Kvpy 2012 Sb Sx Solution1

8/13/2019 Kvpy 2012 Sb Sx Solution1

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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 :

MATHEMATICS

Question No. 1 to 20 consist of ONE (1) mark for each correct response.

PHYSICS


CHEMISTRY


BIOLOGY


PART-II :

MATHEMATICS

Question No. 81 to 90consist of TWO (2) marks for each correct response.

PHYSICS


CHEMISTRY


BIOLOGY


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

KISHORE VAIGYANIK PROTSAHAN YOJANA - 2012

STREAM - SB/SX


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KVPY QUESTION PAPER - STREAM (SB / SX)

PART-IOne Mark Questions

MATHEMATICS

1. 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; dsiz kpk;Z,d ds ckn ,dlHkh 6 O;fDr;ksals bl izdkj lk{kkRdkj djuk pkgrs gS fd fdlh Hkh cPPksdk lk{kkRdkj mlds vfHkHkkod ds igys u gks A ;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=

8

720= 90 ways

2. In the real number system, the equation 1x43x! + 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 1x43x! + 1x68x ""! = 1 dsfy,

(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 1x68x "!

1x" 3 = 1x68x ""!

Ans. (D)Infinite many solutions.

3. The maximum value M of 3x+ 5x9x+ 15x25x, as x varies over reals, satisfies

O;atd 3x+ 5x9x+ 15x25x, 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 = 3xand rFkk b = 5x

Therefore blfy, ,3x+ 5x9x+ 15x25x = a + b a2+ ab b2

=2

1[2 (a 1)2(b 1)2(a b)2]

# 2

1

(2)

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

4. Suppose two perpendicular tangents can be drawn from the origin to the circle x2+ y26x 2py + 17 = 0,for some real p. Then |p| is equal to

;fn fdlh okLrfod p ds fy, x2+ y26x 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


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Sol.

Equation of chord of contact T = 0 w.r.t. origin is

3(x + 0) p (y + 0) + 17 = 03x + py 17 = 0By Homoginization

x2+ y2(6x + 2py) $%

&'(

) !17

pyx3+ 17

2

17

pyx3$%

&'(

) != 0

* For perpendicular coeff of x2+ coeff of y2= 0

1 17

3.6+

.9

17+ 1

17

p2 2+

17

p2= 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 = 2x 3+ 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 bIf a c = 0 & d b -0* Max. (a c)2+ (b d)= 0 + 5 Ans.

6. Consider the conic ex2+ .y22e2x 2.2y + 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+ .y22e2x 2.2y + e3+ .3= .e ij ;fn P dksbZ fcUnq gks rFkk S1, S

2'kkado dsukfHkd gks ] rc

(PS1+ PS

2) dk vf/kdre eku gksxk&

(A) .e (B) e. (C) .2 (D) e2Sol. e(x22 ex + e2) + .(y2 2.y + .2) = .e

/ 0 / 0e

yex22 .

!.

= 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 + 1) -f(x) for any 0 < 1 < 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 < 1 < 2. ds fy, f(x + 1) -f(x)

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


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(D) f ,d fu;r Qyu gSA

Sol. f(x) =asinxsin2

acos.xsin2(x -n., n 2I)

= cot a(D) = constant function

8. The value of tan 813tan 633tan273+ tan93is

O;atd tan 813

tan 633

tan273+ tan93dk eku gS&(A) 0 (B) 2 (C) 3 (D) 4Sol. tan813tan633tan273+ tan93

cot93cot273tan273+ tan93(cot93+ tan93) (cot273+ tan 273)2 cosec1832 cosec543

2 $$%

&''(

)

!" 15

4

15

4

8 $%

&'(

)4

2

= 4Ans. (D)

9. The midpoint 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)4

1(B)

2

3(C)

3

2(D)

5

2"

Sol. 4 5x2 ! 40, 2x + 5 40

0 #2x + 5 #16

5 #2x #11

2

5"#x #

2

11

* mid point is2

2

11

2

5!"

=2.2

6.=

2

3

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 theinterval [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 a

eku ysa fd n ,d izkdr 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 + af 5(x) = (2n + 1)(x2n1)

x =

1 6point of local maximax = 1 6point of local minima


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graph of y = f(x) when a = 0

f(x) is decreasing on [1, 1]. Thus, f(x) will have atmost one root on [1,1] that too will happen if f(1) 40 andf(1) # 0 i.e. for a 2[2n, 2n]Answer (C)

11. Let f : R 6R 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

32 R. Then

f(x) 40 if and only if(A) at least two of a

1, a

2, a

3are equal (B) a

1= a

2= a

3

(C) a1, a

2, a

3are all distinct (D) a

1, a

2, a

3are all positive and distinct

;fn f : R 6R ,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

32 R rc

f(x) 40 ;fn vkSj dsoy ;fn(A) a

1, a

2, a

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

1= a

2= a

3

(C) a1, a

2, a

3rhuksa fofHkUu gSA (D) a

1, a

2, a

3rhuksa /kukRed vkSj fofHkUu gSA

Sol. f(x) = 3x22(a1+ a2+ a3) x + a1a2+ a2a3+ a3a140* D #04(a

1+ a

2+ a

3)24.3.(a

1a

2+ a

2a

3+ a

3a

1) #0

21a +

22a +

23a a1a2a2a3a3a1#0

2

1[(a

1a

2)2+ (a

2a

3)2+ (a

3a

1)2] #0

* a1= a

2= a

3

Ans. (B)

12. The value

/ 2

2 1

0

/ 2

2 1

0

(sinx) dx

(sinx) dx

.

!

."

7

7is

/ 2

2 1

0

/ 2

2 1

0

(sinx) dx

(sin x) dx

.!

."

7

7dk eku gS&

(A)12

12!(B)

12

12

!

"(C)

2

12!(D) 22

Sol. I1= 7

. 2/

0

2)x(sin (sinx)1dx

I1=

2/

0

2)x)(sinx(cos.

$%&'

()" + 2

12

2/

0

2 )x(sin)x(cos ".

