STREAM - SB/SX - · PDF fileŁThe Test Booklet consists of 120 ... interview all the 6 persons one after the other subject to the condition that no child ... KVPY QUESTION PAPER -

GENERAL INSTRUCTIONS

� The Test Booklet consists of 120 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 20 consist of ONE (1) mark for each correct response.

PHYSICSQuestion No. 21 to 40 consist of ONE (1) mark for each correct response.

CHEMISTRYQuestion No. 41 to 60 consist of ONE (1) mark for each correct response.

BIOLOGYQuestion No. 61 to 80 consist of ONE (1) mark for each correct response.

PART-II :MATHEMATICSQuestion No. 81 to 90 consist of TWO (2) marks for each correct response.

PHYSICSQuestion No. 91 to 100 consist of TWO (2) marks for each correct response.

CHEMISTRYQuestion No. 101 to 110 consist of TWO (2) marks for each correct response.

BIOLOGYQuestion No. 111 to 120 consist of TWO (2) marks for each correct response.

Date : 04-11-2012 Duration : 3 Hours Max. Marks : 160

KISHORE VAIGYANIK PROTSAHAN YOJANA - 2012

STREAM - SB/SX

RESONANCE PAGE - 2

KVPY QUESTION PAPER - STREAM (SB / SX)

PART-IOne Mark Questions

MATHEMATICS1. Three children, each accompanied by a guardian, seek admission in a school. The principal wants to

interview all the 6 persons one after the other subject to the condition that no child is interviewed before itsguardian. In how many ways can this be done?rhu cPps] ftuesa izR;sd vius vfHkHkkod ds lkFk gS] ,d fo|ky; esa izos'k ikuk pkgrs gSA fo|ky; ds izkpk;Z ,d ds ckn ,dlHkh 6 O;fDr;ksa ls bl izdkj lk{kkRdkj djuk pkgrs gS fd fdlh Hkh cPPks dk lk{kkRdkj mlds vfHkHkkod ds igys u gksA ;gfdrus rjhds ls fd;k tk ldrk gS ?(A) 60 (B) 90 (C) 120 (D) 180

Sol. Number ways are rjhdksas dh la[;k = !2!2!2

!6 =

8720 = 90 ways

2. In the real number system, the equation 1�x4�3x + 1x68x = 1 has

(A) no solution (B) exactly two distinct solutions(C) exactly four distinct solutions (D) infinitely many solutions

okLrfod la[;k (fudk;) esa lehdj.k 1�x4�3x + 1x68x = 1 ds fy,

(A) dksbZ gy ugha gSA (B) rF;r% nks fofHkUu gy gSA(C) rF;r% pkj fofHkUu gy gSA (D) vuUr gy laHko gSA

Sol. x 3 4 x 1 = x + 8 � 6 1x + 1 � 2 1x68x

1x2 � 6 = �2 1x6�8x

1x �3 = � 1x68x

Ans. (D) Infinite many solutions.

3. The maximum value M of 3x + 5x � 9x + 15x � 25x, as x varies over reals, satisfiesO;atd 3x + 5x � 9x + 15x � 25x, tgk¡ x ,d okLrfod la[;k gS] dk vf/kdre eku (M) fufEufyf[kr dks larq"V djrk gS&(A) 3 < M < 5 (B) 0 < M < 2 (C) 9 < M < 25 (D) 5 < M < 9

Sol. Let ekuk a = 3x and rFkk b = 5x

Therefore blfy, ,3x + 5x � 9x + 15x � 25x = a + b � a2 + ab � b2

= 21

[2 � (a � 1)2 � (b � 1)2 � (a � b)2]

21

(2)

Thus vr%, M = 1 (happens when tc x = 0)Answer (B)

4. Suppose two perpendicular tangents can be drawn from the origin to the circle x2 + y2 � 6x � 2py + 17 = 0,

for some real p. Then |p| is equal to;fn fdlh okLrfod p ds fy, x2 + y2 � 6x � 2py + 17 = 0 ij ewyfcUnq ls nks ijLij yEcor~ Li'kZthok [khaph tk ldrhgS] rc |p| gksxk&(A) 0 (B) 3 (C) 5 (D) 17

RESONANCE PAGE - 3


Sol.

Equation of chord of contact T = 0 w.r.t. origin is� 3(x + 0) � p (y + 0) + 17 = 0

3x + py � 17 = 0

By Homoginization

x2 + y2 � (6x + 2py)

17pyx3

+ 17

2

17pyx3

= 0

For perpendicular coeff of x2 + coeff of y2 = 0

1 � 17

3.6 +

.917

+ 1 � 17p2 2

+ 17p2

= 0

34 � 18 + 9 � p2 = 0 p2 = 25 |p| = 5

5. Let a, b, c, d be numbers in the set {1, 2, 3, 4, 5, 6} such that the curves y = 2x3 + ax + b and y = 2x3 + cx+ d have no point in common. The maximum possible value of (a � c)2 + b � d is

;fn fdlh leqPp; {1, 2, 3, 4, 5, 6} ds in a, b, c, d bl izdkj gS fd y = 2x3 + ax + b rFkk y = 2x3 + cx + d ijdksbZ fcUnq mHk;fu"B ugha gS] rc (a � c)2 + b � d dk vf/kdre laHkkfor eku gksxk&(A) 0 (B) 5 (C) 30 (D) 36

Sol. 2x3 + ax+ b 2x3 + cx + d(a � c)x d � b

If a � c = 0 & d � b 0 Max. (a � c)2 + (b � d)

= 0 + 5 Ans.

6. Consider the conic ex2 + y2 � 2e2x � 22y + e3 + 3 = e. Suppose P is any point on the conic and S1, S

2

are the foci of the conic, then the maximum value of (PS1 + PS

2) is

'kkado (conic) ex2 + y2 � 2e2x � 22y + e3 + 3 = e ij ;fn P dksbZ fcUnq gks rFkk S1, S

2 'kkado ds ukfHkd gks] rc

(PS1 + PS

2) dk vf/kdre eku gksxk&

(A) e (B) e (C) 2 (D) e2Sol. e(x2 � 2 ex + e2) + (y2 � 2y + 2) = e

e�ye�x 22

= 1 (a > b)

Ellipse a = , b = e

Req. PS1 + PS

2 = 2

7. Let f(x) = )axcos(�)axcos()axsin()axsin(

, then

(A) f(x + 2) = f(x) but f(x + ) f(x) for any 0 < < 2 (B) f is a strictly increasing function(C) f is strictly decreasing function (D) f is a constant function

;fn f(x) = )axcos(�)axcos()axsin()axsin(

, rc

(A) f(x + 2) = f(x) ijUrq fdlh Hkh 0 < < 2ds fy, f(x + ) f(x)

(B) f vfuok;Zr% ,d o/kZeku Qyu gSA(C) f vfuok;Zr% ,d á leku Qyu gSA

RESONANCE PAGE - 4


(D) f ,d fu;r Qyu gSA

Sol. f(x) = asinxsin2acos.xsin2

(x n, n I)

= cot a(D) = constant function

8. The value of tan 81 � tan 63 � tan27 + tan9 isO;atd tan 81 � tan 63 � tan27 + tan9 dk eku gS&(A) 0 (B) 2 (C) 3 (D) 4

Sol. tan81 � tan63 � tan27 + tan9cot9 � cot27 � tan27 + tan9(cot9 + tan9) � (cot27 + tan 27)2 cosec18 � 2 cosec54

2

15

4�

15

4

8

42

= 4Ans. (D)

9. The mid�point of the domain of the function f(x) = 5x24 for real x is

okLrfod x ds fy, Qyu f(x) = 5x24 ds izkUr dk e/; fcUnq gS&

(A) 41

(B) 23

(C) 32

(D) 52

Sol. 4 � 5x2 0, 2x + 5 0

0 2x + 5 16�5 2x 11

25

x 211

mid point is 2

211

25

= 2.2

6. =

23

Ans. (B)

10. Let n be a natural number and let a be a real number. The number of zeros of x2n+1 � (2n + 1) x + a = 0 in the

interval [�1, 1] is

(A) 2 if a > 0(B) 2 if a < 0(C) at most one for every vale of a(D) at least three for every value of aeku ysa fd n ,d izkd r la[;k gS rFkk a ,d okLrfod la[;k gSA lehdj.k x2n+1 � (2n + 1) x + a = 0 ds 'kwU;ksa dh la[;kvUrjky [�1, 1] esa gS&(A) 2 ;fn a > 0

(B) 2 ;fn a < 0

(C) izR;sd a ds eku ds fy, vf/kd ls vf/kd ,d(D) izR;sd a ds eku ds fy, de ls de rhu

Sol. f(x) = x2n +1 � (2n + 1) x + a

f (x) = (2n + 1)(x2n � 1)

x = �1 point of local maximax = �1 point of local minima

RESONANCE PAGE - 5


graph of y = f(x) when a = 0f(x) is decreasing on [�1, 1]. Thus, f(x) will have atmost one root on [�1,1] that too will happen if f(�1) 0 andf(1) 0 i.e. for a [�2n, 2n]

Answer (C)

11. Let f : R R be the function f(x) = (x � a1) (x � a

2) + (x � a

2) (x � a

3)+ (x � x

3) (x �x

1) with a

1, a

2, a

3 R. Then

f(x) 0 if and only if(A) at least two of a

1, a

2, a

3 are equal (B) a

1 = a

2 = a

3

(C) a1, a

2, a

3 are all distinct (D) a

1, a

2, a

3 are all positive and distinct

;fn f : R R ,d Qyu f(x) = (x � a1) (x � a

2) + (x � a

2) (x � a

3)+ (x � x

3) (x �x

1) tgk¡ a

1, a

2, a

3 R rc

f(x) 0 ;fn vkSj dsoy ;fn(A) a

1, a

2, a

3 esa de ls de nks cjkcj gSA (B) a

1 = a

2 = a

3

(C) a1, a

2, a

3 rhuksa fofHkUu gSA (D) a

1, a

2, a

3 rhuksa /kukRed vkSj fofHkUu gSA

Sol. f(x) = 3x2 � 2(a1 + a

2 + a

3) x + a

1a

2 + a

2a

3 + a

3a

1 0

D 04(a

1 + a

2 + a

3)2 � 4.3.(a

1a

2 + a

2a

3 + a

3a

1) 0

21a + 2

2a + 23a � a

1a

2 � a

2a

3 � a

3a

1 0

21

[(a1 � a

2)2 + (a

2 � a

3)2 + (a

3 � a

1)2] 0

a1 = a

2 = a

3

Ans. (B)

12. The value

/ 22 1

0/ 2

2 1

0

(sinx) dx

(sin x) dx

is

/ 22 1

0/ 2

2 1

0

(sinx) dx

(sin x) dx

dk eku gS&

(A) 1�2

12 (B)

12

12

(C)

2

12 (D) 2�2

Sol. I1 =

2/

0

2)x(sin (sinx)�1 dx

I1 =

2/

0

2)x)(sinx(cos

+ 2

122/

0

2 )x(sin)x(cos

dx

I1 = 2

dxxsindx)x(sin2/

0

122/

0

12

RESONANCE PAGE - 6


I1 = 2 (I

2 � I

1) Here I

1 =

2/

0

12)x(sin dx

2

1

=

)12(

2

×

)12(

12

= 22 Ans.

13. The value

2012

2012�

53 dx)1x)x(sin( is

2012

2012�

53 dx)1x)x(sin( dk eku gS&

(A) 2012 (B) 2013 (C) 0 (D) 4024

Sol.

2012

2012�

53 dx)xx(sin + 2012

2012�

dx.1

= 0 + 2012 � (� 2012)

= 4024 Ans.

14. Let [x] and {x} be the integer part and fractional part of a real number x respectively. The value of the integral

5

0

dx}x{]x[ is

;fn [x] vkSj {x}, ,d okLrfod la[;k x ds Øe'k% iw.kk±d rFkk fHkUukRed va'k gSA rc lekdy 5

0

dx}x{]x[ dk eku gksxk&

(A) 5/2 (B) 5 (C) 34.5 (D) 35.5

Sol. 5

0

dx}x]{x[

= dx}x{.01

0 +

2

1

dx}x{.1. + 3

2

dx}x{2 + 4

3

dx}x{3 + 5

4

dx}x{.4

= 0 + 1 1

0

dx.x + 2

1

0

3dx.x

1

0

4xdx . 1

0

xdx

= (1 + 2 + 3 + 4) 1

0

dx.x

= 10 .

1

0

2

2x

= 10 .

