42

2x2 � 40x + 400 + (122 � 202) + (82 � 42) = 02x2 � 40x + 144 + 12.4 = 0

RESONANCE PAGE - 6

KVPY QUESTION PAPER - STREAM (SA)x2 � 20x + 72 + 24 = 0

x2 � 20x + 96 = 0

x = 12, 8

9. Three circles each of radius 1 touch one another externally and they lie between two parallel lines. Theminimum possible distance between the lines is

(A) 32 (B) 33 (C) 4 (D) 3

12

Sol. (A)

AP = 2 sin 60° = 3

d = 32

10. The number of distinct prime divisors of the number 5123 � 2533 � 2593 is(A) 4 (B) 5 (C) 6 (D) 7

Sol. (C)5123 � (2533 + 2593)

= 5123 � [(512) (2532 + 2592 � 253.259)]

= 512 (5122 � ((512)2 � 3(253)(259))

= 512 (3.253 � 259)

= 29. 3. 253 . 7.37= 29. 3. (11) . (23). 7.37

So, number of distinct prime divisors are 6.

11. Consider an incomplete pyramid of balls on a square base having 18 layers; and having 13 balls on eachside of the top layer. Then the total number N of balls in that pyramid satisfies(A) 9000 < N < 10000 (B) 8000 < N < 9000 (C) 7000 < N < 8000 (D) 10000 < N < 12000

Sol. (B)Top layer has (13 × 13) balls

Simillary one layer below top layer will have (14 × 14) balls and we have 18 lesens to total number of ball

N = (13)2 + (14)2 + .............+ (30)2

N = 6

613130 �

6251312

N = 8805

12. A man wants to reach a certain destination. One sixth of the total distance is muddy while half the distanceis tar road. For the remaining distance he takes a boat. His speed of traveling in mud, in water on tar roaedis in the ratio 3 : 4 : 5. The ratio of the durations he requires to cross the patch of mud. stream and tar roadis:

(A) 25

:54

:21

(B) 3 : 8 : 15 (C) 10 : 15 : 18 (D) 1 : 2 : 3

RESONANCE PAGE - 7

KVPY QUESTION PAPER - STREAM (SA)Sol. (C)

Let distance is 6dMud : Tar : Stream

Distance d : 3d : 2dSpeed 3V : 5V : 4V

timeV3d

:V5d3

:V4d2

10 : 18 : 15 (Note : order is changed in questions)

13. A frog is presently located at the origin (0, 0) in the xy-plane. It always jumps from a point with integercoordinates to a point with integer coordinates moving a distance of 5 units in each jump. What is theminimum number of jumps required for the frog to go from (0, 0) to (0, 1) ?(A) 2 (B) 3 (C) 4 (D) 9

Sol. (B)3 step

14. A certain 12-hour digital clock displays the hour and the minute of a day. Due to a defect in the clockwhenever the digit 1 is supposed to be displayed it displays 7. What fraction of the day will the clock showthe correct time ?

(A) 21

(B) 85

(C) 43

(D) 65

Sol. (B) The clock will show the incorrect time (between 1 - 2, 11 - 12 day and night both)Therefore incorrect time 4 × 60 = 240.

Remaining 20 hour it will show the incorrect time 20 × 15 = 300

Total incorrect time = 240 + 300 = 540

the fraction day when the clock shows correct time is = 1 � 6024

540

= 1 � 249

= 1 � 83

= 85

(Note: 12-hour Digital clock show first digit only 0 and 1)

RESONANCE PAGE - 8

KVPY QUESTION PAPER - STREAM (SA)15. There are 30 questions in a multiple-choice test. A student gets 1 mark for each unattempted question, 0

mark for each wrong answer and 4 marks for each correct answer. A student answered x questions correctlyand scored 60. Then the number of possible value of x is(A) 15 (B) 10 (C) 6 (D) 5

Sol. (C)

Right Wrong Unattempted 15 15 0 14 4 13 8 12 12 11 16 10 20

6 cases only

PHYSICS16. A simple pendulum oscillates freely between points A and B.