7 dx

I1= 2 $$%

&

''(

)

" 77.

!

.

" dxxsindx)x(sin

2/

0

12

2/

0

12


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I1= 2 (I2I1) Here I1= 7

."

2/

0

12)x(sin dx

2

1

88

=)12(

2

! $

$%

&''(

)

"

"

)12(

12

= 22 " Ans.

13. The value 7 !!2012

2012

53 dx)1x)x(sin( is

7 !!2012

2012

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

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

Sol. 7 !

2012

2012

53 dx)xx(sin

+ 7

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

75

0

dx}x{]x[ is

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

7

5

0

dx}x{]x[ dk eku gksxk&

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

Sol. 75

0

dx}x]{x[

= dx}x{.0

1

0

7 + 72

1

dx}x{.1. + 73

2

dx}x{2 + 74

3

dx}x{3 + 75

4

dx}x{.4

= 0 + 1 71

0

dx.x + 2 7 !1

0

3dx.x 9 7 !1

0

4xdx .

7

1

0

xdx

= (1 + 2 + 3 + 4) 71

0

dx.x

= 10 .

1

0

2

2

x

::;

?

= 10 . $%

&'(

)" 0

2

1 = 5

Ans. (B)


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15. Let Sn= @ A

n

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

1, S

2, S

3, ....S

99are written

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

eku ysaSn= @ A

n

1kk izFke n /kukRed iw.kkdksads ;ks x dksfu:fir djrk gS A la[;k,aS

1, S

2, S

3,...S

99dks99 i=kksaij fy[kk

x;k gSA blesals ,d i=k ftlij le la[;k va fdr gks] ds [khapus dh izkf;drk gksxh&

(A) 2

1

(B) 100

49

(C) 99

49

(D) 99

48

Sol. Sn @

A

n

1k

K =/ 0

2

1nn !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] =99

49Ans.

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 esa4 rkacs vkSj 3 pkanh ds flDds gSA ,d nwljh FkSyh esa 6 rkacsvkSj 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)70

41(B)

70

31(C)

70

27(D)

3

1

Ans. (A)

Sol. req. prob = :

;

?!

10

6

7

4

2

1

=70

41

70

4240

2

1A

!9 Ans.

17. Let H be the orthocentre of an acuteangled 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 dsUnzgS ,oa O bldk ifjdsUnz gSA rc HCHBHA !!

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

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

Sol.

OH = cba"""

!!

HA = )cb( "" !


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HB = )ac(""

!

HC = )ba(""

!

HCHBHA !! = 2( cba"""

!! )

= 2 OH = 2 HOAns. (C)

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

Sol. 6m + 9n

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

19. Suppose a1

, a2

, a3

, ....., a2012

are integers arranged on a circle. Each number is equal to the average of itstwo adjacent numbers. If the sum of all even indeced numbers is 3018, what is the sum of all numbers?

eku ysafd iw.kkd a1, a

2, a

3, ....., a

2012dks,d o k ij lt;k x;k gSA izR;sd la[;k viuh fudVorhZnks la [;kvksadh vkSlr

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

Sol. a2=

2

aa 31!

a3=

2

aa 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 9S . A subset B of A is said to be a good subset if

(x, x) 2B for every x 2S. Then the number of good subsets of A is;fn S = 1, 2, 3, ...., n} rFkk A = {(a, b) | 1 #a, b #n} = S 9S . A ds mileqPp; B dks vPNk mileqPp; rc dgktk,xk tc izR;sd x 2S ds fy, (x, x) 2B gksA rc A ds vPNs mileqPp;ksa dh la[;k gksxh&

(A) 1 (B) 2n (C) 2n(n1) (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 n2n elements.

*Number of subsets =nn

nn2

nn1

nn0

nn2

2222

C......CCC !!!

=nn22


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PHYSICS

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

done by the gas is Wi. If the process is adiabatic the magnitude of work done by the gas is W

a. Which of

the following is true

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

x;k dk;ZWi gSA ;fn izf;k :)ks"e gS rks xSl }kjk fd;k x;k dk;Z WagSA fuEufyf[kr esa dkSu lgh gS \

(A) Wi = Wa> 0 (B) W i> Wa> 0 (C) W i> Wa= 0 (D) Wa> Wi= 0

Sol.

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

So, wi> wa > 0Ans. (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 frjks/k gSA

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

is :

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

(A)2

RC(B) RCln2 (C) 2RCln2 (D) 2

2lnRC

Sol. Q = Q0et/RC

2

Q0= Q

0et/RC

2

1= et/RC Ans. (D)

RC

t

2

1n "A#

2nRC

t#A

t = RC2

2n#


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23. A liquid drop placed on a horizontal plane has a near spherical shape (slightly flattened due to gravity). Let R

be the radius of its largest horizontal section. A small disturbance causes the drop to vibrate with frequency

Babout its equilibrium shape. By dimensional analysis the ratio

3R

c

C

Dcan be (Here Dis surface tension,

Cis 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 vkfr dh gS xq:Rokd"kZ.k ds dkj.k cwan FkksM+h lh QSy tkrh gSxksys dslcls cM+s {kSfrt vuq LFk dh f=kT;k R gSA ,d NksVk fo{kksHk cwan dks blds lkE;koLFkk vkfr ds ifjr% vkofk B

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

3R

c

C

Dvuqikr fuEufyf[kr esa dkSulk gks ldrk gS\ ;gk Di"B ruko] C?kuRo]

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

(A)D

C 2gRk(B)

DCg

Rk 3

(C)D

Cg

Rk 2

(D) DC

g

k

Sol. 3R/CD

B=

DC

B3R

=

2/1

2

331

MT

LMLT

::;

?"

""

= T1T

= 1

KR/ 3

ACD

B

2

3

2 KR ADCB

K2=2

3

TR "

DC

=2

2

RTR "

DC

K = gR2

DC

Ans. (A)

24. Seven identical coins are rigidly arranged on a flat table in the pattern shown below so that each coin touches

its 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 is2

mr 2.