0

21

= 5

Ans. (B)

RESONANCE PAGE - 7


15. Let Sn =

n

1kk denote the sum of the first n positive integers. The numbers S

1, S

2, S

3, ....S

99 are written

on 99 cards. The probability of drawing a cards with an even number written on it is

eku ysa Sn =

n

1kk izFke n /kukRed iw.kk±dksa ds ;ksx dks fu:fir djrk gSA la[;k,a S

1, S

2, S

3,...S

99 dks 99 i=kksa ij fy[kk

x;k gSA blesa ls ,d i=k ftlij le la[;k vafdr gks] ds [khapus dh izkf;drk gksxh&

(A) 21

(B) 10049

(C) 9949

(D) 9948

Sol. Sn

n

1k

K =

21nn

is even (n : 1, 2, ............. 99)

n = multiple of 44, 8, .......... 96Or (=) n is (multiple of 4 � 1)

3, 7 ............... 99Total favourable cases 24 + 25 = 49

P[E] = 9949

Ans.

16. A purse contains 4 copper coins and 3 silver coins. A second purse contains 6 coins and 4 silver coins. Apurse is chosen randomly and a coin is taken out of it. What is the probability that it is a copper coin?,d FkSyh esa 4 rkacs vkSj 3 pkanh ds flDds gSA ,d nwljh FkSyh esa 6 rkacs vkSj 4 pkanh ds flDds gSA bu nks FkSfy;ksa esa ls ,dFkSyh ;knfPNd fudkyh tkrh gS vkSj mlesa ls ,d flDdk fudkyk tkrk gSA bl ckr dh izkf;drk D;k gS fd flDdk rkacsdk gS ?

(A) 7041

(B) 7031

(C) 7027

(D) 31

Ans. (A)

Sol. req. prob =

106

74

21

=7041

704240

21

Ans.

17. Let H be the orthocentre of an acute�angled triangle ABC and O be its circumcenter. Then HCHBHA

(A) is equal to HO (B) is equal to HO3

(C) is equal to HO2 (D) is not a scalar multiple of HO in general

;fn H ,d fuEudks.k f=kHkqt ABC dk yEc dsUnz gS ,oa O bldk ifjdsUnz gSA rc HCHBHA

(A) HO ds cjkcj gksxkA (B) HO3 ds cjkcj gksxkA

(C) HO2 ds cjkcj gksxkA (D) lkekU;r% HO dk vfn'k xq.ku ugha gSA

Sol.

OH = cba

HA = � )cb(

RESONANCE PAGE - 8


HB = � )ac(

HC = � )ba(

HCHBHA = 2( cba

)

= � 2 OH = 2HOAns. (C)

18. The number of ordered pairs (m,n) where m, n {1, 2, 3, ......, 50}, such that 6m + 9n is a multiple of 5 isØfer ;qXeksa (m,n) tgk¡ m, n {1, 2, 3, ......, 50} bl izdkj gS fd 6m + 9n, 5 dk xq.kt gSA bl izdkj ds Øfer ;qXeksadh la[;k gksxh&(A) 1250 (B) 2500 (C) 625 (D) 500

Sol. 6m + 9n

Unit digit of 6m is = 6Unit digit of 9n will be = 9 or 1For multiple of 5 unit digit of 9n must be = 9It occur when n = oddTotal number of ordered pair = 50 × 25 = 1250

19. Suppose a1, a

2, a

3, ....., a

2012 are integers arranged on a circle. Each number is equal to the average of its

two adjacent numbers. If the sum of all even indeced numbers is 3018, what is the sum of all numbers?eku ysa fd iw.kk�±d a

1, a

2, a

3, ....., a

2012 dks ,d oÙk ij lt;k x;k gSA izR;sd la[;k viuh fudVorhZ nks la[;kvksa dh vkSlr

gSA ;fn izR;sd le lwpd la[;k dk ;ksx 3018 gks] rks lHkh la[;kvksa dk ;ksx D;k gksxk ?(A) 0 (B) 1509 (C) 3018 (D) 6036

Sol. a2 =

2

aa 31

a3 =

2aa 42

a1 =

2

aa 20122

a2012

= 2

aa 12011

Now a2 + a

4 + ......... + a

2012 = 3018 .................... (1)

2a2 + 2a

4 + .................... + 2a

2012 = 6036

a1 + a

3 + a

3 + a

5 + .............. + a

2011 + a

1 = 6036

2(a1 + a

3 + ................. + a

2011) = 6036

a1 + a

3 + ................ + a

2011 = 3018 ..................... (2)

By adding (1) and (2) we geta

1 + a

2 + a

3 + ............ + a

2012 = 6036

20. Let S = 1, 2, 3, ...., n} and A = {(a, b) | 1 a, b n} = S S . A subset B of A is said to be a good subset if(x, x) B for every x S. Then the number of good subsets of A is;fn S = 1, 2, 3, ...., n} rFkk A = {(a, b) | 1 a, b n} = S S . A ds mileqPp; B dks vPNk mileqPp; rc dgktk,xk tc izR;sd x S ds fy, (x, x) B gksA rc A ds vPNs mileqPp;ksa dh la[;k gksxh&

(A) 1 (B) 2n (C) 2n(n�1) (D) 2n2

Sol. Number of element in B = n × n = n2

Number of elements of type (x, x) = n after choosing n elements.We can choose any number of elements from remaining n2 � n elements.

Number of subsets = n�n

n�n2

n�n1

n�n0

n�n2

2222C......CCC

= n�n22

RESONANCE PAGE - 9


PHYSICS21. An ideal monatomic gas expands to twice its volume. If the process is isothermal, the magnitude of work

done by the gas is W i . If the process is adiabatic the magnitude of work done by the gas is Wa . Which ofthe following is true

,d ,dijek.kfod vkn'kZ xSl izlkfjr gksdj nks xquh vk;ru dh gks tkrh gSA ;fn izfØ;k lerkih; gS rks xSl }kjk fd;k

x;k dk;Z W i gSA ;fn izfØ;k :)ks"e gS rks xSl }kjk fd;k x;k dk;Z Wa gSA fuEufyf[kr esa dkSu lgh gS \(A) W i = Wa > 0 (B) W i > Wa > 0 (C) W i > Wa = 0 (D) Wa > W i = 0

Sol.

Since area of PV graph under isothermal curve is greater than area under adiabatic curveSo, w

i > w

a > 0

Ans. (B)

22. The capacitor of capacitance C in the circuit shown is fully charged initially, Resistance is R.

ifjiFk esa n'kkZ, C /kkfjrk dk ,d la/kkfj=k izkjEHk ls iw.kZ :i ls vkosf'kr gSA R çfrjks/k gSA

After the switch S is closed, the time taken to reduce the stored energy in the capacitor to half its initial valueis :

fLop (S) ds can gksus ds ckn la/kkfj=k esa lafpr Å tkZ fdrus le; esa izkjfEHkd eku dh vk/kh gks tk;sxh\

(A) 2

RC(B) RC ln 2 (C) 2RC ln 2 (D)

22lnRC

Sol. Q = Q0e�t/RC

2

Q0 = Q

0e�t/RC

2

1 = e�t/RC Ans. (D)

RCt

2

1n

2nRC

t

t = RC 22n

RESONANCE PAGE - 10


23. A liquid drop placed on a horizontal plane has a near spherical shape (slightly flattened due to gravity). Let Rbe the radius of its largest horizontal section. A small disturbance causes the drop to vibrate with frequency

about its equilibrium shape. By dimensional analysis the ratio

3R

c

can be (Here is surface tension,

is density, g is acceleration due to gravity, and k is an arbitrary dimensionless constant)

,d {kSfrt ry ij fLFkr nzo dh ,d cwan yxHkx xksykdkj vkÑfr dh gS ¼xq:Rokd"kZ.k ds dkj.k cwan FkksM+h lh QSy tkrh gS½

xksys ds lcls cM+s {kSfrt vuqçLFk dh f=kT;k R gSA ,d NksVk fo{kksHk cwan dks blds lkE;koLFkk vkÑfr ds ifjr% vkofÙk

ij daiu djkrk gSA foeh; fo'ys"k.k ls

3R

c

vuqikr fuEufyf[kr esa dkSulk gks ldrk gS\ ¼;gk¡ i"B ruko] ?kuRo]

g xq:Roh; Roj.k vkSj k ,d ;knPN foek jfgr fLFkjkad gSA½

(A)

2gRk(B)

gRk 3

(C)

gRk 2

(D)

gk

Sol. 3R/

=

3R

=

2/1

2

331

MT

LMLT

= T�1 T

= 1

KR/ 3

23

2 KR

K2 = 23

TR

= 22

RTR

K = gR2

Ans. (A)

24. Seven identical coins are rigidly arranged on a flat table in the pattern shown below so that each coin touchesits neighbours. Each coin is a thin disc of mass m and radius r. Note that the moment of inertia of an

individual coin about an axis passing through center and perpendicular to the plane of the coin is 2

mr 2

.

,d {kSfrt est ij leku vkdkj ds lkr flDds n<+rkiwoZd bl rjg ls O;ofLFkr gS fd çR;sd flDdk vius utnhdh flDdksa

dks Nwrk gSA çR;sd flDds dk nzO;eku m vkSj f=kT;k r gSA /;ku nsa fd gj flDds dk ¼flDds ds dsUnz ls xqtjrh gqbZ vkSj flDds

ds ry ds yEcor v{k ds lkis{k½ tM+Ro vk?kw.kZ 2

mr 2

gSA

RESONANCE PAGE - 11


The moment of inertia of the system of seven coins about an axis that passes through the point P (the centreof the coin positioned directly to the right of the central coin) and perpendicular to the plane of the coins is

n'kkZ, x,s fp=k esa bu lkrksa flDdksa ds lewg dk fcUnq P ls xqtjrs gq, ,oa flDds ds yacor v{k ds lkis{k tM+Ro vk?kw.kZ D;k

gksxk\ ¼P dsUnzh; flDds ds nkfgus fLFkr flDds dk dsUnz gSA½

(A) 2

55 mr2 (B)

2127

mr2 (C) 2

111 mr2 (D) 55 mr2

Sol.

P =

22

22

222

)r4(m2

mr2)32r(m

2mr

3)r2(m2

mr2

mr

= 2mr2

111 .

25. A planet orbits in an elliptical path of eccentricity e around a massive star considered fixed at one of the foci.The point in space where it is closest to the star is denoted by P and the point where it is farthest is denotedby A. Let vP and vA be the respective speeds at P and A. Then

,d xzg mRdsUnzrk (eccentricity) 'e' okyh ,d nh?kZ oÙkkdkj d{k esa ,d fo'kky rkjs dk pDdj yxkrk gSA fo'kky rkjk nh?kZ

oÙk ds ,d ukfHk ij fLFkj gSA rkjs ds lcls utnhd ds fcUnq dks P ,oa lcls nwj ds fcUnq dks A. ls n'kkZ;k x;k gSA ;fn vP

,oa vA Øe'k% P vkSj A ij xzg dk osx gS rks

(A) e1e1

vv

A

P

(B)

A

P

vv

= 1 (C) e1e1

vv 2

A

P

(D) 2

2

A

P

e1

e1vv

RESONANCE PAGE - 12


Sol.

Angular momentum conservation about Sv

p (a � ae) = v

A(a + ae)

aeaaea

v

v

A

p

e1e1

v

v

A

p

Ans. (A)

26. In a Young's double slit experiment the intensity of light at each slit is 0. Interference pattern is observedalong a direction parallel to the line S1 S2 on screen S.

;ax f}&f>jh (Young's double slit) iz;ksx esa izR;sad f>jh ij izdk'k fdj.k dh rhozrk 0. gSA O;frdj.k izfr:i dks insZ S

ij S1 S2 js[kk ds lekukUrj ns[kk tkrk gSA

The minimum, maximum, and the intensity averaged over the entire screen are respectively.

izdk'k dh U;wure] vf/kdre ,oa iwjs insZ ij ekih xbZ vkSlr rhozrk D;k gksxh\(A) 0, 40 , 20 (B) 0 , 20 , 30/2 (C) 0, 40 , 0 (D) 0, 20 , 0

Sol. max

= 200 )(

= 40

min

= 0)( 200

avg

= 0 +

0 = 2

0

Ans. (A)

27. A loop carrying current has the shape of a regular polygon of n sides. If R is the distance from the centre to

any vertex, then the magnitude of the magnetic induction vector B

at the centre of the loop is

n Hkqtkvksa okys ,d lecgqHkqtkdj can ifjiFk esa /kkjk izokfgr gks jgh gSA ;fn cgqHkqt ds dsUnz ls fdlh 'kh"kZ dh nqjh R gS

rks ifjiFk ds dsUnz ij ;qXedh; izsj.k B

dk ifjek.k gksxkA&

(A) n

tanR2

n 0

(B)

n2

tanR2

n 0

(C)

R20

(D) n

tanR0

RESONANCE PAGE - 13


Sol.

