Kvpy Sb Solutions

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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 90 consist of TWO (2) marks for each correct response.

PHYSICS


CHEMISTRY


BIOLOGY


Date : 30-10-2011 Duration : 3 Hours Max. Marks : 160

KISHORE VAIGYANIK PROTSAHAN YOJANA - 2012

STREAM - SB / SX



PAGE - 2

KVPY QUESTION PAPER - STREAM (SB / SX)

PART-IOne Mark Questions

MATHEMATICS

1. Suppose logab + log

ba= c. The smallest possible integer value of c for all a,b > 1 is

(A) 4 (B) 3 (C) 2 (D) 1

Sol. (C) Let loga

b = x then, logba =

x

1

C = x +x

1, For a, b > 1 log

ab > 1

By A.M. G.M

C =x

1x 2

2. Suppose n is a natural number such that | i + 2i2 + 3i3 + ....+ nin | = 18 2 . Where i is the square root of

–1 . Then n is.(A) 9 (B) 18 (C) 36 (D) 72

Sol. (C)Hint n is multiple of = 4

3. Let P be an m × m matrix such that P2 = P . Then (I+P)n equals(A) I + P (B) I + nP (C) I + 2nP (D) I + (2n –1)P

Sol. (D) Given P2 = P

For n = 2 (I + P)2 = I + P2 + 2PI

= I + P + 2P

= I + 3P

= I + (22

–

1) P

4. Consider the cubic equation x3 + ax2 + bx + c = 0 , where a,b,c are real numbers . Which of the followingstatements is correct ?(A) If a2 – 2b < 0 , then the equation has one real and two imaginary roots

(B) If a2 – 2b 0, then the equation has all real roots(C) If a2 – 2b > 0, then the equation has all real and distinct roots

(D) If 4a3 – 27b2 > 0 , then the equation has real and distinct rootsSol. (A)

f ' (x) = 3x2 + 2ax + bD = 4a2 – 4(3b)

= 4(a2 – 3b)

If a2 – 2b < 0 thena2 – 3b < 0

So, D is negative, so f(x) has one real and two imaginary roots.

5. All the points (x,y) in the plane satisfying the equation x2 + 2x sin(xy) + 1 = 0 lie on(A) a pair of straight lines (B) a family of hyperbolas(C) a parabola (D) an ellipse

Sol. (A) sin (xy) =x2

x1 2=

x

x

1

2

1

sin xy 1 or –1

sin xy = 1 or –1

For x = –1 or 1

–sin y = 1 or sin y = –1



PAGE - 3


y =

2n2 , n I

(x, y) on pair of straight line.

6. Let A = (4,0) , B = (0,12) be two points in the plane . The locus of a point C such that the area of triangleABC is 18 sq.units is(A) (y + 3x + 12)3 = 81 (B) (y + 3x + 81)2 = 12

(C) (y + 3x –

12)2

= 81 (D) (y + 3x –

81)2

= 12

Sol. (C) 18

1120

104

1yx

2

1

| –12x – y(4) + 48| = 36

(3x + y – 12)2 = 92 = 81

7. In a rectangle ABCD , the coordinates of A and B are (1,2) and (3,6) respectively and some diameter of thecircumscribing circle of ABCD has equation 2x – y + 4 = 0. Then the area of the rectangle is

(A) 16 (B) 2 10 (C) 2 5 (D) 20

Sol. (A)

m1

m2

= –1

31t2

66t4

13

26= –1

4t2

t4(2) = –1 4t = –t + 2

t =5

2

then C is

5

38,

5

1

BC =

22

5

8

5

16

=

25

320

=5

8

5

58

Area = xy = 525

8 = 16

8. In the xy –plane , three distinct lines 1,

2,

3concur at a point (,0). Further the lines

1,

2,

3are normals to

the parabola y2 = 6x at the points A = (x1

, y1) , B (x

2, y

2) , C = (x

3, y

3) respectively . Then we have

(A) < – 5 (B) > 3 (C) – 5 < < – 3 (D) 0 < < 3Sol. (B) h > 2a here 4a = 6, 2a = 3

h > 3



PAGE - 4


9. Let f(x) = cos5x + A cos 4x + Bcos3x + Ccos2x + Dcosx + E. and

T = f(0) – f

5+ f

5

2 – f

5

3+ ..............+ f

5

8 – f

5

9Then T

(A) depends on A,B,C,D,E(B) depends on A,C,E, but independent of B and D(C) depends on B,D , but independent of A,C,E(D) is independent of A,B,C,D,E