,d {kSfrt est ij leku vkdkj dslkr flDds n


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PAGE - 11


The moment of inertia of the system of seven coins about an axis that passes through the point P (the centre

of 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 flDdksads lewg dk fcUnq P ls xqtjrs gq, ,oa flDdsds ya cor v{k ds lkis{k tM+Ro vk?kw.kZD;k

gksxk\ P dsUnzh; flDds ds nkfgus fLFkr flDds dk dsUnzgSA

(A)2

55mr2 (B)

2

127mr2 (C)

2

111mr2 (D) 55 mr2

Sol.

8P=

::;

?!!9

::;

?!!9

::;

?!! 2

22

22

22

)r4(m2

mr2)32r(m

2

mr3)r2(m

2

mr

2

mr

=2mr

2

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 denoted

by A. Let vPand vAbe the respective speeds at P and A. Then

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

ok ds,d ukfHk ij fLFkj gSA rkjs ds lcls utnhd ds fcUnq dks P ,oa lcls nwj dsfcUnq dks A.ls n'kkZ;k x;k gSA ;fn vP,oavA e'k% P vkSj A ij xzg dk osx gSrks

(A) e1

e1

v

v

A

P

"!

A (B)A

P

v

v= 1 (C)

e1

e1

v

v 2

A

P

"!

A (D) 2

2

A

P

e1

e1

v

v

"

!A


12/43


13/43

PAGE - 13


Sol.

B0=

:;

? .!

...

E

nsin

nsin

n

cosR

i

40

B0=

ntan

R

i

2

0 ..

E

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 uniform

magnetic 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 LyackbZ 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 energy2

1mv0

2

(A) will be converted fully into heat energy in the resistor

(B) will enable rod to continue to move eith velocity v0since 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 tkZ2

1mv

02

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

(B) NM+ dks v0xfr 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; tkZ esa :ikUrfjr gks tk,xhA(D) pqEcdh; {ks=k ds f[kykQ dk;Z djus esaO;; gks tk,xhA

Sol. Due to the negative work on rod K.E. will decrease and finally become zero.Ans. (A)

29. A steady current 8flows through a wire of radius r, length L and resistivity C. 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 is

independent of

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

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

ij fuHkZj ugha gS\(A) L (B) r (C) 8 (D) C


14/43

PAGE - 14


Sol. R =A

#C

R = 2r

L

.

C

82R = K2.rLdt

dT

2

2

r

L

.

C8= K2.rL

dt

dT

dt

dT= 32

2

r2K

L

.

C8Ans. (A)

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

is close to

300K rkieku ,oa1 atm nkc ij /ofu dh xfr vkSj ok;q ds,d v.kq dh vkSlr xfr dk vuqikr fuEu esa lsfdlds fudV

gS \

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

1(D) 300

Sol.

m

RT8

m

RT

.

C

=85

7.= 0.73

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

31. In one model of the election, tho electron of mass meis thought to be a uniformly charged shell of radius Rand total charge e, whose electrostatic energy E is equivalent to its mass mevia Einstein's mass energy

relation E = mec2 . In this model, R is approximately (me = 9.1 10

31 kg, c = 3 108 m.s1,

04

1

.F = 9 109 Farads m1, magnitude of the electron charge = 1.6 1019C)

bysDVkWu ds ,d fu:i.k esa menzO;eku ds bysDVkWu 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 tkZ E ,oa nzO;eku mevkbaLVhu ds E = mec2.laca/k ls nh tkrh

g SA bl fu:i.k e s a R dk eku fdlds djhc g S\ (m e = 9.1 1031 kg, c = 3 108 m.s1,

04

1

.F = 9 109 Farads m1, bysDVkWu dk vkos'k = 1.6 1019C)

(A) 1.4 1015m (B) 2 1013m (C) 5.3 1011m (D) 2.8 1035mSol. U = mc2

22

mcR2

KQA

R2

)106.1(109 2199 "999= 9.1 1031 (3 108)2

R = 2831

2199

)103(101.92

)106.1(109

9999

999"

"

= 1.4 10

15


15/43

PAGE - 15


Ans. (A)

32. A body is executing simple harmonic motion of amplitude a and period T about the equilibrium position

x = 0. Large numbers of snapshots are taken at random of this body in motion. The probability of the body

being found in a very small interval x to x + |dx| is highest at

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

morion) djrk gSA bl oLrqdh ;knfPNd xfr ds dbZ lkj s QksVk s fp=k snapshots [khpsa tkrs gSA x vkSj x + |dx| ds

chp NksVs varjky dsa fiaM ds ik, tkus dh lclsT;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 = aAns. (A)

33. Two identical bodies are made of a material for which the heat capacity increases with temperature One of

these is held at a temperature of 100C while the other one is kept at 0C. 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 ,slsinkFkZ ls cuk, tkrs gS ftudh "ek & /kkfjrk rkieku ds lkFk c


16/43

PAGE - 16


36. A ball of mass m suspended from a rigid support by an inextensible massless string is released from a

height h above its lowest point. At its lowest point it colIides elastically with a block of mass 2m at rest on a

frictionless surface. Neglect the dimensions of the ball and the block. After the collision the ball rises to a

maximum height of

,d


17/43

PAGE - 17


(A) (B)

(C) (D)

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

Hence, mdt

dv= 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.m1.K1is welded end to end to copper

rod of thermal conductivity 400 W.m1.K1and of the same area of cross section but 0.20 m long. The free

end of the steel rod is maintained at 100C and that of the copper rod at 0C. Assuming that the rods areperfectly insulated from the surrounding the temperature at the junction of the two rods is

LVhy dh ,d csyukdkj NM+ ftldh yackbZ 0.10m rFkk "eh; pkydrk 50 W.m1.K1gS] ,d Nksj ij leku vuqizLFk

dkV ,oa0.20m yach ,d nwljh csyukdkj NM+ftldh "eh; pkydrk 400 W.m1.K1gS] ls tqMh gSA LVhy NM+ds [kqys

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

"ek dk kl ughagks jgk gS ] NMksa ds tksM+ ij rkieku D;k gksxk\(A) 20C (B) 30C (C) 40C (D) 50C