B0 =

nsin

nsin

ncosR

i4

0

B0 =

ntan

Ri

20

Ans. (A)

28. A conducting rod of mass m and length is free to move without friction on two parallel long conducting rails,

as shown below. There is a resistance R acorss the the rails. In the entire space around, there is a uniformmagnetic filed B normal to the plane of the rod and rails. The rod is given an impulsive velocity v0.

n'kkZ, x, fp=k esa ,d m nzO;eku vkSj L yackbZ dh pkyd NM nks yach vkSj lekUrj pkyd iVfj;ksa ij fcuk ?k"kZ.k ds xfr

dj ldrh gSA iVfj;ksa ds chp dk izfrjks/k R gSA fp=k dh lrg ds yacor~] gj txg ,d leku pqEcdh; {ks=k B gSA NM dks

vpkud ,d v0 ifjek.k dk vkosxh osx iznku fd;k tkrk gSA

Finally, the initial energy 21

mv02

(A) will be converted fully into heat energy in the resistor(B) will enable rod to continue to move eith velocity v0 since the rails are frictionless(C) will be converted fully into magnetic energy due to induced current(D) will be converted into the work done against the magnetic field

vUrr% izkjafHkd Å tkZ 21

mv02

(A) iw.kZ :i ls izfrjks/k R esa Å "eh; Å tkZ esa :ikUrfjr gks tk,xhA

(B) NM+ dks v0 xfr ls xfreku j[ksxh D;ksafd NM+ ,oa iVfj;ksa ds chp ?k"kZ.k ugha gSA

(C) mRizsfjr fo|qr /kkjk dh otg ls] iw.kZr;k pqEcdh; Å tkZ esa :ikUrfjr gks tk,xhA

(D) pqEcdh; {ks=k ds f[kykQ dk;Z djus esa O;; gks tk,xhASol. Due to the negative work on rod K.E. will decrease and finally become zero.

Ans. (A)

29. A steady current flows through a wire of radius r, length L and resistivity . The current produces heat in thewire. The rate of heat loss in a wire is proportional its surface area. The steady temperature of the wire isindependent of

,d r f=kT;k] L yackbZ ,oa izfrjks/kdrk ds rkj esa LFkk;h fo|qr /kkjk I izokfgr gksrh gSA /kkjk dh otg ls rkj esa Å "ek dk

fuekZ.k gksrk gSA rkj esa Å "ek ds {k; dh nj rkj ds lrgh {ks=kQy ds vuqikrh gSA rkj dk fLFkj rkieku fuEu esa ls fdlds

Å ij fuHkZj ugha gS\(A) L (B) r (C) (D)

RESONANCE PAGE - 14


Sol. R = A

R = 2r

L

2R = K2rLdtdT

2

2

r

L

= K2rL

dtdT

dtdT

= 32

2

r2K

L

Ans. (A)

30. The ratio of the speed of sound to the average speed of an air molecule at 300K and 1 atmospheric pressureis close to

¼300K rkieku ,oa 1 atm nkc ½ ij /ofu dh xfr vkSj ok;q ds ,d v.kq dh vkSlr xfr dk vuqikr fuEu esa ls fdlds fudV

gS \

(A) 1 (B) 300 (C) 300

1(D) 300

Sol.

mRT8mRT

= 85

7 = 0.73

So, closest ans is 1.Ans. (A)

31. In one model of the election, tho electron of mass me is thought to be a uniformly charged shell of radius Rand total charge e, whose electrostatic energy E is equivalent to its mass me via Einstein's mass energyrelation E = mec2 . In this model, R is approximately (me = 9.1 × 10�31 kg, c = 3 × 108 m.s�1,

041

= 9 × 109 Farads m�1, magnitude of the electron charge = 1.6 × 10�19 C)

bysDVªkWu ds ,d fu:i.k esa me nzO;eku ds bysDVªkWu dks R f=kT;k ds leku :i ls vkosf'kr ,d xksyk] ftldk dqy vkos'k

e, gS] le>k tkrk gSA bl vkosf'kr xksys dh fLFkj oS|qr Å tkZ E ,oa nzO;eku me vkbaLVhu ds E = mec2 . laca/k ls nh tkrh

gSA bl fu: i.k esa R d k eku fd l d s d jhc gS\ (me = 9.1 × 10�31 kg, c = 3 × 108 m.s�1,

041

= 9 × 109 Farads m�1, bysDVªkWu dk vkos'k = 1.6 × 10�19 C)

(A) 1.4 × 10�15 m (B) 2 × 10�13 m (C) 5.3 × 10�11 m (D) 2.8 × 10�35 mSol. U = mc2

22

mcR2

KQ

R2)106.1(109 2199

= 9.1 × 10�31 × (3 × 108)2

R = 2831

2199

)103(101.92

)106.1(109

= 1.4 × 10�15

RESONANCE PAGE - 15

KVPY QUESTION PAPER - STREAM (SB / SX)Ans. (A)

32. A body is executing simple harmonic motion of amplitude a and period T about the equilibrium positionx = 0. Large numbers of snapshots are taken at random of this body in motion. The probability of the bodybeing found in a very small interval x to x + |dx| is highest at

vius lkE; voLFkk x = 0. ds lkis{k ,d fiM+ a vk;ke vkSj T vkorZ dky ls ljy vkorZ xfr (simple harmonic

morion) djrk gSA bl oLrq dh ;knfPNd xfr ds dbZ lkjs QksVks fp=k ¼snapshots½ [khpsa tkrs gSA x vkSj x + |dx| ds

chp NksVs varjky dsa fiaM ds ik, tkus dh lcls T;knk izkf;drk dgk¡ gksxh\

(A) x = ± a (B) x = 0 (C) x = ± a/2 (D) x = ± 2

a

Sol. Probability of being found is maximum where speed is minimum.So, x = ±a

Ans. (A)

33. Two identical bodies are made of a material for which the heat capacity increases with temperature One ofthese is held at a temperature of 100°C while the other one is kept at 0ºC. If the two are brought into contact,

then, assuming no heat loss to the environment, the final temperature that they will reach is

nks ,d & tSls fiaM ,d ,sls inkFkZ ls cuk, tkrs gS ftudh Å "ek & /kkfjrk rkieku ds lkFk c<+rh gSA ,d fiaM dks 100°C

ij j[kk tkrk gS vkSj nqljs dks 0ºC.ijA ;fn nksuks fiaM ,d nqljs ls laidZ esa vkrs gS vkSj ;fn eku fy;k tk, fd ifjos'k

esa Å "ek dk {k; ugha gksrk] rks fiaMksa dk vafre rki D;k gksxk\(A) 50°C (B) more than 50°C (C) less than 50°C (D) 0°C

(A) 50°C (B) 50°C ls vf/kd (C) 50°C ls de (D) 0°C

Sol. Since heat capacity at high temperature is high. So for same amount of heat transfer T is more at lowertemperature then at higher temperature.So, final temperature is more than 50ºC .

34. A particle is acted upon by a force given by F = �x3 � x4 where and are positive constants. At the pointx = 0, the particle is

,d cy F = �x3 � x4 ¼tgka vkSj /kukRed fLFkjkad gS ½ ,d d.k ij yxk;k tkrk gSA fdlh fcUnq x = 0, ij d.k(A) in stable equilibrium (B) in unstable equiibrium(C) in neutral equilibrium (D) not in equilibrium

(A) fLFkj lkE;koLFkk esa gSA (B) vLFkk;h lkE;oLFkk esa gSA

(C) mnklhu lkE;oLFkk esa gSA (D) lkE;koLFkk esa ugha gSASol. As F = �x3 � x4

At x = 0, F = 0Hence particle is in equilibrium.

35. The potential energy of a point particle is given by the expression V(x) = x + sin (x / ). A dimensionlesscombination of the constants , and is

,d fcUnq d.k dh fLFkfrt Å tkZ dks V(x) = x + sin (x / ).ls fu:fir fd;k tkrk gSA , vkSj ds chp ,d foekghu

(dimensionless) laca/k gSA

(A)

(B)

2

(C)

(D)

Sol. It is clear thatdimension x = ML2T�2 = MLT�2

= ML2T�2

and x = L

So,

will be dimensionless.

RESONANCE PAGE - 16


36. A ball of mass m suspended from a rigid support by an inextensible massless string is released from aheight h above its lowest point. At its lowest point it colIides elastically with a block of mass 2m at rest on africtionless surface. Neglect the dimensions of the ball and the block. After the collision the ball rises to amaximum height of

,d << lgkjs ls tqM+h ,d Hkkjghu ,oa f[kpkojfgr rkj ls yVdh m nzO;eku dh ,d xsan dks fuEure fcUnq ls h Å apkbZ

ij mBkdj NksM fn;k tkrk gSA (fp=k nsf[k,) vius fuEure fcUnq ij xsan ,d ?k"kZ.k jfgr lrg ij j[ks 2m nzO;eku ds ,d

CykWd ls izR;kLFk VDdj (elastic collision) djrh gSA xsan ,oa CykWd ds vkdkj dh vuns[kh djrs gq,] Vdjkus ds ckn xsan

dh vf/kdre Å apkbZ D;k gksxk\

(A) 3h

(B) 2h

(C) 8h

(D) 9h

Sol. gh2

From conservation of momentum

0gh2m = mv1 + 2mv

2

gh2 = v1 + 2v

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

From equation of e

= 0gh2

vv 12

gh2 = v

2 � v

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

From (i) & (ii)

v1 =

3

gh2

Hence after collision maximum height

hmax

= g2

)3/gh2( 2

hmax

= gh

Ans. (D)

37. A particle released from rest is falling through a thick fluid under gravity. The fluid exerts a resistive force onthe particle proportional to the square of its speed. Which one of the following graphs best depicts thevariation of is speed v with time t?

fLFkj voLFkk ls xq:Roh; Roj.k esa fxjrk gqvk ,d d.k ,d xk<sa rjy inkFkZ ls xqtjrk gSA rjy nzO;d.k ij izfrjks/kd cy

yxrk gS tks d.k ds osx ds oxZ ds lekuqikrh gSA d.k ds osx (v) dk le; (t) ds lkis{k ifjorZu dks fuEu vkjs[kksa esa ls dkSu

lgh lgh fu:fir djrk gSs\

RESONANCE PAGE - 17


(A) (B)

(C) (D)

Sol. Let ball is moving with speedv at anytime t hence

Hence, mdtdv

= mg � Kv2

On the basis of equation we can rays that speed first increase and become constant when mg = kv2

Hence, Ans. (A)

38. A cylindrical steel rod of length 0.10m and thermal conductivity 50 W.m�1.K�1 is welded end to end to copperrod of thermal conductivity 400 W.m�1.K�1 and of the same area of cross section but 0.20 m long. The freeend of the steel rod is maintained at 100°C and that of the copper rod at 0°C. Assuming that the rods are

perfectly insulated from the surrounding the temperature at the junction of the two rods is

LVhy dh ,d csyukdkj NM+ ftldh yackbZ 0.10m rFkk Å "eh; pkydrk 50 W.m�1.K�1 gS] ,d Nksj ij leku vuqizLFk

dkV ,oa 0.20m yach ,d nwljh csyukdkj NM+ ftldh Å "eh; pkydrk 400 W.m�1.K�1 gS] ls tqMh gSA LVhy NM+ ds [kqys

fljs dks 100°C ,oa rkcsa dh NM+ ds [kqys fljs dks 0°C.rkieku ij fLFkj j[kk x;k gSA ;g ekurs gq, fd NMksa ls ifjos'k esa

Å "ek dk âkl ugha gks jgk gS] NMksa ds tksM+ ij rkieku D;k gksxk\(A) 20°C (B) 30°C (C) 40°C (D) 50°C

Sol. Resistance of steel = A50

1.0

= A500

1

Resistance of copper = A400

2.0 =

A2001

Both are connected in series hence heat current in both will be same. So

A5001

T100 =

A20001

0T T = 20ºC

Ans. (A)

39. A parent nucleus X is decaying into daughter nucleus Y which in turn decays to Z. The half lives of X and Yare 4000 years and 20 years respectively. In a certain sample, it is found that the number of Y nuclei hardlychanges with time. If the number of X nuclei in the sample is 4 × 1020 , the number of Y nuclei present in itis

,d ewy ukfHkd X nwljs ukfHkd Y esa {kf;r gksrk gS] tks fQj Z. esa {kf;r gksrk gSA X vkSj Y dh v/kZ vk;q Øe'k% 4000 o"kZ

,oa 20 o"kZ gSA ,d uewus esa ;g ns[kk x;k gS fd Y ukfHkdksa dh la[;k esa le; ds lkFk dksbZ [kkl ifjorZu ugha gks jgk gSA

;fn bl uewus esa X ukfHkdksa dh la[;k 4 × 1020 rks ml le; Y ukfHkdksa dh la[;k D;k gksxh\(A) 2 × 1017 (B) 2 × 1020 (C) 4 × 1023 (D) 4 × 1020

Sol. FromN

1

1 = N

2

2

RESONANCE PAGE - 18


4 × 1020 = 40000

)2(n =

20)2(n

N2

N2 = 2 × 107

Ans. (A)

40. An unpolarized beam of light of intensity 0 passes through two linear polarizers making an angle of 30° with

respect to each other. The emergent beam will have an intensity.

nks js[kh; /kqzodksa ds chp] tks ijLij 30° ij fLFkj gS ,d 0 rhozrk dh v/kzqfor izdk'k fdj.k xqtjrh gSA ckgj vkus okyh

fdj.k dh rhozrk D;k gksxh\

(A) 4

3 0 (B) 4

3 0 (C) 8

3 0 (D) 80

Sol.