We now put a peg (nail) at some point C as shown. As the pendulum moves from A to the right, the string willbend at C and the pendulum will go to its extreme point D. Ignoring friction, the point D.(A) will lie on the line AB (B) will lie above the line AB(C) will lie below the line AB (D) will coincide with B

Sol. (A)By mechanical energy conservationKE

i + U

i = KE

f + U

f

O + Ui = 0 + U

f

Ui = U

f

hi = h

f

So D will lie on line AB

17. A small child tries to move a large rubber toy placed on the ground. The toy does not move but gets deformed

under her pushing force )F(

which is obliquely upward as shown . Then

(A) The resultant of the pushing force )F(

, weight of the toy, normal force by the ground on the toy and

the frictional force is zero.(B) The normal force by the ground is equal and oipposite to the weight of the toy.

(C) The pushing force )F(

of the child is balanced by the equal and opposite frictional force

(D) The pushing force )F(

of hte child is ballanced by the total internal force in the toy generated due to

deformationSol. (A)

Since toy is not accelerating so net external force on toy is zero. So (A)

RESONANCE PAGE - 9

KVPY QUESTION PAPER - STREAM (SA)18. A juggler tosses a ball up in the air with initial speed u. At the instant it reaches its maximum height H, he

tosses up a second ball with the same initial speed. The two balls will collide at a height.

(A) 4H

(B) 2H

(C) 4H3

(D) H43

Sol. (C)|S

1| + |S

2| = H

2gt21

+ ut � 2gt

21

= H

ut = H . ....(I)

H = g2

u2....(II)

t = g2u

S2 = ut �

2gt21

= g8u3

g4

ug

21

g2u

42

2

2

= H43

19. On a horizontal frictional frozen lake, a girl (36 kg) and a box (9kg) are connected to each other by means ofa rope. Initially they are 20 m apart. The girl exerts a horizontal force on the box, pulling it towards her. Howfar has the girl travelled when she meets the box ?(A) 10 m (B) Since there is no friction, the girl will not move(C) 16 m (D) 4m

Sol. (D)

By concept of centre of mass36x = 9(20�x)

36x = 180 � 9x

45x = 180x = 4m

20. The following three objects (1) a metal tray, (2) a block of wood, and (3) a wooden cap are left in a closedroom overnight. Next day the temperature of each is recorded as T

1 , T

2 and T

3 respectively. The likely

situation is(A) T

1 = T

2 = T

3(B) T

3 > T

2 > T

1(C) T

3 = T

2 > T

1(D) T

3 > T

2 = T

1

Sol. (A)All three will be in thermal equilibrium with air of room. so temperature of the three will be same

21. We sit in the room with windows open. Then(A) Air pressure on the floor of the room equals the atmospheric pressure but the air pressure on the

ceiling is negligible(B) Air pressure is nearly the same on the floor, the walls and ceiling(C) Air pressure on the floor equals the weight of the air coloumn inside the room (from floor to ceiling)

per unit area(D) Air pressure on the walls is zero since the weight of air acts downward

Sol. (B)Pressure of gas is same everywhere in the vessel.

RESONANCE PAGE - 10

KVPY QUESTION PAPER - STREAM (SA)22. A girl standing at point P on a beach wishes to reach a point Q in the sea as quickly as possible. She can

run at 6 kmh�1 on the beach and swim at 4 km-h�1 in the sea. She should take the path

(A) PAQ (B) PBQ (C) PCQ (D) PDQSol. (C)

To travel from P to Q in minimum time, she should travel on path PCQ.

23. Light enters an isosceles right triangular prism at normal incidence through face AB and undergoes totalinternal reflection at face BC as shown below :

The minimum value of the refractive index of the prifsm is close to :(A) 1.10 (B) 1.55 (C) 1.42 (D) 1.72

Sol. (C)i = 45° CC = 45° for minimum

sin 45 = 1

= 2 = 1.42

24. A convex lens is used to form an image of an object on a screen. If the upper half of the lens is blackened sothat it becomes opaque. Then :(A) Only half of the image will be visible (B) the image position shifts towards the lens(C) the image position shifts away from the lens (D) the brightness of the image reduces

Sol. (D)

In this case only half part of lens is used to form the image so intensity will reduce

25. A cylindrical copper rod has length L and resistance R. If it is melted and formed into another rod of length 2L.the resistance will be :(A) R (B) 2R (C) 4R (D) 8R