Sol. Resistance of steel =A50

1.0

9=

A500

1

Resistance of copper =A400

2.0=

A200

1

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

$%

&'(

)

"

A500

1

T100 =

$

%

&'

(

)

"

A2000

1

0T, T = 20C

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 Y

are 4000 years and 20 years respectively. In a certain sample, it is found that the number of Y nuclei hardly

changes 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 nwljsukfHkd Y esa {kf;r gksrk gS] tks fQj Z. esa {kf;r gksrk gSA X vkSj Y dh v/kZvk;q e'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 ughagks jgk gSA

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

Sol. FromN

1J

1= N

2J

2


18/43

PAGE - 18


4 1020=40000

)2(n#=

20

)2(nN2

#

, N2= 2 107

Ans. (A)

40. An unpolarized beam of light of intensity 80passes through two linear polarizers making an angle of 30 withrespect to each other. The emergent beam will have an intensity.

nks js[kh; /kqzodksa ds chp] tks ijLij 30 ij fLFkj gS ,d 80rhozrk dh v/kzqfor izdk'k fdj.k xqtjrh gSA ckgj vkus okyhfdj.k dh rhozrk D;k gksxh\

(A)4

3 08(B)

4

3 08 (C)8

3 08(D)

808

Sol.

$%&'

()8A$

%&'

()8

4

3

230cos

2

020

Final intensity will be =8

3 08

Ans. (C)

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

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

(A) O2 (B) F2 (C) O2+ (D) F2

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

2, F

2, O

2+and F

2is respectively 2, 1, 2.5 and 0.5.

gy MOT dsvuqlkj O2, F

2, O

2+rFkk F

2dk ca/k e e'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

2and CCl

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

3is pyrami-

dal due to presence of lone pair.

gy BCl3, BeCl

2o CCl

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

3dh vkfr fijsfefM;y

gksrh gSD;ksafd blesa ,dkdh bysDVkWu ;qXe mifLFkr gksrk gS A

43. For a one-electron atom, the set of allowed quantum numbers is :

,d&bysDVkWu ijek.kqds fy, vuq er DokaVe la[;k,(allowed quantum numbers) gS \

(A) n = 1, # = 0, #m = 0, ms= + (B) n = 1, # = 1, #m = 0, ms= +

(C) n = 1, # = 0, #m = 1, ms= (D) n = 1, # = 1, #m = 1, ms=

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 .


19/43

PAGE - 19


44. In the reaction of benzene with an electrophile E+, the structure of the intermediate D-complex can berepresented as :

csathu dh ,d bysDVkWuLusgh (electrophile) E+dslkFk vfHkf;k esa e/;orhzD-ladj dh lajpuk dksbl izdkj iznf'kZ r djldrsgSa \

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

Ans. Option DClosest, if () is change to #$fodYi Dlgh 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.kqvksads bysDVkWfud tkZ&varjky (energy gaps)dk eku yxHkx 1.0 eV gSA rkieku dslanHkZesa;g va rj buesals fdldsfudV gS \(A) 102K (B) 104K (C) 103K (D) 105K

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

* G(KE) = G(T.E.)

2

3 8.3 (GT) = 1.6 1019 6 1023

(GT) =3.8

106.9 49

3

2

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

47. The major final product in the following reaction is :

fuEu vfHkf;k dk eq[; vafre mRikn D;k gS \

CH3CH

2CN

!KKKKK 6K

OH)2

MgBrCH)1

3

3

(A) (B)

(C) (D)


20/43

PAGE - 20


Ans. (C)

Sol.

48. A zero-order reaction, A 6Product, with an initial concentration [A]0has a half-life of 0.2 s. If one starts with

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

,d 'kwU; dksfV vfHkf;k] A 6mRikn (product), dh v)Z&vk;q 0.2 s gS] tc izkFkfed lkanzrk [A]0gSA ;fn vfHkf;k

2[A]0dh 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

1/2L(a)1 n

II

I

)t(

)t(

2/1

2/1=

01

2

1

a

a "

$$%

&''(

)

II)t(

2.0

2/1=

0

0

A2

A

(t1/2

)II= 0.4 s

49. The isoelectronic pair of ions is :

fuEu vk;fud ;qXe lebysDVkWfud (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 esa23 bysDVkWu gSA

50. The major product in the following reaction is :

fuEu vfHkf;k dk eq[; mRikn D;k gS \

KKKK 6K 2NaNH

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

Ans. (A)

Sol.

51. The major product of the following reaction is :

fuEu vfHkf;k dk eq[; mRikn D;k gS \

KKKKK 6K HBr.Conc

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

Ans. (B)

Sol.


21/43

PAGE - 21


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 pKaof 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 pKadk eku 5.85 gSA pH dsfdl 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

athe concenteration of acid and its conjugate base are equal.

pH = pKaij 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]2ds fy, dsoy izp.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 Clis weak field effect ligand. In [CoCl

4]2oxidation 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 p.k ladqy gS D;ksafd Cl,d nqcZy {ks=k fyxs.M gSA [CoCl4]2esa Co dh vkWDlhdj.k voLFkk +2 gSA blesa rhu

v;qfXer bysDVkWu mifLFkr gS tks fd 3.83 BM ds cjkcj dsoy p.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 esavfHkf;k nj ds ifjorZ u (k)dksrkieku (T) dslkFk n'kkZ ;k x;k gSA iwjsrki ifjlj esavkgsZfu;l (Arrhenius)O;ogkj dks buesa lsfdl vkjs[k }kjk n'kkZ;k x;k gS \

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

Ans. (D)

Sol. K = RT/E aAe

ln K =RT

E a+ ln AA

56. The reaction that gives the following molecule as the major product is

fuEufyf[kr v.kq fdl vfHkf;k dk eq[; mRikn gS \


22/43

PAGE - 22


(A) (B)

(C) (D)

Ans. (B)

Sol.