43

2º30cos

2020

Final intensity will be = 8

3 0

Ans. (C)

CHEMISTRY41. Among the following, the species with the highest bond order is :

fuEufyf[kr esa fdldk vf/kdre vkcU/k Øe (bond order) gS \(A) O

2(B) F

2(C) O

2+ (D) F

2�

Ans. (C)Sol. According to MOT bond order of O

2, F

2, O

2+ and F

2� is respectively 2, 1, 2.5 and 0.5.

gy MOT ds vuqlkj O2, F

2, O

2+ rFkk F

2� dk ca/k Øe Øe'k% 2, 1, 2.5 o 0.5 gSA

42. The molecule with non-zero dipole moment is :v.kq ftldk f}/kzqo vk?kw.kZ (dipole moment) v'kwU; gS \(A) BCl

3(B) BeCl

2(C) CCl

4(D) NCl

3

Ans. (D)Sol. BCl

3, BeCl

2 and CCl

4 has zero dipole moment due to symmetric structure where as shape of NCl

3 is pyrami-

dal due to presence of lone pair.gy BCl

3, BeCl

2 o CCl

4 dk f}/kzqo vk?kw.kZ 'kwU; gksrk gS D;ksafd budh lefer lajpuk gksrh gS] tcfd NCl

3 dh vkÑfr fijsfefM;y

gksrh gS D;ksafd blesa ,dkdh bysDVªkWu ;qXe mifLFkr gksrk gSA

43. For a one-electron atom, the set of allowed quantum numbers is :,d&bysDVªkWu ijek.kq ds fy, vuqer DokaVe la[;k,¡ (allowed quantum numbers) gS \

(A) n = 1, = 0,

m = 0, ms = +½ (B) n = 1, = 1,

m = 0, m

s = +½

(C) n = 1, = 0,

m = �1, ms = �½ (D) n = 1, = 1,

m = 1, m

s = �½

Ans. (A)Sol. The value of n > and m should have values from � to + .

n dk eku > o m ds eku � ls + rd gksus pkfg,A .

RESONANCE PAGE - 19

KVPY QUESTION PAPER - STREAM (SB / SX)44. In the reaction of benzene with an electrophile E+, the structure of the intermediate -complex can be

represented as :csathu dh ,d bysDVªkWuLusgh (electrophile) E+ ds lkFk vfHkfØ;k esa e/;orhz -ladj dh lajpuk dks bl izdkj iznf'kZr djldrs gSa \

(A) (B) (C) (D)

Ans. Option �D� Closest, if (�) is change to fodYi �D� lgh gS] D;ksafd (�) dks esa ifjofrZr fd;k tk ldrk gSA

Sol.

45. The most stable conformation of 2, 3-dibromobutane is :2, 3-MkbczkseksC;wVsu dk vf/kdre LFkk;h la:i.k D;k gS \

(A) (B) (C) (D)

Ans. (Bonus)46. Typical electronic energy gaps in molecules are about 1.0 eV. In terms of temperature, the gap is closest to:

v.kqvksa ds bysDVªkWfud ÅtkZ&varjky (energy gaps) dk eku yxHkx 1.0 eV gSA rkieku ds lanHkZ esa ;g varj buesa ls fdldsfudV gS \(A) 102 K (B) 104 K (C) 103 K (D) 105 K

Ans. (B)Sol. KE = | T.E. |

(KE) = (T.E.)

23

× 8.3 × (T) = 1.6 × 10�19 × 6 × 1023

(T) = 3.8106.9 4

× 32

(T) = 7.6 × 103 K i.e. close to 104 K.(T) = 7.6 × 103 K vFkkZr 104 K ds fudV gSA

47. The major final product in the following reaction is :fuEu vfHkfØ;k dk eq[; vafre mRikn D;k gS \

CH3CH

2CN

OH)2

MgBrCH)1

3

3

(A) (B)

(C) (D)

RESONANCE PAGE - 20

KVPY QUESTION PAPER - STREAM (SB / SX)Ans. (C)

Sol.

48. A zero-order reaction, A Product, with an initial concentration [A]0 has a half-life of 0.2 s. If one starts with

the concentration 2[A]0, then the half-life is :

,d 'kwU; dksfV vfHkfØ;k] A mRikn (product), dh v)Z&vk;q 0.2 s gS] tc izkFkfed lkanzrk [A]0 gSA ;fn vfHkfØ;k

2[A]0 dh lkUnzrk ls 'kq: gksrh rks bldh v)Z vk;q D;k gksxh \

(A) 0.1 s (B) 0.4 s (C) 0.2 s (D) 0.8 sAns. (B)Sol. t

1/2 (a)1 � n

II

I

)t()t(

2/1

2/1 =

01

2

1aa

II)t(2.0

2/1 =

0

0A2

A

(t1/2

)II = 0.4 s

49. The isoelectronic pair of ions is :fuEu vk;fud ;qXe lebysDVªkWfud (isoelectronic) gS %(A) Sc2+ and V3+ (B) Mn3+ and Fe2+ (C) Mn2+ and Fe3+ (D) Ni3+ and Fe2+

Ans. (C)Sol.

25Mn2+ and

26Fe3+ both has 23 electrons.

25Mn2+ o

26Fe3+ nksuksa esa 23 bysDVªkWu gSA

50. The major product in the following reaction is :fuEu vfHkfØ;k dk eq[; mRikn D;k gS \

2NaNH

(A) (B) (C) (D)

Ans. (A)

Sol.

51. The major product of the following reaction is :fuEu vfHkfØ;k dk eq[; mRikn D;k gS \

HBr.Conc

(A) (B) (C) (D)

Ans. (B)

Sol.

RESONANCE PAGE - 21

KVPY QUESTION PAPER - STREAM (SB / SX)52. The oxidation state of cobalt in the following molecule is :

fuEufyf[kr v.kq esa dksckYV dh vkWDlhdj.k voLFkk (oxidation state) D;k gS\

(A) 3 (B) 1 (C) 2 (D) 0Ans. (D)Sol. In metal carbonyls oxidation state of metal is equal to zero.

/kkrq dkcksZfuy esa /kkrq dh vkWDlhdj.k voLFkk 'kwU; ds cjkcj gksrh gSA

53. The pKa of a weak acid is 5.85. The concentrations of the acid and its conjugate base are equal at a pH of:

,d nqcZy vEy ds pKa dk eku 5.85 gSA pH ds fdl eku ds fy, vEy dh vkSj blds la;qXe (conjugate) {kkj dh lkUnzrk

leku gksxh \(A) 6.85 (B) 5.85 (C) 4.85 (D) 7.85

Ans. (B)Sol. At pH = pK

a the concenteration of acid and its conjugate base are equal.

pH = pKa ij vEy o blds l;qfXer {kkj dh lkUnzrk cjkcj gksrh gSA

54. For a tetrahedral complex [MCl4]2�, the spin-only magnetic moment is 3.83 BM. The element M is :

prq"Qydh; ladqy (tetrahedral complex) [MCl4]2� ds fy, dsoy izpØ.k pqEcdh; vk?kw.kZ (spin-only magnetic

moment) 3.83 BM gSA rRo M D;k gS \(A) Co (B) Cu (C) Mn (D) Fe

Ans. (A)Sol. It is high spin complex as Cl� is weak field effect ligand. In [CoCl

4]2� oxidation state of Co is +2 in which 3

unpaired electrons are present which gives the spin-only magnetic moment equal to 3.83 BM.;g mPp pØ.k ladqy gS D;ksafd Cl� ,d nqcZy {ks=k fyxs.M gSA [CoCl

4]2� esa Co dh vkWDlhdj.k voLFkk +2 gSA blesa rhu

v;qfXer bysDVªkWu mifLFkr gS tks fd 3.83 BM ds cjkcj dsoy pØ.k pqEcdh; vk?kw.kZ n'kkZrs gSaA

55. Among the following graphs showing variation of rate (k) with temperature (T) for a reaction, the one thatexhibits Arrhenius behavior over the entire temperature range is :fuEu vkjs[k esa vfHkfØ;k nj ds ifjorZu (k) dks rkieku (T) ds lkFk n'kkZ;k x;k gSA iwjs rki ifjlj esa vkgsZfu;l (Arrhenius)

O;ogkj dks buesa ls fdl vkjs[k }kjk n'kkZ;k x;k gS \

(A) (B) (C) (D)

Ans. (D)

Sol. K = RT/E� aAe

ln K = RTE� a + ln AA

56. The reaction that gives the following molecule as the major product isfuEufyf[kr v.kq fdl vfHkfØ;k dk eq[; mRikn gS \

RESONANCE PAGE - 22


(A) (B)

(C) (D)

Ans. (B)

Sol.

57. The C�O bond length in CO, CO2 and CO

32� follows the order :

CO, CO2 vkSj CO

32� esa C�O vkca/k dh yEckbZ dk Øe D;k gS \

(A) CO < CO2 < CO

32� (B) CO

2 < CO

32� < CO (C) CO > CO

2 > CO

32� (D) CO

32� < CO

2 < CO

Ans. (A)Sol. Greater is the bond order shorter will be bond length. In CO

32� bond order in between 1 and 2, bond order of

CO2 is 2 where as bond order for CO is in between 2 and 3.

gy ftldk ca/k Øe vf/kd gksxk mldh ca/k yEckbZ de gksxhA CO3

2� esa ca/k Øe 1 o 2 ds e/; gS] CO2 dk ca/k Øe 2 gS tcfd

CO dk ca/k Øe 2 o 3 ds e/; gSA

58. The equilibrium constant for the following reactions are K1 and K

2, respectively.

2P(g) + 3Cl2(g) 2PCl

3(g)

PCl3(g) + Cl

2(g) PCl

5(g)

The the equilibrium constant for the reaction2P(g) + 5Cl

2(g) 2PCl

5(g)

is :fuEufyf[kr vfHkfØ;kvksa ds fy, lkE; fLFkjkad Øe'k% K

1 rFkk K

2 gSA


3(g)

PCl3(g) + Cl

2(g) PCl

5(g)

rc 2P(g) + 5Cl2(g) 2PCl

5(g) vfHkfØ;k ds fy, lkE; fLFkjkad D;k gksxk \

(A) K1K

2(B) K

1K

22 (C) K

12K

22 (D) K

12K

2

Ans. (B)Sol. 2P(g) + 3Cl

2(g) 2PCl

3(g) ........(i)

PCl3(g) + Cl

2(g) PCl

5(g) ........(ii)


5(g) ........(iii)

On multipling equation (ii) by 2 and adding in (i) we obtain equation (iii)Therefore, K

3 = K

1K

22

lehdj.k (ii) dks 2 ls xq.kk dj lehdj.k (i) esa tksM+us ij gesa lehdj.k (iii) izkIr gksrh gSAblfy,, K

3 = K

1K

22

RESONANCE PAGE - 23

KVPY QUESTION PAPER - STREAM (SB / SX)59. The major product of the following reaction is :

fuEufyf[kr vfHkfØ;k dk eq[; mRikn D;k gS \

+ (CH3C)

2CHCH

2Cl 3AlCl

(A) (B)

(C) (D)

Ans. (A)Sol. The reagent given the question has an addition �C� within bracket.

fn;s x;s iz'u ds vfHkdeZd ds czsdsV esa ,d vfrfjDr �C� fn;k x;k gSA(CH

3)2CH�CH

2�Cl + AlCl

3(CH

3)

2CH�CH

2�Cl - - - - AlCl

3 (CH

3)

2CH�CH

2+ + AlCl

4�

(CH3)2CH�CH

2+

shifftingH2,1 � (CH3)3C+

60. Doping silicon with boron produces a :(A) n-type semiconductor (B) metallic conductor(C) p-type semiconductor (D) insulatorflfydu ds cksjkWu vifeJ.k (Doping) ls dkSu&lk mRikn curk gS \(A) n-izdkj ds v)Zpkyd (B) /kkfRod pkyd(C) p-izdkj ds v)Zpkyd (D) fo|qr jks/kh

Ans. (C)Sol. As boron is trivalent impurity it will proudce p-type semiconductor.