Sol. (C)

R = A

R =

A

RESONANCE PAGE - 11


R = V

2

R� = V

)2( 2

= 4R

26. Two charges +Q and _2Q are located at points A and B on a horizontal line as shown below :

The electric field is zero at a point which is located at a finite distance :(A) On the perpendicular bisector of AB (B) left of A on the line(C) between A and B on the line (D) right of B on the line

Sol. (B)

E1 = Electric field due to +Q

E2 = Electric field due to �2Q

There resultant is 0 at this point

27. A 750 W motor drives a pump which lifts 300 litres of water per minute to a height of 6 meters. The efficiencyof the motor is nearly (take acceleration due to gravity to be 10 m/s2)(A) 30% (B) 40% (C) 50% (D) 20%

Sol. (B)

= 100750

t/mgh

= 10075060

610300

= 100750300

= 40%

28. Figure below shows a portion of an electric circuit with the currents in ampreres and their directions. Themagnitude and direction of the current in the portion PQ is :

(A) 0A (B) 3A from P to Q (C) 4A from Q to P (D) 6A from Q to P

RESONANCE PAGE - 12

KVPY QUESTION PAPER - STREAM (SA)Sol. (D)

6A from Q to P

29. A nucleus of lead 31482pb emits two electrons followed by an alpha particle. The resulting nucleus will have

(A) 82 protons and 128 neutrons (B) 80 protons and 130 neutrons(C) 82 protons and 130 neutrons (D) 78 protons and 134 neutrons

Sol. (A)31482pb 2e�1 + 4

2He + 21082X

So 82 proton and 128 Neutron

30. The number of air molecules in a (5m × 5m × 4m) room at standard temperature and pressure is of the order

of(A) 6 × 1023 (B) 3 × 1024 (C) 3 × 1027 (D) 6 × 1030

Sol. (C)PV = NKT105 × 100 = N × 1.38 × 10�23 × 273

N 3 × 1027

CHEMISTRY

31. Two balloons A and B containing 0.2 mole and 0.1 mole of helium at room temperature and 2.0 atm.respectively, are connected. When equilibrium is established, the final pressure of He in the system is(A) 0.1 atm (B)1.5 atm(C) 0.5 atm (D) 2.0 atm

Sol. (D)Since pressures of the gases are same in both the containers. So, the final pressure will not change

32. In the following set of aromatic compounds

(i) (ii) (iii) (iv)

the correct order of reactivity toward friedel-crafts alkylations is(A) i > ii > iv (B) ii > iv > iii > i(C) iv > iii >i (D) iii > i > iv > ii

Sol. (C)Reativity towards Friedel-Crafts alkylation is proportional to electron density in the benzene ring.

RESONANCE PAGE - 13


33. The set of principal (n). azimuthal (l) and magnetic (m1) quantum numbers that is not allowed for the

electron in H-atom is(A) n = 3, l = 1, m

= -1 (B) n = 3, l = 0, m

= 0

(D) n = 2, l = 1, m

= 0 (D) n = 2, l = 2, m = -1

Sol. (D)n = 2 = 0, 1

m

= 0,

1

1

0

34. At 298 K, assuming ideal behaviour, the average kinetic energy of a deuterium molecule is :(A) two times that of a hydrogen molecule (B) four times that of a hydrogen molecule(C) half of that of a hydrogen molecule (D) same as that of a hydrogen molecule

Sol. (D)Average Kinetic Energy depends only on temperature

(KE)avg

= 2kT3

per molecule

35. An isolated box, equally partitioned contains two ideal gases A and B as shown

When the partition is removed, the gases mix. The changes in enthalpy (H) and entropy (S) in theprocess, respectively, are :(A) zero, positive (B) zero, negative (C) positive, zero (D) negative, zero

Sol. (A)Ideal gas H = 0

S > 0(randomness increases)

36. The gas produced from thermal decomposition of (NH4)

2Cr

2O

7 is :

(A) oxygen (B) nitric oxide (C) ammonia (D) nitrogenSol. (D)

(NH4)

2 Cr

2O

7

Cr2O

3 + N

2 + 4H

2O

37. The solubility curve of KNO3 in water is shown below.

The amount of KNO3 that dissolves in 50 g of water at 40°C is closest to :

(A) 100 g (B) 150 g(C) 200 g (D) 50 g

RESONANCE PAGE - 14


Sol. (A)According to the graph solubility 40° is approx. 200 g per 100 ml. For 50 ml, amount is 100 g approx.