57. The CO bond length in CO, CO2and CO

32follows the order :

CO, CO2vkSj CO

32esaCO vkca/k dh yEckbZ dk e 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 CO32

bond order in between 1 and 2, bond order ofCO

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

gy ftldk ca/k e vf/kd gksxk mldh ca/k yEckbZ de gksxhA CO32esa ca/k e 1 o 2 ds e/; gS] CO

2dk ca/k e 2 gS tcfd

CO dk ca/k e 2 o 3 ds e/; gSA

58. The equilibrium constant for the following reactions are K1and K

2, respectively.

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

3(g)

PCl3(g) + Cl

2(g) PCl

5(g)

The the equilibrium constant for the reaction


5(g)

is :

fuEufyf[kr vfHkf;kvksa ds fy, lkE; fLFkjkad e'k% K1rFkk K2gSA2P(g) + 3Cl

2(g) 2PCl

3(g)

PCl3(g) + Cl

2(g) PCl

5(g)

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

5(g) vfHkf;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) + 3Cl2(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, K3= K1K22lehdj.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


23/43

PAGE - 23


59. The major product of the following reaction is :

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

+ (CH3C)

2CHCH

2Cl KKK 6K 3

AlCl

(A) (B)

(C) (D)

Ans. (A)

Sol. The reagent given the question has an addition Cwithin bracket.

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

3)2CHCH

2Cl + AlCl

3K6 (CH

3)

2CHCH

2Cl - - - - AlCl

3K6(CH

3)

2CHCH

2++ AlCl

4

(CH3)2CHCH

2+ KKKKKK 6K " shifftingH2,1

(CH

3)3C+

60. Doping silicon with boron produces a :(A) n-type semiconductor (B) metallic conductor(C) p-type semiconductor (D) insulator

flfydu ds cksjkWu vifeJ.k (Doping) ls dkSu&lk mRikn curk gS \

(A) n-izdkj ds v)Zpkyd (B) /kkfRod pkyd(C) p-izdkj dsv)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 lsdkSu 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) IgEAns (A)


24/43

PAGE - 24


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 gksukAAns (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)

VkUlQj (tRNA)(A) jkbckslkse esafLFkr gS vkSj lajpukRed v[kaMrk nsrk gSA(B) lkekU;r% frifr;k (clover leaf-like) vkdkj dh gksrh gSA(C) DNA lsjkbckslkse 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 okysiz k.kh dks ;wfjdksVsfyd dgrs gSA ;g izf;k ikuh ds laj{k.k dsfy, vuqdwy gSA buesals dkSu lk izk.kh ;wfjdksVsfyd gS\(A) eNfy;k (B) i{kh (C) mHk;pj (D) Lru/kkjh

Ans (C)


25/43

PAGE - 25


68. A ripe mango, kept with unripe mangoes causes their ripening. This is due to the release of a gaseous planthormone(A) auxin (B) gibberlin (C) cytokinine (D) ethylene

Ans (D)

,d ids vke dks dPpsvkeksa ds lkFk j[kus ls dPPks vke idus 'kq: gks tkrs gSA ;g fdl xS lh; 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 p ds nkSjku ksekslkse dh lajpuk esa cnyko gksrk gSA bl cnyko dks Li"V :i ls ns[kus ds fy, pdh fdl izkoLFkk esa ksekslkse 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 izf;k ds }kjk xqnsZ ds fuL;aan }kjk Xykses:yj fQYVsV ls Xyqdkst iqu% 'kksf"kr gksrk gS\(A) ijklj.k (osmosis) (B) folj.k (diffusion)(C) lf; xeu (active transport) (D) fuf"; 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 p ds ckjs esa fuEufyf[kr dkSu lk dFku lgh gS\

(A) ekbVksdkfUM;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)


26/43

PAGE - 26


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 mutations

Ans (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/kksadh tM+ esa DyksjksfQy ugha gksrk gS] ftlds dkj.k os izdk'k la'ys"k.k (photosynthesis)ugha dj ikrsA buesals 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'kdkvksads jksMkW fIlu 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 lae.k ls dbZjksfx;ksa esa vkUrfjd jDrlzko gksrk gS A 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 DNAdssence strand dk vuqe, 5'-GTTCATCG-3, 5'-GTTCATCG-3 5'-GTTCATCG-3 gS] rksmlds RNAdh izfrfyfi (transcript) dk vuqe 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)


27/43

PAGE - 27


fdlh mhiu ds vuSfPNd vuqf;k dks izfrorhZ f;k dgrs gSA buesa ls dkSu lk mnkgj.k vizkuqdwyh vkSj izkuqdwyh izfrorhZf;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 6 deer 6lion, 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 tkZ ds vuqikr esa izR;sd Lrj ij 'olu 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 roots

eku ysa fd a, b, c okLrfod la[;k,agS 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= 4a

2

4b

2

> 0 , 4 (a2

b

2

) > 0a2b2> 0 .......... (1)D

2= 4b24c2> 0 , b2c2> 0 ............. (2)

now D = 4c24a2= 4(c2b2+ b2a2)= 4 (b2c2+ a2b2)= 4 (D

1+ D

2)

= Negative

* Equation have no real root.

82. The coefficient of x2012in)x1)(x1(

x12 "!

!is

)x1)(x1(

x12

"!

!esax2012dk ,d xq.kkad gS&

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

(1+x) @M

A

"0r

rr2 )1(x @M

A0r

rx = @M

A

"0r

rr2 )1(x @M

A0r

rx +@M

A

"0r

rr2 )1(x @M

A

!

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


28/43

PAGE - 28


83. Let (x, y) be a variable point on the curve 4x2+ 9y28x 36y + 15 = 0. Thenmin (x22x + y24y + 5) + max(x22x + y24y + 5) is

;fn o 4x2+ 9y28x 36y + 15 = 0 ij (x, y) ,d pjfcUnq gS] rcmin (x22x + y24y + 5) + max(x22x + y24y + 5) dk eku gksxk&

(A)36

325(B)

325

36(C)

25

13(D)

13

25

Sol. 4x2

+ 9y2

8x

36y + 15 = 04(x2 2x + 1) + 9(y2 4y + 4) 25 = 04(x 1)2+ 9 (y 2)2= 25

2

2

2

2

3

5

)2y(

2

5

)1x(

$%

&'(

)

"!