Sol. (D) f(x) = cos5x + A cos 4x + Bcos3x + Ccos2x + Dcosx + Ef( – x) = – cos5x + Acos4x – B cos3x + C cos2x – D cos x + E

f (x) + f( – x) = 2 [A cos4x + C cos2x + E) and f(x) – f( – x) = 2 [cos 5x + B cos 3x + D cos x]

for x = 0f(0) – f() = 2(1 + B + D) ...... (i)

at x =5

, f

5

– f

5

4= 2 ( –1 + B cos

5

3 + D cos

5

) ...... (ii)

at x =5

2, f

5

2 – f

5

3= 2

5

2cosD

5

6cosB1 ..... (iii)

f( + x) = – cos5x + A cos4x – B cos5x + C cos2x – D cosx + E

f ( –

x) = f ( + x) ...... (iv)

T = f(0) – f() –

5

9f

5f +

5

8f

5

2f –

5

7f

5

3f +

5

6f

5

4f

from equation IInd f

5

2= f

5

8

f

5= f

5

9

f

5

3= f

5

7and f

5

4= f

5

6

So T = f(0) – f() – 2

5

3f –

5

2f

5

4f –

5f

from equation 1, 2 & 3

T = 2(1 + B + D) – 2 [ –1 + B cos5

3 + Dcos

5

] + 2 [ 1 + B cos

5

6 + D cos

5

2] = 10

is independent of A, B, C, D, E

10. In triangle ABC , we are given that 3sinA + 4 cosB = 6 and 4sinB + 3cosA = 1. Then the measured of theangle C is(A) 30º (B) 150º (C) 60º (D) 75º

Sol. (A) 3sinA + 4cosB = 6 ...... (i)3 cosA + 4sinB = 1 .......(ii)

Squaring and adding , we get9 + 16 + 24 sin(A+B) = 37

sin(A+B) =2

1

sin( – C) =2

1

sin C =2

1



PAGE - 5


11. Which of the following intervals is possible domain of the function f(x) = log{x}

[x] + log[x]

{x} , where [x] isthe greatest integer not exceeding x and {x} = x – [x] ?

(A) (0,1) (B) (1,2) (C) (2,3) (D) (3,5)Sol. (C) [x] > 1

[x] 2

x 2But x cannot integers

Possible domain = (2,3)

12. If f(x) = (2011 + x)n , where x is a real variable and n is a positive integer , then the value of f(0) + f(0) +

!2

)0("f+ ......+

!)1 –n(

)0(f )1 –n(

is

(A) (2011)n (B) (2012)n (C) (2012)n – 1 (D) n(2011)n

Sol. (C) f(0) + f(0) + !2

)0("f+ ......+

!)1 –n(

)0(f )1 –n(

(2011) +

!1 –n

)'2011(!2...)1 –n(n......

!2

)2011)(1 –n(n,

!1

)2011(n 2 –n1 –n

= nC02011 + nC12011n –

1 + .... + nCn –1(2011) + nCn (2011)0 – 1= (2011 + 1)n – 1

= 2012n – 1

13. The minimum distance between a point on the curve y = ex and a point on the curve y = loge

x is

(A)2

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

Sol. (B)