38. A compound that shows positive iodoform test is :(A) 2-pentanone (B) 3-pentanone (C) 3-pentanol (D) 1-pentanol

Sol. (A)

Aldehyde, ketones with acetyl group CH3 � � show Iodoform test.

39. After 2 hours the amount of a certain radioactive substance reduces to 1/16th of the original amount (thedecay process follows first-order kinetics). The half-life of the radioactive substance is :(A) 15 min (B) 30 min (C) 45 min (D) 60 min

Sol. (B)

161

2

1n n, number of half-lives = 4 = 2 hrs. half-life = 30 min.

40. In the conversion of a zinc ore to zinc metal, the process of roasting involves.(A) ZnCO

3 ZnO (B) ZnO ZnSO

4(C) ZnS ZnO (D) ZnS ZnSO

4

Sol. (D)ZnS ZnSO

4 during roasting, sulphide ore is converted into sulphate.

41. The number of P-H bond(s) in H3PO

2, H

3PO

3 and H

3PO

4, respectively, is :

(A) 2, 0, 1 (B) 1, 1, 1(C) 2, 0, 0 (D) 2, 1, 0

Sol. (D)

42. When chlorine gas is passed through an aqueous solution of KBr, the solution turns orange brown dueto the formation of :(A) KCl (B) HCl (C) HBr (D) Br

2

Sol. (D)Cl

2 + 2KBr 2KCl + Br2

reddish brown

43. Among

the compound which is not aromatic is :(A) i (B) ii (C) iii (D) iv

Sol. (B)(i) and (iv) are hetro aromatic and the resonance form of azulene (iii) is aromatic.

(ii) Is nonaromatic.

RESONANCE PAGE - 15


44. Among the following compounds :

(i) (ii) (iii) (iv)

2, 3�dimethylhexane is :

(A) i (B) ii (C) iii (D) ivSol. (B)

Simple nomenclature of alkane.

45. The major product formed in the reaction :

Product is :

(A) i (B) ii(C) iii (D) iv

Sol. (C)The given reaction is S

N2 which occurs at sp3 carbon with good leaving group.

BIOLOGY46. If parents have free ear lobes and the offspring has attached ear lobes, then the parents must be

(A) homozygous (B) heterozygous(B) co-dominant (D) nullizygous

Sol. (B)heterozygous

47. During meiosis there is(A) one round of DNA replication and one division(B) two round of DNA replication and one division(C) two round of DNA replication and two division(D) one round of DNA replication and two division

Sol. (D)one round of DNA replication and two division

48. Blood clotting involves the conversion of(A) prothrombin to thromboplastin (B) thromboplastin to prothrombin(C) fibrinogen to fibrin (D) fibrin to fibrinogen

Sol. (C)fibrinogen to fibrin

49. The gall bladder is involved in(A) synthesizing bile (B) storing and secreting bile(C) degrading bile (D) producing insulin

Sol. (B)Storing and secreting bile

RESONANCE PAGE - 16

KVPY QUESTION PAPER - STREAM (SA)50. Which one of the following colors is the LEAST useful for plant life ?

(A) red (B) blue (C) green (C) violetSol. (C)

Green

51. At rest the volume of air that moves in and out per breath is called(A) resting volume (B) vital capacity (C) lung capacity (D) tidal volume

Ans. (D)

52. How many sex chromosomes does a normal human inherit from father ?(A) 1 (B) 2 (C) 23 (D) 46

Sol. (A)1

53. In the 16th century, sailors who travelled long distances had diseases related to malnutrition, becausethey were not able to eat fresh vegetables and fruits for months at a time scurvy is a result of deficiencyof(A) carbohydrates (B) proteins (C) vitamin C (D) vitamin D