$%

&'(

)

"= 1

min / 022 )2y()1x( "!" + max. / 0/ 022 )2y(1x "!"

=

2

3

5$%

&'(

)+

2

2

5$%

&'(

)=

36

225100!=

36

325

Ans. (A)

84. The sum of all x 2[0, .] which satisfy the equation sinx +2

1cosx = sin2(x +

4

.) is

;fn x 2[0, .] lehdj.k sinx +2

1cosx = 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 +

2

1cos x = sin2 $

%

&'

(

) .!

4

x

sinx +2

1cosx =

/ 02

2/x2cos1 .!"

sinx +2

1cosx =

2

1

2

1! 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 =

2

1

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 55(x) -0 for all x. Suppose P(0) = 1 andP5(0) = 1. What can you say about P(1)?cgqin P(x) ds okLrfod xq.kkad bl izdkj gS fd lHkh x ds fy, P55(x) - 0. eku yssa fd P(0) = 1 rFkkP5(0) = 1 rc vki P(1) ds ckjs esa D;k dg ldrs gS ?

(A) P(1) 40 (B) P(1) -0 (C) P(1) #0 (D)2

1< P(1) 1. Then

1n

n

n a

alim

"M6

(A) equals2

1(B) equal 1 (C) equals

5

2(D) does not exists

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

1= 5, an= a

1a

2... a

n1+ 4 rc

1n

n

n a

alim

"M6dk eku

(A)2

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

5

2ds cjkcj gSA (D) vfLrRo ugha gSA

Sol. a1

= 5a

n= a

1a

2a

3.......a

n 1+ 4

=n

n1n321

a

aa......aaa+ 4

an=

n

1n

a

4a !+ 4

an2= a

n + 14 + 4a

n

an+1

= an24a

n+ 4 = (a

n2)2

1na ! = an2

n

1n

aa ! = 1

na2 ,

1n

n

aa = 1

1na1

M6nlim n

n 1

a

a "= M6n

limn 1

21

a "

) &' $( %

= 1 ( ! M6nlim a

n= M)

87. The value of the integral dxa1

xcos

x

2

7.

. ! , where a > 0, is

lekdydx

a1

xcos

x

2

7

.

. !,tgk

a > 0dk eku gS&

(A) . (B) a. (C)2

.(D) 2.

Sol. I= 7.

. !

x

2

dxa1

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

I= 7.

. !

x

2

dxa1

)x(cos

I= 7.

. !

x

2xdx

a1

xcosa ...(2)


30/43

PAGE - 30


(1) + (2)

2I= dx)a1(

)a1(xcos

x

x2

7.

. !

!

2I= dxxcos2

0

27.

I=00

1 cos2x 1 sin2xdx x

2 2 2

. .! ) &

A !' $( %7

I= / 0 / 0/ 001 sin2 0 sin02

. ! . ! =2

..

88. Consider

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

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

and I= dxx33012

2012

7 .

eku ysa fd

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

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

rFkk I= dxx33012

2012

7 rc

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

= I

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

Area of trapezoid AA5B5B < 72013

2012

3/1 dxx

i.e.2

)2013()2012( 3/13/1 !< 7

2013

2012

3/1 dxx

Similarly,2

)2014()2013( 3/13/1 !< 7

2014

2013

3/1 dxx

2

)2015()2014( 3/13/1 !< 7

2015

2014

3/1 dxx

...................................

and so on


31/43

PAGE - 31


2

)3012()3011( 3/13/1 !< 7

3012

3011

3/1 dxx

Adding we get

2

RL!< 7

3012

2012

3/1 dxx

2

RL!< 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) >2

1is

,d vkneh ,d flDds dks 10 ckj mNkyrk gSA mls izR;sd fpk 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 djusdh izkf;drk P(K) gks] rc P(K) >2

1ds 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 xKin (x1+ x2)10

= x10(1 + x)10

coeff xk10in (1 + x)10

= 10CK10

K 410

Now P(K) >2

1

10

10K10

210

110

010

2

C......CCC "!!!!4

2

1

10C0+ 10C

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

K10> 29

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

0+ 10C

1+ 10C

2+ 10C

3+ 10C

4+ 10C

5> 512

K 10 = 5K = 15Ans. (B)

90. Let f(x) =1x

1x

"!

for all x -1. Let f1(x) = f(x), f2(x) = f (f(x)) and generally fn(x) = f(fn1(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 lHkhx -1

ds fy,f(x) =

1x

1x

"

!gSA ;fn

f1(x) = f(x), f2(x) = f (f(x))vkSj

n > 1ds fy, lkekU;r%

fn(x) = f(fn1(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)

= $%

&'(

)1

3 $$

%

&''(

)$%

&'(

)2

4f f $$

%

&''(

)$%

&'(

)3

5f

$$

%

&

''

(

)$$%

&''(

)$$%

&''(

)$%

&'(

)4

6fff

= (3) $%

&'(

)1

3 $

$

%

&

''

(

)

$$

%

&

''

(

)

"

!

1

1f

35

35

f $$

%

&

''

(

)

$$

%

&

''

(

)

"

!

1

1f

23

23

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


32/43

PAGE - 32


= 9 $%

&'(

)3

5 $$

%

&''(

)$%

&'(

)4

6f

= (15)$$$$

%

&

''''

(

)

"

!