D;ksafd cksjksu f=kla;ksth v'kqf) gS ;g p-izdkj ds v)Zpkyd cukrk gSA

BIOLOGY61. The disorders that arise when the immune system destroys 'self' cells are called autoimmmune disorders.

Which of the following would be classified under this(A) rheumatoid arthritis (B) asthma (C) rhintis (D) eczema

Ans (A)

ftl fodkj jksx ds dkj.k izfrj{kd ra=k (immune system) viuh gh dksf'kdkvksa dk fouk'k djrk gS] mls vkReizfrj{kdjksx dgrs gSA fuEufyf[kr esa ls dkSu lk ,d jksx dgrs gSA fuEufyf[kr esa ls dkSulk ,d ,slk jksx gSA(A) xfB;k laf/k'kksFk (rheumatoid archritis) (B) nek (asthma)

(C) uklk & 'kksFk(rhintis) (D) Nktu (eczema)Ans (A)

62. Which of the following class of immunoglobulins can trigger the complement cascade ?(A) IgA (B) IgM (C) IgD (D) IgE

Ans (A)dkWfIyesUV dSLdsM dks buesa ls dkSu lk bEequksXykscqyhu (immunoglobulins) izofrZr djrk gS\(A) IgA (B) IgM (C) IgD (D) IgE

Ans (A)

RESONANCE PAGE - 24

KVPY QUESTION PAPER - STREAM (SB / SX)63. Diabetes insipidus is due to

(A) hypersecretion of vasoperssin(B) Hyposecretion of insulin(C) hypersecretion of insulin(D) hyposecretion of vasopressin

Ans (D)MkbfCkVht buflfiMl (Diabetes insipidus) dk dkj.k D;k gS\(A) oSlksxzsflu dk vf/kd ek=kk esa lzo.k gksukA (B) bUL;wyhu dk lzo.k de gksukA(C) bUL;wyhu dk vf/kd ek=kk esa lzo.k gksukA (D) oSlksizsflu dk lzo.k de gksukA

Ans (D)

64. Fossils are most often found in which kind of rocks ?(A) meteorites (B) sedimentary rocks(C) igneous rocks (D) metamorphic rocks

Ans (B)fdl izdkj dh f'kykvksa esa T;knkrj thok'e ik, tkrs gS\(A) mYdkfi.M (meteorites) (B) volknh 'kSy (sedimentary rocks)

(C) vkXus; 'kSy (igneous rocks) (D) dk;krj.kh 'kSy (metamorphic rocks)Ans (B)

65. Peptic ulcers are caused by(A) a fungus, Candida albicans(B) a virus, cytomegalo virus(C) a parasite, Trypanosoma brucei(D) a bacterium, Helicobacter pylori

Ans (D)isV ds oz.k (Peptic ulcers) fdl tho ds dkj.k gksrs gS\(A) ,d dodh] (Candida albicans) (B) ,d ok;jl (cytomegalo virus)

(C) ,d ijtho] (Trypanosoma bruci) (D) ,d thok.kq (Helicobacter pylori)Ans (D)

66. Transfer RNA (tRNA)(A) is present in the ribosomes and provides structural integrity(B) usually has clover leaf-like structure(C) carries genetic information from DNA to ribosomes(D) codes for proteins

Ans (B)VªkUlQj (tRNA)

(A) jkbckslkse esa fLFkr gS vkSj lajpukRed v[kaMrk nsrk gSA(B) lkekU;r% frifr;k (clover leaf-like) vkdkj dh gksrh gSA(C) DNA ls jkbckslkse rFkk vkuqoaf'kd lwpuk igqapkrk gSA(D) izksVhu cukrk gS

Ans (B)

67. Some animals excrete uric acid in urine (uricotelic) as it requires very little water. This is an adaptation toconserve water loss. Which animals among the following are most likely to be uricotelic?(A) fishes (B) amphibians (C) birds (D) mammals

Ans (C)ew=k esa ;wfjd vEy folftZr djus okys izk.kh dks ;wfjdksVsfyd dgrs gSA ;g izfØ;k ikuh ds laj{k.k ds fy, vuqdwy gSA buesals dkSu lk izk.kh ;wfjdksVsfyd gS\(A) eNfy;k¡ (B) i{kh (C) mHk;pj (D) Lru/kkjh

Ans (C)

RESONANCE PAGE - 25

KVPY QUESTION PAPER - STREAM (SB / SX)68. A ripe mango, kept with unripe mangoes causes their ripening. This is due to the release of a gaseous plant

hormone(A) auxin (B) gibberlin (C) cytokinine (D) ethylene

Ans (D),d ids vke dks dPps vkeksa ds lkFk j[kus ls dPPks vke idus 'kq: gks tkrs gSA ;g fdl xSlh; ikni gkeksZu ds fudyusdk dkj.k gS\(A) vkDlhu (B) lkbVksdkbuhu (C) ftCcjyhau (D) ,Fkhyhu

Ans (D)

69. Human chromosomes undergo structural changes during the cell cycle. Chromosomal structure can be bestvisualized if a chromosome is isolated from a cell at(A) G1 phase (B) S phase (C) G2 phase (D) M phase

Ans (D)euq"; ds dksf'kdk pØ ds nkSjku Øksekslkse dh lajpuk esa cnyko gksrk gSA bl cnyko dks Li"V :i ls ns[kus ds fy, pØdh fdl izkoLFkk esa Øksekslkse dk ijh{k.k fd;k tkuk pkfg,\(A) G1 izkoLFkk (B) S izkoLFkk (C) G2 izkoLFkk (D) M izkoLFkk

Ans (D)

70. By which of the following mechanisms is glucose reabsorbed from the glomerular filtrate by the kidney tubule(A) osmosis (B) diffusion (C) active transport (D) passiver transport

Ans (C)fuEufyf[kr fdl izfØ;k ds }kjk xqnsZ ds fuL;aan }kjk Xykses:yj fQYVªsV ls Xyqdkst iqu% 'kksf"kr gksrk gS\(A) ijklj.k (osmosis) (B) folj.k (diffusion)

(C) lfØ; xeu (active transport) (D) fuf"Ø; vfHkxeu (passiver transport)

Ans (C)

71. In mammals, the hormones secreted by the pituitary, the master gland, is itself regulated by(A) Hypothalamus (B) median cortex (C) pineal gland (D) cerebrum

Ans (A)efLr"d ds fdl Hkkx }kjk Lru/kkjh ih;w"k ( pituitary gland) ds gkeksZu dk lzo.k & fu;a=k.k gksrk gSA(A) v/k'psrd (Hypothalamus) (B) ehfM;u dkVsZDl (median cortex)

(C) ihfu;e xzfUFk (pineal gland) (D) izefLr"d (cerebrum)Ans (A)

72. Which of the following is true for TCA cycle in eukaryotes(A) takes place in mitochondrion(B) produces no ATP(C) takes place in Golgi complex(D) independent of electron transport chain

Ans (A);wdSfj;ksV ds TCA pØ ds ckjs esa fuEufyf[kr dkSu lk dFku lgh gS\(A) ekbVksdkfUMª;k esa gksrk gSA (B) ATP ugha cukrkA(C) xkWYth dkEIysDl esa gksrk gSA (D) izefLr"d (cerebrum)

Ans (A)

73. A hormone molecule binds to a specific protein on the plasma membrane inducing a signal. The protein itbinds to it called(A) ligand (B) antibody (C) receptor (D) histone

Ans (C)dksf'kdk f>Yyh ij fLFkr ml izksVhu dks D;k dgrs gS tks gkeksZu ls tqMrk gS vkSj ladsr izsfjr djrk gS\(A) layXuh (ligand) (B) xzkgh (receptor) (C) izfrj{kh (antibody) (D) fgLVksu (histone)

Ans (C)

RESONANCE PAGE - 26

KVPY QUESTION PAPER - STREAM (SB / SX)74. DNA mutations that do not cause my functional change in the protein product are known as

(A) nonsense mutations (B) missense mutations (C) deletion mutations (D) silent mutationsAns (D)

DNA esa mRifjorZu] tks izksVhu ds izdk;Z esa cnyko ugha ykrs] mUgsa D;k dgrs gSa\(A) vuFkZd mRifjorZu (nonsense mutations) (B) vikFkZd mRifjorZu (missense mutations)

(C) foyksih mRifjorZu (deletion mutations) (D) uhjo mRifjorZu (silent mutations)Ans (D)

75. Plant roots are usually devoid of chlorophyll and cannot perform photosynthesis. However, three are excep-tions. Which of the following plant root can perform photosynthesis(A) Arabidopsis (B) Tinospora (C) Rice (D) Hibiscus

Ans (B)ikS/kksa dh tM+ esa DyksjksfQy ugha gksrk gS] ftlds dkj.k os izdk'k la'ys"k.k (photosynthesis) ugha dj ikrsA buesa ls dkSulk vlk/kkj.k foijhr mnkgj.k gS\(A) vjschMkIlhl (Arabidopsis) (B) VhuksLiksjk (Tinospora)

(C) pkoy (Rice) (D) fgfcLdl (Hibiscus)Ans (B)

76. Vitamin A deficiency leads to night-blindness. Which of the following is the reason for the disease ?(A) rod cells are not converted to cone cells(B) rhodopsin pigment of rod cells is defective(C) melanin pigment is not synthesized in cone cells(D) cornea of eye gets dried

Ans (B)foVkfeu , dh deh ls jrkSa/kh (night-blindness) gksrh gSA bldk dkj.k D;k gS\(A) 'kykdk dksf'kdkvksa (rod cells) dk 'kadq dksf'kdkvksas (cone cells) esa ifjofrZr ugha gksukA(B) 'kykdk dksf'kdkvksa ds jksMkWfIlu o.kZd dk (pigment) dk nks"kiw.kZ gksukA(C) 'kadq dksf'kdk esa esykuhu o.kZd dk ugha cuukA(D) vka[kksa ds LoPN eaMy (cornea) dk lw[k tkukA

Ans (B)

77. In Dengue virus infection, patients often develop haemorrhagic fever due to internal bleeding. This happensdue to the reduction of(A) platelets (B) RBCs (C) WBCs (D) lymphocytes

Ans (A)MsUX;q ok;jl ds laØe.k ls dbZ jksfx;ksa esa vkUrfjd jDrlzko gksrk gSA buesa ls fdl dksf'kdk ds de gksus ls ,slk gksrk gS\(A) jDr foEck.kq (platelets) (B) yky :f/kj df.kdk (RBCs)

(C) lQsn :f/kj df.kdk (WBCs) (D) ylhdk.kq (lymphocytes)Ans (A)

78. If the sequence of bases in sense strand of DNA is 5'-GTTCATCG-3, then the sequence ofd bases in its RNAtranscript would be(A) 5'-GTTCATCG-3' (B) 5'GUUCAUCG-3 (C) 5'CAAGTAGC-3' (D) 5'CAAGUAGC -3

Ans (B)vxj DNA ds sence strand dk vuqØe, 5'-GTTCATCG-3, 5'-GTTCATCG-3 5'-GTTCATCG-3 gS] rks mlds RNA

dh izfrfyfi (transcript) dk vuqØe D;k gksxk\(A) 5'-GTTCATCG-3' (B) 5'GUUCAUCG-3 (C) 5'CAAGTAGC-3' (D) 5'CAAGUAGC -3

Ans (B)

79. A refilex aciton is a quick involuntary response to stimulus. Which of the following is an example of BOTH,unconditioned and conditioned reflex(A) knee jerk reflex(B) secretion of saliva in response to the aroma of food(C) sneezing reflex(D) contration of the pupil in response to bright light

Ans (A)

RESONANCE PAGE - 27


fdlh míhiu ds vuSfPNd vuqfØ;k dks izfrorhZ fØ;k dgrs gSA buesa ls dkSu lk mnkgj.k vizkuqdwyh vkSj izkuqdwyh izfrorhZfØ;k,¡ gS\(A) uh&tdZ (knee jerk) izfrorhZ (B) [kkus dh [kq'kcq ls eaqg esa ikuh vkuk(C) Nhad izfrorhZ (D) pedhys izdk'k esa iqrfy;ksa dk fldqMUkk

Ans (A)

80. In a food chain such as grass deer lion, the energy cost of respiration as a proportion of total assimi-lated energy at each level would be(A) 60%- 30%-20% (B) 20%- 30%-60% (C) 20%- 60%-30% (D) 30%- 30%-30%

Ans (B)?kkl & fgj.k & 'ksj] fd vkgkj Ja[kyk esa dqy laxzghr Å tkZ ds vuqikr esa izR;sd Lrj ij 'olu Å tkZ dh ykxr fdruhgksaxh\(A) 60%- 30%-20% (B) 20%- 30%-60% (C) 20%- 60%-30% (D) 30%- 30%-30%

Ans (B)

PART-IITwo Mark Questions

MATHEMATICS81. Suppose a, b, c are real numbers, and each of the equations x2 + 2ax + b2 = 0 and x2 + 2bx + c2 = 0 has two

distinct real roots. Then the equation x2 + 2cx + a2 = 0 has(A) two distinct positive real roots (B) two equal roots(C) one positive and one negative root (D) no real rootseku ysa fd a, b, c okLrfod la[;k,a gS rFkk izR;sd lehdj.k x2 + 2ax + b2 = 0 ,oa x2 + 2bx + c2 = 0 ds nks fofHkUuokLrfod ewy gSA rc lehdj.k x2 + 2cx + a2 = 0 ds(A) nks fofHkUu /kukRed okLrfod ewy gksxsaA (B) nks leku ewy gksxsaA(C) ,d /kukRed o ,d _ .kkRed ewy gksxkA (D) dksbZ okLrfod ewy ugha gksxkA

Sol. D1 = 4a2 � 4b2 > 0 4 (a2 � b2) > 0

a2 � b2 > 0 .......... (1)D

2 = 4b2 � 4c2 > 0 b2 � c2 > 0 ............. (2)

now D = 4c2 � 4a2 = 4(c2 � b2 + b2 � a2)= � 4 (b2 � c2 + a2 � b2)= � 4 (D

1 + D

2)

= Negative Equation have no real root.