Sol. (C)vitamin C

54. The following structure is NOT found in plant cells(A) vacuole (B) nucleus (C) centriole (D) vitamin D

Sol. (C)centriole

55. The cell that transfers information about pain to the brain is called a(A) neuron (B) blastocyst (C) histoblast (D) vitamins

Sol. (A)Neuron

56. The presence of nutrient sin the food can be tested. Benedict's's test is used to detect.(A) sucrose (B) glucose (C) fatty acid (D) vitamins

Sol. (B)Glucose

57. Several mineral such as iron, iodine, calcium and phosphorous are important nutrients. Iodine is foundin(A) thyroxine (B) adrenaline (C) insulin (D) testosterone

Sol. (A)thyroxine

58. The principle upon which a lactometer works is(A) viscosity (B) density (C) surface tension (D) presence of protein

Sol. (B)Density

59. Mammalian liver cells will swell when kept in(A) hypertonic solution (B) hypotonic solutions(C) isotonic solution (C) isothermal solutions

Ans. (B)Hypotonic solutions

60. The form of cancer called 'carcinoma' is associated with(A) lymph cells (B) mesodermal cells(C) blood cells (D) epithelial cells

Ans. (D)Epithelial cells

RESONANCE PAGE - 17


PART-IITwo Mark Questions

MATHEMATICS

61. Let f(x) = ax2 + bx + c, where a, b, c are integers. Suppose f(1) = 0, 40 < f(6) < 50, 60 < f(7) < 70, and 1000t< f(50) < 1000 (t + 1) for some integer t. Then the value of t is

(A) 2 (B) 3 (C) 4 (D) 5 or more

Sol. (C)

f(x) = ax2 + bx + c, f(1) = 0

given, f(1) = 0

a + b + c = 0

and 40 < f(6) < 50

40 < 36a + 6b + c < 50

40 < 35a + 5b < 50

8 < 7a + b < 10

7a + b = integer = 9 .......(1)

and 60 < f(7) < 70

60 < 49a + 7b + c < 70

60 < 48a + 6b < 70

10 < 8a + b < 11.6

8a + b = 11 .......(2)

solving equation (1) and (2)

a = 2, b = �5, so c = 3

f(x) = 2x2 � 5x + 3

f(50) = 4753

t = 4

62. The expression

1)2011(

1)2011(.............

14

14

13

13

12

122

2

2

2

2

2

2

2

lies in the interval

(A) (2010, 2010 21

) (B)

2012

12011,

2011

12011

(C) (2011, 2011 21

) (D) (2012, 2012 21

)

Sol. (C)

We can write the expression as

2011

2r 2

2

1r

1r =

1r1r2

1

=

1r

1

1r

11

RESONANCE PAGE - 18


Putting r = 2, 3, ............, 2011

= 2010 + 1 + 2011

12012

121

= 2011 +

20121

20111

21

this lies between (2011, 2011 21

)

63. The diameter of one of the bases of a truncated cone is 100 mm. If the diameter of this base is increased by21% such that it still remains a truncated cone with the height and the other base unchanged, the volumealso increases by 21%. The radius the other base (in mm) is

(A) 65 (B) 55 (C) 45 (D) 35

Sol. (B)

Let initially 2 bases have radii 5 & r. and finally bases have radii (1.21 × 5) & r.

Ratios of volumes = 21.1VV

1

2

V2 =

3h

((6.05)2 + 6.05 r + r2)

V1 =

3h

(52 + 5r + r2)

21.1VV

1

2 21.1rr55

rr05.6)05.6(22

22

36.6025 + 6.05 r + r2 = 30.25 + 6.05r + 1.21 r2

.21r2 = 6.3525

r2 = 21.

3525.6

r = 211

cm. = 55 mm

64. Two friends A and B are 30 km apart and they start simultaneously on motorcycles to meet each other. Thespeed of A is 3 times that of B. The distance between them decreases at the rate of 2 km per minute. Tenminutes after they start A's vehicle breaks down and A stops and waits for B to arrive. After how much time(in minutes) A started riding, does B meet A ?

(A) 15 (B) 20 (C) 25 (D) 30

Sol. (B)

Let speed of B = V km/hr.

Let speed of A = 3V km/hr.

Given 4r = 2 × 60 km/hr V = 30 km/hr

Distance covered by then after 10 min. = 2 × 10 = 20 km

So, remaining distance = (30 � 20) km = 10 km.