12

3

12

3

= (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. The

temperature of the water increases from 10.0C to 11.0C. The absolute temperature of the black body isdoubled and its surface area halved and the experiment repeated for the same time. Which of the following

statements would be most nearly correct?

df".kdk fofdj.k ls mRlftZr dqy tkZ dks5 feuV rd ,df=kr djds ikuh xeZ fd;k tkrk gSA ikuh dk rki 10.0C ls

11.0C gkstkrk gSA df".kdk dk ije rki nq xquk djds ,oa {ks=kQy vk/kk djdsiz;ks x dksmrus gh le; (5 feuV) ds fy,

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

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

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

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

ikuh dk rki 10.0C ls 11C gkstk,xkASol. H

1= DAT4 5 = MsGN

1

H2= D

2

A(2T)4 5 = msGq

2

GN2= 8GN

1

GN2= 8C

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

92. A small asteroid is orbiting around the sun in a circular orbit of radius r0with speed V

0. A rocket is launched

from the asteroid with speed V = 1V0where V is the speed relative to the sun. The highest value of 1forwhich the rocket will remain bound to the solar system is (ignoring gravity due to the asteroid and effect of

other planets)

,d xzfgdk V0osx ls lw;Zds pkjksa vks j r0f=kT;k dso kh; iFk esapDdj yxkrh gS A bl xzfgdk lsV = 1V0osx ls ,d jkdsV

iz{ksfir fd;k tkkrk gS] tgkW V lw;Z ds lkis{k osx gSA 1dk mPpre~ eku D;k gksxk tksjkdsV dks lkSj eaMy ls ca/k j[ksxh \(xzfgdk ,oa vU; xzgksa dsxq:Roh; iz Hkko dks vuns[kk dhft,A)

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

Sol. Mechanical energy of asteroid

=2

1mv2

0r

GMm

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

0mv2

1

r

GMm 2

0

A!"

2

0

mv2

1

r

GMmA


33/43

PAGE - 33


20

2

0

vm2

1

r

GMm1A

mv02=

2

1m12v

02

1= 2 .

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

were absent, the half life of B would have been 3 OA. If the actual half life of B is OB, then the ratioA

B

OO

is

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

,d nwljs jsfM;ksa lf; 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

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

vuqikrA

B

O

Odk D;k eku gksxk \

(A)7

3(B)

2

7(C)

3

7(D) 1

Sol. JB= J1+ J2

AAAB 3

2n7

3

2n2n22n

OA

O!

OA

O####

= 7

3

A

B AOO

.

94. A stream of photons having energy 3 eV each impinges on a potassium surface. The work function of

potassium 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 move

away from the potassium surface before being turned back is

3 eV 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 QksVksbysDVkWuksadks 5 mm nwj j[ksrkacsds IysV ls/khek fd;k tkrk gS A ;fn nksuksaIys Vksa dschp dk foHkoka rj

1V gks rks] ihNs ykSVusds igys bysDVkWu 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

P= 2.3 eVK

max= 3 2.3 = 0.7 eV

5 mm Q1V1V Q5mm0.7 V Q5 0.7 mm = 3.5 mm.

95. Consider three concentric metallic spheres A, B and C of radii a, b, c respectively where a


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Sol. V =c

KQ

b

Kq!

0c

)Qq(kA

!

q + Q = 0q = Q

Vc

KQ)Q(bK A!"

Vb

1

c

1KQ A$

%

&'(

)"

Q = 04)cb(

bcV2.

" .

96. On a bright sunny day a diver of height h stands at the bottom of a lake of depth H. Looking upward, he can

see objects outside the lake in a circular region of radius R. Beyond this circle he sees the images of objects

lying 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 ij dh vksj ns[kus ij og

rkykc ds ckg j dk n'; R f=kT;k ds ok ds nk;jsesa ns[k ldrk gSA bl ok 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)

3

7

)hH( "(D)

3

5

)hH( "

Sol.

3

4(sinC) = 1 , sinC =

4

3

tanC =17

3=

hH

R

"

R = )hH(7

3" .


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97. As shown in the figure below, a cube is formed with ten indentical resistance R (thick lines) and two shorting

wires (dotted lines) along the arms AC and BD.

uhpsn'kkZ , fp=k esa ,d ?ku 10 leku ifjek.k okysiz frjks/k Reks Vh js[kk,a rFkk Hkqtkvksa BD ,oa ACdsvuq fn'k nksy|q ifFkr

rkjksa ls cuk gqvk gSA

Resistance between point A and B is

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

(A)2

R(B)

6

R5(C)

4

R3(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 :;

?

$%

&'(

) .xL

2

sin :;

? .

!$%

&'(

) .4xL

2

Based on above information, which one of the following statement is incorrect.

(Speed of sound in air is 300 ms1)

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

ms1gSA)(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


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

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

J.2

=L

4., J=

2

L= 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 k1and k2such that

k1> 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,oak2gS (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 after

the release the possible value of the magnitudes of the acceleration of the blocks a1and a

2can be

'kq: esa nksuksadekfu;ka esa dksbZf[ka pko rukoughagS A xqVds1 dks /khjs ls uhpsx nwjh rd [khapdj NksM+fn;k tkrk gSA Nks M+us

ds rRdky ckn nksuksa xqVdksa ds Roj.k a1,oa a2dsifjek.k buesa ls dkSu gks ldrs gS \

(A) either $%

&'(

) !AA

m2

x)kk(aa 2121 orvFkok $

%

&'(

)!A"A g

m

xkaandg

m

xka 22

11

(B) $%

&'(

) !AA

m2

x)kk(aa 2121 only dsoy

(C) $%

&

'(

) "

AA m2

x)kk(

aa21

21 only dsoy

(D) either $%

&'(

) "AA

m2

x)kk(aa 2121 orvFkok $$

%

&''(

)"

!AA g

m)kk(

x)kk(aa

21

2121


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Sol.

T + k1x mg = ma

k2x + mg T = ma

T + k1x mg = k

2x + mg T

2T = (k2k

1)x + 2mg

T = mg2

x)kk( 12 !"

a =m2

x)kk( 21!If T is not zero

If T is 0 then

a1=

m

mgxk1 "= g

m

xk1 "

a2=

m

mgxk2 != g

m

xk2 !

100. A simple pendulum is released from rest at the horizontally stretched position. When the string makes an

angle Nwith the vertical, the angle Pwhich 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 /oZ ls Ndks.k cukrh gS] mlle; yksyd ds Roj.k lfn'k ,oa jLlh dh chp dk dks.k PD;k gksxk \

(A) P= 0 (B) P= tan1

$

%

&'(

) N

2

tan

(C) P= tan1

(2 tan N) (D) P= 2

.