82. The coefficient of x2012 in )x1)(x1(

x12

is

)x1)(x1(

x12

esa x2012 dk ,d xq.kkad gS&

(A) 2010 (B) 2011 (C) 2012 (D) 2013Sol. (1+x) (1+x2)�1 (1-x)�1

(1+x)

0r

rr2 )1(x

0r

rx =

0r

rr2 )1(x

0r

rx +

0r

rr2 )1(x

0r

1r )x(

for Coffi. of x2012 = (-1)0 + (-1)1 + (-1)2 + ....... (-1)1006) + ((-1)0 + (-1)1+ ..... +(-1)1005)= 1 + 0= 1No answer

RESONANCE PAGE - 28


83. Let (x, y) be a variable point on the curve 4x2 + 9y2 � 8x � 36y + 15 = 0. Then

min (x2 � 2x + y2 � 4y + 5) + max(x2 � 2x + y2 � 4y + 5) is

;fn oØ 4x2 + 9y2 � 8x � 36y + 15 = 0 ij (x, y) ,d pjfcUnq gS] rcmin (x2 � 2x + y2 � 4y + 5) + max(x2 � 2x + y2 � 4y + 5) dk eku gksxk&

(A) 36

325(B)

32536

(C) 2513

(D) 1325

Sol. 4x2 + 9y2 � 8x � 36y + 15 = 0

4(x2 � 2x + 1) + 9(y2 � 4y + 4) � 25 = 0

4(x � 1)2 + 9 (y � 2)2 = 25

2

2

2

2

35

)2y(

25

)1x(

= 1

min 22 )2y()1x( + max. 22 )2y(1x

= 2

35

+

2

25

=

36225100

= 36

325

Ans. (A)

84. The sum of all x [0, ] which satisfy the equation sinx + 21

cosx = sin2(x + 4

) is

;fn x [0, ] lehdj.k sinx + 21

cosx = sin2(x + 4

) dks larq"V djrk gS] rc x ds ,sls lHkh ekuksa dk ;ksx gS&

(A) 6

(B) 6

5(C) (D) 2

Sol. sinx + 21

cos x = sin2

4x

sinx +21

cosx = 2

2/x2cos1

sinx + 21

cosx = 21

21 sin2x

2sinx + cosx = 1 + sin2x2sinx + cosx = 1 + 2 sinx cosx2sinx (1 � cosx) � 1 (1 � cosx) = 0

(1 � cosx) (2sinx � 1) = 0

cos x = 1 or sinx = 21

x = 0 x =6

,6

5

sum of roots = 0 + 6

+ 6

5 =

Ans. (C)

85. A polynomila P(x) with real coefficients has the property that P(x) 0 for all x. Suppose P(0) = 1 andP(0) = � 1. What can you say about P(1)?

cgqin P(x) ds okLrfod xq.kkad bl izd kj gS fd lHkh x ds fy, P(x) 0. eku yssa fd P(0) = 1 rFkkP(0) = � 1 rc vki P(1) ds ckjs esa D;k dg ldrs gS ?

(A) P(1) 0 (B) P(1) 0 (C) P(1) 0 (D) 21�

< P(1) < 21

RESONANCE PAGE - 29


Sol. P(x) = ax2 + bx + c a 0P(x) = 2ax + bP(x) = 2aP(0) = c = 1P(0) = b = �1

P(x) = ax2 � x + 1

P(1) = a � 1 + 1 = a 0Ans. (C)

86. Define a sequence (an) by a

1 = 5, an = a

1a

2 ... a

n�1 + 4 for n > 1. Then

1n

n

n a

alim

(A) equals 21

(B) equal 1 (C) equals 52

(D) does not exists

,d vuqØe (an) dks bl izdkj ifjHkkf"kr djssa fd n > 1 ds fy, a

1 = 5, an = a

1a

2 ... a

n�1 + 4 rc

1n

n

n a

alim

dk eku

(A) 21

ds cjkcj gSA (B) 1 ds cjkcj gSA (C) 52

ds cjkcj gSA (D) vfLrRo ugha gSA

Sol. a1 = 5

an = a

1a

2a

3 .......a

n � 1 + 4

= n

n1�n321

a

aa......aaa+ 4

an =

n

1n

a4�a

+ 4

an

2 = an + 1

� 4 + 4an

an+1

= an

2 � 4an + 4 = (a

n � 2)2

1na = an � 2

n

1n

a

a = 1 �

na2

1�n

n

a

a= 1 �

1�na1

nlim n

n 1

a

a

= n

limn 1

21�

a

= 1 ( n

lim an = )

87. The value of the integral dxa1

xcos

�

x

2

, where a > 0, is

lekdy dxa1

xcos

�

x

2

, tgk¡ a > 0 dk eku gS&

(A) (B) a (C) 2

(D) 2

Sol. I =

�

x

2

dxa1

xcos, (a > 0) ...(1)

I =

�

x�

2

dxa1

)x(�cos

I =

�

x

2x

dxa1

xcosa ...(2)

RESONANCE PAGE - 30

KVPY QUESTION PAPER - STREAM (SB / SX)(1) + (2)

2I = dx)a1(

)a1(xcos

�

x

x2

2I = dxxcos20

2

I = 00

1 cos2x 1 sin2xdx x

2 2 2

I = 01sin2 � 0 sin0

2 =

2

.

88. Consider

L = 3 2012 + 3 2013 + ... + 3 3011

R = 3 2013 + 3 2014 + ... + 3 3012

and I = dxx33012

2012 .

eku ysa fd

L = 3 2012 + 3 2013 + ... + 3 3011

R = 3 2013 + 3 2014 + ... + 3 3012

rFkk I = dxx33012

2012 rc

(A) L + R < 2I (B) L + R = 2I (C) L + R > 2I (D) LR = I

Sol. Since y = x1/3 is concave downwards for x > 0

Area of trapezoid AABB < 2013

2012

3/1 dxx

i.e. 2

)2013()2012( 3/13/1

< 2013

2012

3/1 dxx

Similarly, 2

)2014()2013( 3/13/1

< 2014

2013

3/1 dxx

2)2015()2014( 3/13/1

<

2015

2014

3/1 dxx

...................................and so on

RESONANCE PAGE - 31


2)3012()3011( 3/13/1

<

3012

3011

3/1 dxx

Adding we get

2RL

< 3012

2012

3/1 dxx

2RL

< I

L + R < 2IHence Answer is (A)

89. A man tosses a coin 10 times, scoring 1 point for each head and 2 points for each tail. Let P(K) be the

probability of scoring at least K points. The largest value of K such that P(K) > 21

is

,d vkneh ,d flDds dks 10 ckj mNkyrk gSA mls izR;sd fpÙk ds fy, 1 vad vkSj izR;sd iV~V ds fy, 2 vad izkIr gksrs

gSA eku ysa fd de ls de K vad izkIr djus dh izkf;drk P(K) gks] rc P(K) > 21ds fy, K dk vf/kdre eku D;k gksxk?

(A) 14 (B) 15 (C) 16 (D) 17Sol. P(K) = P (at least K points)

x1 + x

2 + x

3 + ....... x

10 = K

Coeff xK in (x1 + x2)10

= x10(1 + x)10

coeff xk�10 in (1 + x)10

= 10CK�10

K 10

Now P(K) > 21

1010K

102

101

100

10

2

C......CCC

21

10C0 + 10C

1 + ............. + 10C

K�10 > 29

1 + 10 + 45 + 120 + 210 + 252 + ...... > 51210C

0 + 10C

1 + 10C

2 + 10C

3 + 10C

4 + 10C

5 > 512

K � 10 = 5

K = 15Ans. (B)

90. Let f(x) = 1x1x

for all x 1. Let f1(x) = f(x), f2(x) = f (f(x)) and generally fn(x) = f(fn�1(x)) for n > 1. Let P = f1(2)

f2(3) f3(4) f4(5) which of the following is a multiple of P

eku ysa fd lHkh x 1 ds fy, f(x) = 1x1x

gSA ;fn f1(x) = f(x), f2(x) = f (f(x)) vkSj n > 1 ds fy, lkekU;r%

fn(x) = f(fn�1(x)) gS vkSj P = f1(2) f2(3) f3(4) f4(5). buesa ls P dk xq.kt D;k gksxk?(A) 125 (B) 375 (C) 250 (D) 147

Sol. P = f(2) . f(f(3)) f(f(f(4)) ff(f(f(5)

=

13

24

f f

35

f

46

fff

= (3)

13

1

1f

3535

f

1

1f

2323

= (3) (3) (f(4)) f(f(5))

RESONANCE PAGE - 32


= 9

35

46

f

= (15)

123

123

= (15) (5) = 75375 is multiple of 75.Ans. (B)

PHYSICS

91. The total energy of a black body radiation source is collected for five minutes and used to heat water. Thetemperature of the water increases from 10.0°C to 11.0°C. The absolute temperature of the black body is

doubled and its surface area halved and the experiment repeated for the same time. Which of the followingstatements would be most nearly correct?

d f".kdk fofdj.k ls mRlftZr dqy Å tkZ dks 5 feuV rd ,df=kr djds ikuh xeZ fd;k tkrk gSA ikuh dk rki 10.0°C ls

11.0°C gks tkrk gSA d f".kdk dk ije rki nqxquk djds ,oa {ks=kQy vk/kk djds iz;ksx dks mrus gh le; (5 feuV) ds fy,

nksgjk;k trk gSA buesa ls dkSulk dFku lcls lgh yxrk gS \(A) The temperature of the water would increase from 10.0°C to a final temperature of 12°C

ikuh dk rki 10.0°C ls 12°C gks tk;sxkA(B) The temperature of the water would increase from 10.0°C to a final temperature of 18°C

ikuh dk rki 10.0°C ls 18°C gks tk,xkA(C) The temperature of the water would increase from 10.0°C to a final temperature of 14°C

ikuh dk rki 10.0°C ls 14°C gks tk,xkA(D) The temperature of the water would increase from 10.0°C to a final temperature of 11°C

ikuh dk rki 10.0°C ls 11°C gks tk,xkASol. H

1 = AT4 × 5 = Ms

1

H2 =

2A

(2T)4 × 5 = msq2

2 = 8

1

2 = 8ºC

The temperature of water would increase from 10ºC to a final temperature of 18ºC

92. A small asteroid is orbiting around the sun in a circular orbit of radius r0 with speed V0. A rocket is launchedfrom the asteroid with speed V = V0 where V is the speed relative to the sun. The highest value of forwhich the rocket will remain bound to the solar system is (ignoring gravity due to the asteroid and effect ofother planets)

,d xzfgdk V0 osx ls lw;Z ds pkjksa vksj r0 f=kT;k ds oÙkh; iFk esa pDdj yxkrh gSA bl xzfgdk ls V = V0 osx ls ,d jkdsV

iz{ksfir fd;k tkkrk gS] tgkW V lw;Z ds lkis{k osx gSA dk mPpre~ eku D;k gksxk tks jkdsV dks lkSj eaMy ls ca/k j[ksxh \

(xzfgdk ,oa vU; xzgksa ds xq:Roh; izHkko dks vuns[kk dhft,A)

(A) 2 (B) 2 (C) 3 (D) 1

Sol. Mechanical energy of asteroid

= 21

mv2 � 0r

GMm

Rocket will remain band to the solar system its B moving energies negative.