Time table by B to cover 10 km = 3010

= 20 min

RESONANCE PAGE - 19


65. There taps A, B, C fill up a tank independently in 10 hr, 20 hr, 30 hr, respectively. Initially the tank is emptyand exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour.What is the minimum number of hours required to fill the tank ?

(A) 8 (B) 9 (C) 10 (D) 11

Sol. (A)A B C

10 hr 20 hr 30 hr

Exactly one pair of taps is open during each hour and every pair of taps is open at least for one hour.

So, first we say, A and B are open for 1 hour, then B & C and then C & A.

201

101

+

301

201

+

101

301

= 6022

First then second then Third then

In three hours the tank will be filled th

60

22

part

Now, for minimum time, the rest tank must be filled with A and B taps.

609

201

101

So, the rest th

60

38

Part of tank will take 5 hour more

So, the tank will be filled in 8th hour.

PHYSICS

66. An object with uniform desity is attached to a spring that is known to stretch linearly with applied force asshown below

When the spring object system is immersed in a liquid of density 1 as shown in the figure, the spring

stretches by an amount x1 ( >

1). When the experiment is repeated in a liquid of density

2 <

1 . the spring

strethces by an amound x2. Neglecting any buoyant force on the spring, the density of the object is :

(A) 21

2211xx

xx

(B)

12

1221xx

xx

(C)

21

1221xx

xx

(D)

21

2211xx

xx

Sol. (B)

RESONANCE PAGE - 20

KVPY QUESTION PAPER - STREAM (SA)kx

1 +

1vg = vg

v = ggkx

1

1

kx2 +

2vg = vg

v = ggkx

2

2

g)(kx

g)(kx

2

2

1

1

x1 �

x

1 = x

2 �

x

2

x1 � x

2) =

x

1 �

x

2

= 21

2112xx

xx

67. A body of 0.5 kg moves along the positive x - axis under the influence of a varying force F (in Newtons) asshown below :

If the speed of the object at x = 4m is 3.16 ms�1 then its speed at x = 8 m is :(A) 3.16 ms�1 (B) 9.3 ms�1 (C) 8 ms�1 (D) 6.8 ms�1

Sol. (B)

2i

2f

8

4

mv21

mv21

dx.F

21

× 3 × 8 � 21

× 1.5 × 4 = 21

× 0.5 (vf2 � 3.16)2

24 � 6 = 0.5 (vf2 � 3.162)

36 = vf2 � 3.162

vf 6.8 m/s

68. In a thermally isolated system. Two boxes filled with an ideal gas are connected by a valve. When the valveis in closed position, states of the box 1 and 2. respectively, are (1 atm, V, T) and (0.5 atm, 4V, T). When thevalve is opened, the final pressure of the system is approximately.(A) 0.5 atm (B) 0.6 atm (C) 0.75 atm (D) 1.0 atm

RESONANCE PAGE - 21

KVPY QUESTION PAPER - STREAM (SA)Sol. (B)

vi = v

f

25

×1 × V + 25

× 0.5 × 4v = 25

P 5V

3V = 5PV

P = 53

= 0.6

69. A student sees the top edge and the bottom centre C of a pool simultaneously from an angle above thehorizontal as shown in the figure. The refractive index of water which fills up to the top edge of the pool is 4/3. If h/x = 7/4 then cos is :

(A) 72

(B) 453

8(C)

533

8(D)

218

Sol. (C)

sin(90 � ) = sinr

cos =

4x

h

2/x

3

4

22

RESONANCE PAGE - 22


cos = 41

x

h

132

2

2 =

164

1649

1

3

2

= 53

432

= 533

8

70. In the following circuit, the 1 resistor dissipates power P. If the resistor is replaced by 9. the powerdissipated in it is

(A) P (B) 3P (C) 9P (D) P/3Sol. (A)

Pi = i2R

P = 16

1001

410 2

Pf = i2R

= 16

1009

1212100

91210

2

= P

CHEMISTRY

71. An aqueous buffer is prepared by adding 100 ml of 0.1 mol 1�1 acetic acid to 50 ml of 0.2 mol 1�1 ofsodium acetate. If pKa of acetic acid is 4.76, the pH of the buffer is :(A) 4.26 (B) 5.76 (C) 3.76 (D) 4.76