Point where slope of tangent is 1 of the curve y = ex isy = ex = 1

x = 0

(0,1)and for y = lnx point is (1,0)

required minimum distance is 2

14. Let f : (2,) N be defined by f(x) = the largest prime factor of [x] , Then 8

2

dx)x(f is equal to

(A) 17 (B) 22 (C) 23 (D) 25

Sol. (B) 8

2

dx)x(f = 4

3

3

2

dx)x(fdx)x(f + ...... + 8

7

dx)x(f

= 2 + 3 + 2 + 5 + 3 + 7 = 22



PAGE - 6


15. Let [x] denote the largest integer not exceeding x and {x} = x – [x]. Then

2012

0})x{cos( –})x{cos(

})x{cos(

ee

edx is

equal to(A) 0 (B) 1006 (C) 2012 (D) 2012

Sol. (B) I = 2012

1

0xcosxcos

xcos

dxee

e

again I = 2012

1

0xcosxcos

xcos

dxee

e

2I = 2012 1

0

dx

I = 1006

16. The value of

2222n n –n4

1............

4 –n4

1

1 –n4

1lim is

(A)4

1(B)

12

(C)

4

(D)

6

Sol. (D)

2222n n –n4

1............

4 –n4

1

1 –n4

1lim

= nlim

n

1r 22 r –n4

1

=n

lim

n

1r 2

n2

r –1n2

1

=2

1 1

02

4

x –1

dx

= 1

02x –4

dx

=

1

0

1

2

xsin

17. Two players play the following game : A writes 3,5,6 on three different cards ; B writes 8 ,9,10 on threedifferent cards . Both draw randomly two cards from their collections . Then A computes the product of twonumbers he/she has drawn , and B computes the sum of two numbers he/she has drawn . The playergetting the larger number wins. What is the probability that A wins ?

(A)3

1(B)

9

5(C)

9

4(D)

9

1



PAGE - 7


Sol. (C)

E1

: A wins when A takes out 3 & 6 and B takes out 8 & 9

P (E1) =

3

1×

3

1=

9

1

E2 : A wins when A takes out 5 & 6 and B takes out any two no.

P (E2) =

3

1× 1 =

3

1

Required probability =9

1+

3

1=

9

4

18. Let c,b,a

be three vectors in the xyz space such that ba

= cb

= ac

0

. If A,B,C are points with

position vectors c,b,a

respectively , then the number of possible positions of the centriod of triangle ABC

is(A) 1 (B) 2 (C) 3 (D) 6

Sol. (A) accbba

........(i)

0cbba

0bcba

0bca

bca

........(ii)

ca

b

( cb

) = ac

cca

= ac

acca

0caca

(1 + ) ( ca

) = 0 = –1

From (ii) bca

0cba

centroid of ABC = 03

cba

19. The sum of (12 – 1 + 1) (1!) + (22 – 2 + 1)(2!) + .... + (n2 – n + 1)(n!) is

(A) (n + 2) ! (B) ( n – 1) ((n + 1) !) + 1

(C) (n + 2)! – 1 (D) n((n + 1)!) – 1

Sol. (B)

n

1r

2 !r1rr

Tr= [(r + 1) (r – 1) – (r – 2)] r!

Tr= (r – 1) (r + 1)! – (r – 2) (r!)

but r = 1, 2, .............., n

Tr

= 1 + (n + 1)! (n – 1)



PAGE - 8


20. Let X be a nonempty set and let P(X) denote the collection of all subsets of X. Define f : X × P(X) R by

f(x,A) =

Ax,0

Ax,1.Then f(x , AA B) equals

(A) f(x, A) + f(x ,B)(B) f(x ,A) + f (x ,B) – 1

(C) f (x , A) + f(x , B) – f (x ,A) f( x ,B)

(D) f( x ,A) + | f( x ,A) – f( x ,B) |

Sol. (C) If f (x , A B) =

Ax,0

Ax,1

Case (i) x A, x Bthen,f(x, A) + f(x, B) – f(x, A) f(x, B) = 1 + 0 – 0 = 1

Case (ii) x A, x Bf(x, A) + f(x, B) – f(x, A) f(x, B) = 0 + 1 – 0 = 1

Case (iii) x A, x Bf(x, A) + f(x, B) – f(x, A) f(x, B) = 0 + 0 – 0 = 0

Case (iv) x A and x Bf(x, A) + f(x, B) – f(x, A) f(x, B) = 1 + 1 – 1 = 1

f(x, A B) = f(x, A) + f(x, B) – f(x, A) f(x, B)

Aliter

If f (x , A) =

BAx,0

BAx,1

Option A is rejected because x (AB) ,then f(x,A) + f(x,B) = 1+ 1 = 2 which can’t be attained by f.