0mv21

rGMm 2

0

2

0

mv21

rGMm

RESONANCE PAGE - 33


20

2

0

vm21

rGMm

mv02 =

21

m2v02

= 2 .

93. A radioactive nucleus A has a single decay mode with half life A. Another radioactive nucleus B has twodecay modes 1 and 2. If decay mode 2 were absent, the half life of B would have been A/2. If decay mode 1

were absent, the half life of B would have been 3 A. If the actual half life of B is B , then the ratio A

B

is

,d jsfM;ksa lfØ; ukfHkd A ,dy&{k; iz.kkyh single decay mode) ls {kf;r gksrk gS vkSj bl izfØ;k dh v/kZ vk;q A gSA

,d nwljs jsfM;ksa lfØ; ukfHkd B dh nks {k; iz.kkfy;kW 1 vkSj 2 gSaA ;fn {k; iz.kkyh 2 miyC/k ugha gksrh rks B dh v/kZ vk;q

A/2 gksrh gSA ;fn {k; iz.kkyh 1 miyC/k ugha gksrh rks B dh v/kZ vk;q 3 A gksrh gSA ;fn B dh okLrfod v/kZ vk;q B gS rc

vuqikr A

B

dk D;k eku gksxk \

(A) 73

(B) 27

(C) 37

(D) 1

Sol. B = 1 + 2

AAAB 32n7

32n2n22n

= 73

A

B

.

94. A stream of photons having energy 3 eV each impinges on a potassium surface. The work function ofpotassium is 2.3 eV. The emerging photo-electrons are slowed down by a copper plate placed 5 mm away.If the potential difference between the two metal plates is 1V, the maximum distance the electrons can moveaway from the potassium surface before being turned back is

3 eV Å tkZ ds QksVku d.k ,d iksVsf'k;e ds IysV dh lrg ij iM+rs gSaA iksVsf'k;e dk dk;Z Qyu (work function) 2.3

eV gSA mRlftZr QksVks bysDVªkWuksa dks 5 mm nwj j[ks rkacs ds IysV ls /khek fd;k tkrk gSA ;fn nksuksa IysVksa ds chp dk foHkokarj

1V gks rks] ihNs ykSVus ds igys bysDVªkWu iksVksf'k;e IysV ls vf/kdre~ fdruh nwjh rd tk ldrs gSa \(A) 3.5 mm (B) 1.5 mm (C) 2.5 mm (D) 5.0 mm

Sol. 3eV = 2.3 eVK

max = 3 � 2.3 = 0.7 eV

5 mm 1V1V 5mm0.7 V 5 × 0.7 mm

= 3.5 mm.

95. Consider three concentric metallic spheres A, B and C of radii a, b, c respectively where a<b<c. A and B areconnected whereas C is grounded. The potential of the middle sphere B is raised to V then the charge on thesphere C is

/kkrq ds rhu ladsUnzh xksys A, B ,oa C dh f=kT;k,a a, b, c gSa tgkW a<b<cA A ,oa B tqM+s gSa] tcfd C dks HkwlaifdZr fd;k x;k

gSA ;fn chp okys xksys B dk foHko c<+kdj V fd;k tkrk gS rc C xksys ij vkos'k D;k gksxk \

(A) � 40V bcbc

(B) + 40V bcbc

(C) � 40V acac

(D) zero 'kwU;

RESONANCE PAGE - 34


Sol. V = c

KQb

Kq

0c

)Qq(k

q + Q = 0q = �Q

Vc

KQ)Q(

bK

Vb

1

c

1KQ

Q = 04)cb(

bcV

.

96. On a bright sunny day a diver of height h stands at the bottom of a lake of depth H. Looking upward, he cansee objects outside the lake in a circular region of radius R. Beyond this circle he sees the images of objectslying on the floor of the lake. If refractive index of waler is 4/3, then the value of R is :

,d mTtoy fnu esa h yackbZ dk ,d xksrk[ksj H xgjkbZ okys ,d rkykc ds ry ij [kM+k gSA Å ij dh vksj ns[kus ij og

rkykc ds ckgj dk n'; R f=kT;k ds oÙk ds nk;js esa ns[k ldrk gSA bl oÙk ds ckgj og rkykc ds ry ij fLFkr oLrqvks

ds izfrfcEc gh ns[k ikrk gSA ;fn ty dk viorZukad 4/3 gS rks R dk eku D;k gksxk \

(A) 7

)hH(3 (B) 7h3 (C)

37

)hH( (D)

35

)hH(

Sol.

34

(sinC) = 1 sinC = 43

tanC = 17

3 =

hHR

R = )hH(7

3 .

RESONANCE PAGE - 35


97. As shown in the figure below, a cube is formed with ten indentical resistance R (thick lines) and two shortingwires (dotted lines) along the arms AC and BD.

uhps n'kkZ, fp=k esa ,d ?ku 10 leku ifjek.k okys izfrjks/k R ¼eksVh js[kk,a½ rFkk Hkqtkvksa BD ,oa AC ds vuqfn'k nks y|qifFkr

rkjksa ls cuk gqvk gSA

Resistance between point A and B is

A ,oa B ds chp dk izfrjks/k D;k gksxk \

(A) 2R

(B) 6R5

(C) 4R3

(D) R

Sol.

W.S. BridgeRAB = R/2.

98. A standing wave in a pipe with a length L = 1.2 m is described y

,d L = 1.2 ehVj yackbZ ds ikbi eas ,d vizxkkeh rjax dks bl Qyu ls fu:fir fd;k tkrk gSA

y (x, t) = y0 sin

x

L2

sin

4x

L2

Based on above information, which one of the following statement is incorrect.(Speed of sound in air is 300 ms�1)

Å ijh nh xbZ tkudkjh ds vk/kkj ij fuEufyf[kr esa ls dkSu lk dFku xyr gS \ ¼/;ku ns fd ok;q esa /ofu dh xfr 300

ms�1 gSA)(A) A pipe is closed at both ends

ikbi nksuksa fljksa ij can gSA(B) The wavelength of the wave could be 1.2 m

rjax dk rjaxnS/;Z 1.2 m gks ldrk gSA(C) There could be a nods at x = 0 and antinode at x = L/2

x = 0 ij fuLian (node) ,oa x = L/2 ij izLian (antinode) gks ldrk gSA

RESONANCE PAGE - 36


(D) The frequency of the fundamental mode of vibrations is 137.5 Hz

daiu dh ewy fo/kk (fundamental mode of vibrations) dh vkofÙk 137.5 Hz gSASol. From equation

2 =

L4

= 2L

= 0.6 m.

Ans. (B)

99. Two block (1 and 2) of equal mass m are connected by an ideal string (see figure below) over a frictionlesspulley. The blocks are attached to the ground by springs having spring constants k1 and k2 such thatk1 > k2 .

fp=k esa ,d vkn'kZ jLlh ls ca/ks leku nzO;eku m ds nks xqVds (1 ,oa 2) ,d ?k"kZ.k jfgr f?kjuh ls yVds gSaA nksuksa xqVdksa

dks nks dekuh (spring) ftudk dekuh fu;rkad k1 ,oa k2 gS (k1 > k2) ds ek/;e ls tehu ls tksM+ fn;k x;k gSA

Initially, both springs are unstretched. The block 1 is slowly pulled down a distance x and released. Just afterthe release the possible value of the magnitudes of the acceleration of the blocks a1 and a2 can be

'kq: esa nksuksa dekfu;ka esa dksbZ f[kapko ¼ruko½ ugha gSA xqVds 1 dks /khjs ls uhps x nwjh rd [khapdj NksM+ fn;k tkrk gSA NksM+us

ds rRdky ckn nksuksa xqVdksa ds Roj.k a1 ,oa a2 ds ifjek.k buesa ls dkSu gks ldrs gS \

(A) either

m2

x)kk(aa 21

21 or vFkok

g

m

xkaandg

m

xka 2

21

1

(B)

m2

x)kk(aa 21

21 only dsoy

(C)

m2

x)kk(aa 21

21 only dsoy

(D) either

m2

x)kk(aa 21

21 or vFkok

g

m)kk(

x)kk(aa

21

2121

RESONANCE PAGE - 37


Sol.

T + k1x � mg = ma

k2x + mg � T = ma

T + k1x � mg = k

2x + mg � T

2T = (k2 � k

1)x + 2mg

T = mg2

x)kk( 12

a = m2

x)kk( 21 If T is not zero

If T is 0 then

a1 =

mmgxk1

= gm

xk1

a2 =

mmgxk2

= gm

xk2

100. A simple pendulum is released from rest at the horizontally stretched position. When the string makes anangle with the vertical, the angle which the acceleration vector of the bob makes with the string is givenby

,d ljy yksyd dks {kSfrt ls [khaph gqbZ fLFkfr ls fLFkj voLFkk esa NksM+k tkrk gSA tc jLlh Å /oZ ls dks.k cukrh gS] ml

le; yksyd ds Roj.k lfn'k ,oa jLlh dh chp dk dks.k D;k gksxk \

(A) = 0 (B) = tan�1

2tan

(C) = tan�1 (2 tan ) (D) = 2

Sol.

at = g sin ........... (i)

ac = v2/ ........... (ii)

RESONANCE PAGE - 38


0 + 0 = 21

mv2 � mg cos ........... (iii)

aC = v2/ = 2g cos

at = g sin

tan = at/a

c

= g sin/2g cos

= 2

tan

=

2

tantan 1

Ans. (B)

CHEMISTRY

101. The final major product obtained in the following sequence of reactions is :fuEufyf[kr vuqØfed vfHkfØ;kvksa dk eq[; vafre mRikn D;k gS \

(A) (B) (C) (D)

Ans. (D)

Sol.

102. In the DNA of E. Coli the mole ratio of adenine to cytosine is 0.7. If the number of moles of adenine in the DNAis 350000, the number of moles of guanine is equal to :E Coli ds DNA esa ,sMsuhu vkSj lkbVkslhu dk eksyj vuqikr 0.7 gSA ;fn DNA esa ,sMsuhu ds eksyksa dh la[;k 350000 gSrc Xokuhu ds eksyksa dh la[;k D;k gksxh \(A) 350000 (B) 500000 (C) 225000 (D) 700000

Ans. (B)

Sol.G

3500007.0

CA

GT

G = 7.0

350000 = 5,00,000 Ans.

103. (R)-2-bromobutane upon treatment with aq. NaOH gives(R)-2-czkseksC;qVsu vkSj tyh; NaOH vfHkfØ;k dk mRikn D;k gksxk \

(A) (B)

(C) (D)

Ans. (C)

RESONANCE PAGE - 39


Sol.

104. Phenol on treatment with dil. HNO3 gives two products P and Q. P is steam volatile but Q is not. P and Q are

respectivelyQhukWy dk ruq HNO

3 (dil. HNO

3) ds lkFk foospu djus ij nks mRikn P vkSj Q curs gSaA P Hkki }kjk ok"i'khy (steam

volatile) gS tc fd Q ,slk ugha gSA P vkSj Q Øe'k% D;k gksaxs \

(A) and (rFkk) (B) and (rFkk)

(C) and (rFkk) (D) and (rFkk)

Ans. (A)Sol. Phenol on treatment with dil HNO

3 gives o�nitrophenol and p-nitrophenol.

o�nitrophenol has intra molecular H�bonding hence steam volatile ie P, where as �Q� is p-nitrophenol, Q has

inter molecular H-bonding.