Sol. (A)100 mL 0.1 M CH

3COOH + 50 mL 0.4 M CH

3COONa

COOHCh

COOCH

3

3

= 1

pH = pKa + log 1= pKa = 4.76

RESONANCE PAGE - 23

KVPY QUESTION PAPER - STREAM (SA)72. The maximum number of stgructural isomers possible for the hydrocarbon having the molecular formuka

C4H

6, is :

(A) 12 (B) 3 (C) 9 (D) 5Sol. (C)

9-structural isomers are possible.CH

3CH

2CCH, CH

3 �CC�CH

3

CH2 =CH�CH=CH

2 , CH

2 =C=CH�CH

3

73. In the following reaction sequence, X and Y, respectively, are :

X

Y OH

(A) H2O

2 ; LiAlH

4(B) C

6H

5 COOOH ; LiAlH

4

(C) C6H

5 COOOH ; Zn/Hg HCl (D) Alkaline KMnO

4 ; LiAlH

4

Sol. (B)

74. Among (i) [Co(NH3)

6]Cl

3, (ii) [Ni(NH

3)

6]Cl

2 , (iii) [Cr(H

2O

6)]Cl

3, (iv) [Fe(H

2O)

6]Cl

2 the complex which is

diamagnetic is :(A) i (B) ii (C) iii (D) iv

Sol. (A)Co+3 : 3d6 45° strong field ligand (NH

3)

Ni+2 : 3d8 45° SFL (NH3)

Cr+3 : 3d3 45° WFL (H2O)

Fe+2 : 3d6 45° WFL (H2O)

Co+3 will be diamagnetic (i)

75. At 783 K in the reaction H2 (g) + I

2 (g) 2Hl(g), the molar concentrations (mol 1�1) of H

2, I

2 and HI at

some instant of time are 0.1, 0.2 and 0.4, respectively. If the equilibrium constant is 46 at the sametemperature, then as the reaction proceeds.(A) the amount of HI will increase (B) the amount of HI will decrease(C) the amount of H

2 and I

2 will increases (D) the amount of H

2 and I

2 will not change

Sol. (A)

Reaction quotrient Q = ][]H[]H[

22

2

=

2.01.0)4.0( 2

= 8

Q < Kreaction proceeds in the forward direction.

BIOLOGY76. You removed four fresh tobacco leaves of similar size and age. Leave "leaf 1" as it is , smear " leaf 2"

with vaseline on the upper surface, "leaf 3" on the lower surface and "leaf 4" on both the surface . Hangthe leaves for a few hours and you observe that leaf 1 wilts the most, leaf 2 has wilted. leaf 3 wilted lessthan leaf 2 and leaf 4 remains fares. Which of the following conclusion is most logical ?(A) tobacco leaf has more stomata on the upper surface(B) tobacco leaf has more stomata on the lower surface(C) stomata are equally distributed in upper and lower surface(D) no conclusion on stomatal distribution can be drawn from this experiment

Ans. (B)

RESONANCE PAGE - 24

KVPY QUESTION PAPER - STREAM (SA)77. Vestigial organs such as the appendix exist because

(A) they had and important function during development which is not needed in the adult(B) they have a redundant role to play if an organ with similar functions fails.(C) nature cannot get rid of structures that have already formed(D) they were inherited from an evolutionary ancestor in which they were functional

Ans. (D)

78. Mendel showed that unit factors, now called alleles, exhibit a dominant recessive relationship. In amonohybrid cross. the .............................trait disappears in the first filial generation(A) dominant (B) co-dominant(C) recessive (D) semi-dominant

Ans. (C)

79. If a man with an X-linked dominant disease has six sons with a woman having a normal complement ofgenes, then the sons will(A) not show any symptoms of the disease(B) show strong symptoms of the disease(C) three will show a disease symptom, while three will not(D) five will show a disease symptom, while one will no

Ans. (A)80. In evolutionary terms, an Indian school boy is more closely related to

(A) an Indian frog (B) an American snake(C) a Chinese horse (D) an African shark

Ans. (C)

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