Option B is rejected baecuse if x BA , then f(x,A) + f(x,B) –1 = 0 + 0 –1 = –1 which can’t be attained

by f .Also option D is rejected because if x A , then f(x,A) + |f(x,A) – f(x,B)| = 1 + |1 – 0| = 2

Which is also unattainble

PHYSICS21. A narrow but tall cabin is falling freely near the earth’s surface. Inside the cabin, two small stones A and B are

released from rest (relative to the cabin). Initially A is much above the centre of mass and B much below thecentre of mass of the cabin. A close observation of the motion of A and B will reveal that :(A) both A and B continue to be exactly at rest relative to the cabin(B*) A moves slowly upward and B moves slowly downward relative to the cabin(C) both A and B fall to the bottom of the cabin with constant acceleration due to gravity(D) A and B move slightly towards each other vertically.

Ans. (B)

Sol. Cabin is tall due to this graviation acceleration g = 2)hR(

GM

is different for A and B particle. Because A and B

are on different height. gB

> gA

so that acceleration of A less than of acceleration B. A moves slowly upwardand B moves slowly downward relative to the cabin

22. Two plates each of mass m are connected by a massless spring as shown :

A weight W is put on the upper plate which compresses the spring further. When W is removed, the entireassembly jumps up. The minimum weight W needed for the assembly to jump up when the weight isremoved is just more than :(A) mg (B) 2mg (C) 3mg (D) 4mg



PAGE - 9


Ans. (B)

Sol. kk

mg

k

W

= mg W = 2mg

23. If the speed (v) of the bob in a simple pendulum is plotted against the tangential acceleration (a), the correctgraph will be represented by :

(A) (B)

(C) (D)

(A) I (B) II (C) III (D) IVAns. (A)

Sol. v = 22 rA ...(i) a = 2r ...(ii)

from (i) and (ii)

1A

a

A

v

24

2

22

2

It is the equation of ellipse

24. A container with rigid walls is convered with perfectly insulating material. The container is divided into twoparts by a partition. One part contains a gas while the other is fully evacuated (vacuum). The partition issuddenly removed. The gas rushes to fill the entire volume and comes to equilibrium after a little time. If thegas is not ideal ,(A) the initial internal energy of the gas equals its final internal energy(B) the initial temperature of the gas equals its final temperature(C) the initial pressure of the gas equals its final pressure(D) the initial entropy of the gas equals its final entropy

Ans. (A)Sol. Work done in this process is zero so that

(Ui= U

f)

25. Two bulbs of identical volumes connected by a small capillary are initially filled with an ideal gas at tempera-ture T. Bulb 2 is heated to maintain a temperature 2T while bulb 1 remains at temperature T. Assumethroughout that the heat conduction by the capillary is negligible. Then the ratio of the final mass of the gasin bulb 2 to the initial mass of the gas in the same bulb is close to :(A) 1/3 (B) 2/3 (C) 1/3 (D) 1

Ans. (B)Sol. Using mole conservation

2n = n1

+ n2

...(i)PV = n

1RT ...(ii)

PV = n2R(2T) ...(iii)

n1

= 2n2

2n = 2n2

+ n2

n2

=

3

n2; 3

2

n

3 / n2

M

M

i2

f2



PAGE - 10


26. Two rods, one made of copper and the other steel of the same length and cross sectional area are joinedtogether. (Ther thermal conductivity of copper is 385 J.s –1. m –1. K –1 and steel is 50 J.s –1.m –1.K –1.) If the copperend is held at 100°C and the steel end is held at 0°C, what is the junction temperature (assuming no other

heat losses) ?(A) 12°C (B) 50°C (C) 73°C (D) 88°C

Ans. (D)