105. A metal is irradiated with light of wavelength 660 nm. Given that the work function of the metal is 1.0 eV, thede-Broglie wavelength of the ejected electron is close to :,d /kkrq dks 660 nm rjax}S/;Z ds izdk'k ls fdjf.kr (irradiated) fd;k tkrk gSA Kkr gS fd /kkrq dk dk;Z&Qyu (Work

function) 1.0 eV gSA fu"dkflr bysDVªku dh Mh&czksXyh (de-Broglie) rjax}S/;Z buesa ls fdl ds fudV gS :(A) 6.6 × 10�7 m (B) 8.9 × 10�11 m (C) 1.3 × 10�9 m (D) 6.6 × 10�13 m

Ans. (C)Sol. Kinetic energy xfrt Å tkZ = h � work function dk;Z Qyu

KE = 660

1240 � 1

KE = 0.878 eV

= V

150Å

= 878.0

150

= 13.07 Å = 1.3 × 10�9 m

106. The inter-planar spacing between the (2 2 1) planes of a cubic lattice of length 450 pm is :,d ?ku tkyd (Cubic lattice) dh yEckbZ 450 pm gSA blds (2 2 1) ryksa dh varj&ry nwjh D;k gS \(A) 50 pm (B) 150 pm (C) 300 pm (D) 450 pm

Ans. (B)

Sol. 222 Kh

a

= 222 )1()2()2(

450

=

3450

= 150 pm

RESONANCE PAGE - 40

KVPY QUESTION PAPER - STREAM (SB / SX)107. The H for vaporization of a liquid is 20 kJ/mol. Assuming ideal behaviour, the change in internal energy for

the vaporization of 1 mole of the liquid at 60ºC and 1 bar is close to :

,d nzo ds ok"iu dh ,UFkSYih] H dk eku 20 kJ/mol gSA ;fn vkn'kZ O;ogkj ekus rks] 60ºC rkieku vkSj 1 bar nkc ijnzo ds 1 mol ds ok"iu ds nkSjku vkarfjd Å tkZ esa gq, ifjorZu dk eku fuEu ds fudV gS :(A) 13.2 kJ/mol (B) 17.2 kJ/mol (C) 19.5 kJ/mol (D) 20.0 kJ/mol

Ans. (B)Sol. H = 20 kJ/mol

U = H � ng RT

U = 20 × 103 � 1 × 8.314 × (273 + 60)

U = 20 � 2.768

U = 17.2 kJ/mol

108. Among the following, the species that is both tetrahedral and diamagnetic is :buesa ls dkSu prq"Qydh; vkSj izfrpqEcdh; nksuksa gh gS \(A) [NiCl

4]2� (B) [Ni(CN)

4]2� (C) Ni(CO)

4(D) [Ni(H

2O)

6]2+

Ans. (C)Sol. As in Ni(CO)

4 hybridisation of Ni is sp3 and CO is strong field effect ligand therefore, it is diamangetic.

Ni(CO)4 esa Ni dk ladj.k sp3 gS rFkk CO izcy {ks=k fyxs.M gS blfy, ;g izfrpqEcdh; gSA

109. Three moles of an ideal gas expands reversibly under isothermal condition from 2 L to 20 L at 300 K. Theamount of heat-change (in kJ/mol) in the process is :,d vkn'kZ xSl ds rhu eksy 300 K ij lerkih voLFkk esa 2 L ls 20 L rd mRØe.kh; rjhds ls izlkfjr gksrs gSaA bl izfØ;kesa Å "ek ifjorZu dk ifjek.k (kJ/mol bdkbZ esa) D;k gS :(A) 0 (B) 7.2 (C) 10.2 (D) 17.2

Ans. (D)

Sol. W = � nRT ln1

2VV

W = � 3R × 300 ln 10

= 1000

314.8900� × 2.3 = � 17.2 kJ/mol.

q = � W = 17.2 kJ/mol. (For isothermal process lerkih; izØe ds fy, E = 0).

110. The following data are obtained for a reaction, X + Y Products.Expt. [X

0]/mol [Y

0]/mol rate/mol L�1 s�1

1 0.25 0.25 1.0 × 10�6

2 0.50 0.25 4.0 × 10�6

3 0.25 0.50 8.0 × 10�6

The overall order of the reaction is :(A) 2 (B) 4 (C) 3 (D) 5vfHkfØ;k X + Y mRIkkn ds fy, fuEufyf[kr vkadM+s izkIr gq,Aiz;ksx vuqØe [X

0]/eksy [Y

0]/eksy nj/eksy L�1 s�1

1 0.25 0.25 1.0 × 10�6

2 0.50 0.25 4.0 × 10�6

3 0.25 0.50 8.0 × 10�6

bl vfHkfØ;k dh lexz dksfV (overall order of the reaction) D;k gS \(A) 2 (B) 4 (C) 3 (D) 5

Ans. (D)Sol. Rate nj = K . [X]p [Y]q

From expt. 1 and 2, p = 2 and from expt. 1 and 3, q = 3. Therefore, over all order = 5.iz;ksx 1 o 2 ls p = 2 rFkk iz;ksx 1 o 3 ls q = 3Ablfy, lEiw.kZ dksfV = 5 gSA

RESONANCE PAGE - 41


BIOLOGY111. When hydrogen peroxide is applied on the wound as a disinfectant, there is frothing at the site of injury,

which is due to the presence of an enzyme in th skin that uses hydrogen peroxide as a substrate to produce(A) hydrogen (B) carbon dioxide (C) water (D) oxygen

Ans (D)tc gkbMªkstu ijvkDlkbM ?kko ij jksxk.kq & uk'kd ds rkSj ij yxk;k tkrk gS] rc >kx curk gSA bldh otg Ropk esamifLFkr fd.od (enzyme) gS tks gkbMªkstu ijvkDlkbM ls fuEufyf[kr dks eqDr djrk gSS &(A) gkbZMªkstu (B) ikuh (C) dkCkZuMkbZ vkDlkbM (D) vkDlhtu

112. Persons suffering from hypertension (high blood pressure) are advised a low-salt diet because(A) more salt is absorbed in the body of a patient with hypertension(B) high salt leads to water retention in the blood that further increases the blood pressure(C) high salt increases nerve conduction and increases blood pressure(D) high salt causes adrenaline release that increases blood pressure

Ans (B)mPp jDrpki ls ihfMr O;fDr dks ued de ysus dh lykg nh tkrh gSA D;ksafd](A) mPp jDrpki ds dkj.k ued dk vo'kks"k.k T;knk gksrk gS

(B) T;knk ued ds dkj.k [kqu esa ikuh dk izfr/kkj.k gksrk gS ftlds dkj.k jDrpki c<+rk gSA

(C) T;knk ued ls raf=kdk mÙksftr gksarh gS vkSj jDrpki c<+rk gSA

(D) T;knk ued ls ,WfMªufyu dk lzko.k gksrk gS] ftlds dkj.k jDrpkii c<+rk gSAAns (B)

113. Insectivorous plants that mostly grow on swampy soil use insects as a source of(A) carbon (B) nitrogen (C) phosphorous (D) magnesiumdhV&Hk{kh ikS/ks nyny esa ik, tkrs gSA bu ikS/kksa dks dhMksas ls D;k feyrk gS\(A) dkCkZu (B) ukbVªkstu (C) QkLQksjl (D) eSXuhf'k;e

Ans (B)

114. In cattle, the coat colour red and white are two dominant traits, which express equally in F1 to produce roan(red and white clour in equal proportion). If F1 progeny are self-bred, the resulting progeny in F2 will havephenotypic ratio (red:roan:white) is -(A) 1 : 1 : 1 (B) 3 : 9 : 3 (C) 1 : 2 : 1 (D) 3 : 9 : 4,d izdkj ds i'kqvksa dh [kky ds nks izcy fo'ks"kd jax] Hkwjk vkSj lQsn gSa tks F1 larfr esa cjkcj vuqikr esa izdV gkrs gS]tks fd gYds Hkwjs jax ds gksaxsA vxj F1 larfr dk Loiztuu fd;k tk, tc F2 larfr ds Hkwjk % gYdk Hkwjk% lQsn jaxksa dsley{k.kh; (phenotypic) vuqikr D;k gksxk\(A) 1:1:1 a (B) 3:9:3 (C) 1:2:1 (D) 3:9:4

Ans (C)

115. The restriction endonuclease EcoR-I recognises and cleaves DNA sequence as shown below -5� � G A A T T C � 3�

3� � C T T A A G � 5�

What is the probable number of cleavage sites that can occur in a 10 kb long random DNA sequence?(A) 10 (B) 2 (C) 100 (D) 50

Ans (B)

EcoR-1 ,d fjfLVªD'ku ,UMksU;wDyh,l fd.od gS tks fd fuEufyf[kr DNA vuqØe dks igpku dj mldk foHkktu djrkgS5� � G A A T T C � 3�

3� � C T T A A G � 5�

,d 10 kb yEcsa la;ksfxd DNA vuqØe Ecor-I }kjk fdrus foHkktuksa ds gksus dh laHkkouk gS\(A) 10 (B) 2 (C) 100 (D) 50

Ans (B)

RESONANCE PAGE - 42

KVPY QUESTION PAPER - STREAM (SB / SX)116. Which one of the following is true about enzyme catalysis ?

(A) the enzyme changes at the end of the reaction(B) the activation barrier of the process is lower in the presence of an enzyme(C) the rate of the reaction is retarded in the presence of an enzyme(D) the rate of the reaction is independent of substrate concentration

Ans (B)

fd.od (enzyme) mRizsj.k (catalysis) ds ckjs esa dkSu lk fuEufyf[kr dFku lgh gS\

(A) vfHkfØ;k ds ckn fd.od esa cnyko vkrk gSSA

(B) fd.od ds tfj;s lfØ;.k jks/k (activation barrier) de gks tkrk gS

(C) fd.od ds tfj, vfHkfØ;k dh xfr de gks tkrh gSA

(D) vfHkfØ;k dh xfr dk;ZnzO; (substrate) lkaUnzrk ij fuHkZj ugha djrhAAns (B)

117. Vibrio cholerae causes cholera in humans. Ganga water was once used successfully to combat the infec-tion. The possible reason could be -(A) high salt content of Ganga water(B) low salt content of Ganga water(C) presence of bacteriophages in Ganga water(D) presence of antibiotics in Ganga water

Ans (C)euq";ksa esa dkWysjk jksx fofcz;ksa dkWysjs dhVk.kq ls gksrk gSA xaxkty ds lsou ls bl laØe.k dks jksdk tk ldrk FkkA bldk D;kdkj.k gk ldrk gS\(A) xaxkty esa ued dh ek=kk vf/kd gSA (B) xaxkty esa ued dh ek=kk de gS

(C) xaxkty esa thok.kqHkksth (bacteriophages) gSA (D) xaxkty esa izfrtSfodh (antibiotics ) gSaAAns (C)

118. When a person begins to fast, after some time glycogen stored in the liver is mobilized as a source ofglucose. Which of the following graphs best represents the change of glucose level (y-axis) in his blood,starting from the time (x-axis) when he begins to fast ?

(A) (B)

(C) (D)

Ans (C)

RESONANCE PAGE - 43


miokl 'kq: djus ds dqN le; (time x v{k½ ds ckn ;d r ds Xykbdkstu ds HkaMkj dk Xywdkst (glucose y v{k½

esa ifjorZu 'kq: gksrk gSA fuEufyf[kr fdl xzkQ esa miokl djus okys ds jDr esa Xywdkst dh ek=kk dks lgh n'kkZ;k

x;k gS\

(A) (B)

(C) (D)

Ans (C)

119. The following sequence contains the open reading frame of a polypeptide. How many amino acids will thepolypeptide consist of ?5� � AGCATATGATCGTTTCTCTGCTTTGAACT�3

(A) 4 (B) 2 (C) 10 (D) 7Ans (D)

fuEufyf[krcc DNA ds vuqØe esa ,d isiVkbM dk vksiu jhfMax Ýse gSA ;g isiVkbM fdrus vehuksa ,flM dk gksxk\ 5'AGCATATGATCGTTTCTCTOGCTTTGAACT - 3'(A) 4 (B) 2 (C) 10 (D) 7

Ans (D)

120. Insects constitute the largest animal group on earth. About 25-30% of the insect species are known to beherbivores. In spite of such huge herbiore perssure, globally, green plants have persisted. One possiblereason for this persistence is :(A) food preference of insects has tended to change with time(B) herbivore insects have become inefficient feeders of green plants(C) herbivore population has been kept in control by predators(D) decline in reproduction of herbivores with time

Ans (C)iFoh ds izk.kh lewg esa dhVksa dh ek=kk lcls T;knk gSA 25% - 30% dhVd tkfr 'kkdkgkjh gSA fQj Hkh ouLifr dh ek=kk esacnyko ugha vk;k gSA bldk dkj.k D;k gks ldrk gS\(A) dhVksa dh [kk| izkf;drk le; ds lkFk ifjofrZr gksrh jgrh gSA(B) 'kkdkgkjh dhV gjs ikniksa ds vi;kZIr Hk{kd gksrs gSaA(C) 'kkdkgkjh;kjsa ds leqnk; dks ijHkf{k;ksa }kjk fu;af=kr j[kk tkrk gSA(D) 'kkdkgkjh dhVksa ds iztuu esa le; ds lkFk deh gksrh tkrh gSA

Ans (C)

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STREAM - SB/SX - · PDF fileŁThe Test Booklet consists of 120 ... interview all the 6 persons one after the other subject to the condition that no child ... KVPY QUESTION PAPER -