Sol.

at= g sinN ........... (i)a

c= v2/# ........... (ii)


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0 + 0 =2

1mv2mg # cosN ........... (iii)

aC= v2/# = 2g cosN

at= g sinN

tanP= at/a

c

= g sinN/2g cosN

=2

tanN

P= $%

&'(

) N"2

tantan 1

Ans. (B)

CHEMISTRY

101. The final major product obtained in the following sequence of reactions is :

fuEufyf[kr vuqfed vfHkf;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 DNA

is 350000, the number of moles of guanine is equal to :E Coli ds DNAesa ,sMsuhu vkSj lkbVkslhu dk eks yj vuqikr 0.7gSA ;fn DNAesa ,sMsuhu ds eksyksa dh la[;k 350000gSrc Xokuhu ds eksyksa dh la[;k D;k gksxh \(A) 350000 (B) 500000 (C) 225000 (D) 700000

Ans. (B)

Sol.G

3500007.0

C

A

G

TAAA

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 vfHkf;k dk mRikn D;k gksxk \

(A) (B)

(C) (D)

Ans. (C)


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Sol.

104. Phenol on treatment with dil. HNO3gives two products P and Q. P is steam volatile but Q is not. P and Q are

respectivelyQhukWy dk ruq HNO3(dil. HNO

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

volatile) gS tc fd Q ,slk ugha gSA PvkSj Q e'k% D;k gksaxs \

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

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

Ans. (A)

Sol. Phenol on treatment with dil HNO3gives onitrophenol and p-nitrophenol.

onitrophenol has intra molecular Hbonding hence steam volatile ie P, where as Qis p-nitrophenol, Q hasinter 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 nmrjax}S/;Z ds izdk'k ls fdjf.kr (irradiated) fd;k tkrk gSA Kkr gS fd /kkrq dk dk;Z&Qyu (Workfunction)1.0 eVgSA fu"dkflr bysDVku dh Mh&czksXyh (de-Broglie)rjax}S/;Z buesa ls fdl ds fudV gS :(A) 6.6 107m (B) 8.9 1011m (C) 1.3 109m (D) 6.6 1013m

Ans. (C)

Sol. Kinetic energy xfrt tkZ = hRwork function dk;Z Qyu

KE =660

12401

KE = 0.878 eV

J=V

150

J=878.0

150

J= 13.07 = 1.3 109m

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 pmgSA 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

!!=

3

450= 150 pm


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107. The GH for vaporization of a liquid is 20 kJ/mol. Assuming ideal behaviour, the change in internal energy forthe vaporization of 1 mole of the liquid at 60C and 1 bar is close to :

,d nzo ds ok"iu dh ,UFkSYih] GH dk eku 20 kJ/molgSA ;fn vkn'kZ O;ogkj ekus rks] 60C rkieku vkSj 1 barnkc ijnzo ds 1molds ok"iu ds nkS jku vkarfjd tkZ esagq, 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. GH = 20 kJ/mol

GU = GH GngRT

GU = 20 1031 8.314 (273 + 60)GU = 20 2.768GU = 17.2 kJ/mol

108. Among the following, the species that is both tetrahedral and diamagnetic is :

buesa ls dkSu prq"Qydh; vkSj izfrpqEcdh; nksuksagh 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)

4hybridisation of Ni is sp3and CO is strong field effect ligand therefore, it is diamangetic.

Ni(CO)4esa Ni dk ladj.k sp3gS 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. The

amount of heat-change (in kJ/mol) in the process is :,d vkn'kZxSl dsrhu eks y 300 Kij lerkih voLFkk esa 2 Lls 20 Lrd mRe.kh; rjhds ls iz lkfjr gksrsgSa A bl izf;kesa "ek ifjorZu dk ifjek.k (kJ/molbdkbZ esa ) D;k gS :(A) 0 (B) 7.2 (C) 10.2 (D) 17.2

Ans. (D)

Sol. W = nRT ln1

2

V

V

W = 3R 300 ln 10

=1000

314.8900 9 2.3 = 17.2 kJ/mol.

q = W = 17.2 kJ/mol. (For isothermal process lerkih; ize ds fy, GE = 0).

110. The following data are obtained for a reaction, X + Y 6Products.Expt. [X

0]/mol [Y

0]/mol rate/mol L1s1

1 0.25 0.25 1.0 106

2 0.50 0.25 4.0 106

3 0.25 0.50 8.0 106

The overall order of the reaction is :(A) 2 (B) 4 (C) 3 (D) 5

vfHkf;k X + Y 6mRIkkn ds fy, fuEufyf[kr vkadM+s izkIr gq,Aiz;ksx vuqe [X

0]/eksy [Y

0]/eksy nj/eksy L1s1

1 0.25 0.25 1.0 106

2 0.50 0.25 4.0 106

3 0.25 0.50 8.0 106

bl vfHkf;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


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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 gkbMkstu 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 gkbMkstu ijvkDlkbM ls fuEufyf[kr dks eqDr djrk gSS &

(A) gkbZMkstu (B) ikuh (C) dkCkZuMkbZvkDlkbM (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 dksued 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


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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) vfHkf;k dsckn fd.od esa cnyko vkrk gSS A

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

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

(D)vfHkf;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.kqls gksrk gS A xaxkty dslsou ls bl lae.k dksjksdk 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 esathok.kqHkksth (bacteriophages)gS A (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)


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miokl 'kq: djusds dqN le; (time x v{k ds ckn ;dr ds Xykbdkstu ds HkaMkj dk Xywdkst (glucose y v{k

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 ?5AGCATATGATCGTTTCTCTGCTTTGAACT3(A) 4 (B) 2 (C) 10 (D) 7

Ans (D)

fuEufyf[krcc DNA ds vuqe esa ,d isiVkbM dk vksiu jhfMax se 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 dhVk sa 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 dsleqnk; dks ijHkf{k;ksa }kjk fu;af=kr j[kk tkrk gSA(D) 'kkdkgkjh dhVksa ds iztuu esa le; dslkFk deh gksrh tkrh gSA

Ans (C)

Kvpy 2012 Sb Sx Solution1

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