Sol.dx

dTKA

dt

dQ

)T0(AK)100T(AK 21

K1(T – 100) = K

2( –T)

K1T – K

1 × 100 = –K

2T

T(K1

+ K2

) = K1 × 100

T =

435

100385= 88°C

27. Jet aircrafts fly at altitudes above 30,000 ft where the air is very cold at –40°C and the pressure is 0.28 atm.

The cabin is maintained at 1 atm pressure by means of a compressor which exchanges air from outside

adiabatically. In order to have a comfortable cabin temperature of 25°C, we will require in addition :(A) a heater to warm the air injected into the cabin(B) an air conditionaer to cool the air injected into the cabin(C) neither a heater nor an air -conditioner : the compressor is sufficient(D) alternatively heating and cooling in the two halves of the compressor cycle.

Ans. (B)Sol. P1- T= constant

1

1

2

1

2P

P

T

T

1)5 / 7(5 / 7

1

28.0

T

K233

T = J466)28.0(

233

5 / 2

So that we will required in addition an air conditionaer to cool the air injected into the cabin

28. A speaker emits a sound wave of frequency f0. When it moves towards a stationary observer with speed u,

the observer measures a frequency f1. If the speaker is stationary and the observer moves towards it with

speed u, the measured frequency is f2. Then :

(A) f1

= f2

< f0

(B) f1> f

2(C) f

1< f

2(D) f

1= f

2> f

0

Ans. (B)

Sol. f1

= 0fuv

v

...(i)

f2

= 0fv

uv

...(ii)

22

2

2

1

uv

v

f

f

f1

= f2

22

2

uv

v

f1 > f2



PAGE - 11


29. A plane polarized light passed through successive polarizers which are rotated by 30° with respect to each

other in the clockwise direction. Neglecting absorption by the polarizers and given that the first polarizer’saxis is parallel to the plane of polarization of the incident light, the intensity of light at the exit of the fifthpolarizer is closest to :(A) same as that of the incident light (B) 17.5% of the incident light(C) 30% of the incident light (D) zero

Ans. (C)

Sol. = 0 (cos

2

)

4

= 0 ×

4

4

3

; = 30% of

0

30. At 23C, a pipe open at both ends resonates at a frequency of 450 hertz. At what frequency does the samepipe resonate on a hot day when the speed of sound is 4 percent higher than it would be at 23 C ?(A) 446 Hz (B) 454 Hz (C) 468 Hz (D) 459 Hz

Ans. (C)

Sol. f =

2

nv

f ’ = 2

'vn

v04.1

v

'v

v

'f

f

f ’ = (1.04)(450) = 468 Hz

31. In a Young’s double slit set-up, light from a laser source falls on a pair of very narrow slits separated by 1.0

micrometer and bright fringes separated by 1.0 millimeter are observed on a distant screen. If the frequencyof the laser light is doubled, what will be the separation of the bright fringes ?(A) 0.25 mm (B) 0.5 mm (C) 1.0 mm (D) 2.0 mm

Ans. (B)Sol. ×

2

1

2

1

f1

1= f

2

2

1

2

2

1

f

f

1

2

1

1

f2

f

1

2

2

1

1

2

2

1

2

=21

1= 0.5 mm

32. For a domestic AC supply of 220 V at 50 cycles per second, the potential difference between the terminalsof a two -pin electric outlet in a room is given by :

(A) V(t) = 220 2 Cos(100t) (B) V(t) = 220Cos(50t)

(C) V(t) = 220Cos(100t) (D) V(t) = 220 2 Cos(50t)

Ans. (A)Sol. V = V

0cost

Vrms

=2

V0

V0

= 2220 Volt

= 50 × 2 rad/sec = 100 rad/sec.



PAGE - 12


v = 220 2 Cos(100t)

33. In the circuit shown below the resistances are given in ohms and the battery is assumed ideal with emf equalto 3.0 volts. The resistor that dissipates the most power is

(A) R1

(B) R2

(C) R3

(D) R4

Ans. (A)Sol. Current through R

1is maximum so power dissipates at R

1is maximum.

34. An electron collides with a free molecules initially in its ground state. The collision leaves the molecule in aexcited state that is metastable and does not decay to the ground state by radiation. Let K be the sum of the

initial kinetic energies of the electron and the molecule and P

the sum of their initial momenta. Let K’ and

'P

represent the same physical quantities after the collision. Then

(A) K = K’, 'PP (B) K’ < K, 'PP (C) K = K’, 'PP (D) K’ < K, 'PP Ans. (B)

Sol. Momentum is conserved during collision so 'PP

. But some part of kinetic energy is stored in excited state

So K’ < K

35. In the circuit shown, the switch is closed at time t = 0

Which of the graphs shown below best reprsents the voltage across the inductor, as seen on an oscillo-scope?

(I) (II)

(III) (IV)

(A) I (B) II (C) III (D) IVAns. (D)Sol. Potential difference across inductor during current growth is



PAGE - 13


v = v0e –t/

36. Given below are three schematic graphs of potential energy V(r) versus distance r for three atomic particlesselectron (e), poton (p+) and neutron (n), in the presence of a nucleus at the origin O. The radius of the nucleusis r

0. The scale on the V –axis my not be the same for all figures. The correct pairing of each with the

corresponding atomic particle is :

(A) (1, n), (2, p+), (3, e –) (B) (1, p+), (2, e –), (3, n)(C) (1, e –), (2, p+), (3, n) (D) (1, p+), (2, n), (3, e –)

Ans. (A)Sol. Neutron is chargeless so graph is (1) Proton is positive charge so graph is (2) and for e – graph is (3)

37. Due to transitions among its first three energy levels, hydrogenic atom emits radiation at three discretewavelength

1,

2and

3(

1<

2<

3). Then :

(A) 1

= 2

+ 3

(B) 1

+ 2

= 3

(C) 1/ 1

+ 1/ 2

= 1/ 3

(D) 1/ 1

= 1/ 2

+ 1/ 3

Ans. (D)

Sol.

9

8R

1

1

4

R3

4

11R

1

2

36R5

49R5

91

41R1

3

321

111

38. The total radiative power emitted by spherical blackbody with radius R and temperature T is P. If the radius isdoubled and the temperature is halved then the radiative power will be :(A) P/4 (B) P/2 (C) 2P (D) 4P

Ans. (A)

Sol.

4

4

2TA4

AT

'P

P

4'P

P

P’ =4

P

39. The Quantum Hall Resistance RH

is a fundamental constant with dimensions of resistance. If h is Planck’s

constant and e the electron charge, then the dimension of RH

is the same as :(A) e2 /h (B) h/e2 (C) h2 /e (D) e/h2

Ans. (B)



PAGE - 14


Sol. RH

=i

V

222 e

h

i

P

i

i.V

40. Four students measure the height of a tower. Each student uses a different method and each measures theheight many different times. The data for each are plotted below. The measurement with highest precision is :

(A) I (B) II (C) III (D) IVAns. (A)Sol. From given graph highest precision that means sharpness is maximum in graph (I)

CHEMISTRY

41. Ther hydridizations of Ni (CO)4and Cr(H

2O)

62+, respectively.

(A) sp3 and d3sp2 (B) dsp2 and d2sp3

(C) sp3 and d2sp3 (D) dsp2 and sp3 d2

Sol. (C)

Ni(CO)4

Ni : [Ar] 3d8 4s2 =2,2,2

g2t 2,2

ge

CO is a strong filed ligand

3d 4s 4p

Cr (H2O)

6+2 Cr+2 : [Ar] 3d4 45°

3d 4s 4p

octahedral complex : d2sp3

42. Extraction of silver is achieved by initial complexation of the ore (Argentite) with X followed by reaction

with Y, X and Y, respectively, are :

(A) CN – and Zn (B) CN – and Cu

(C) Cl – and Zn (D) Br – and Zn

Sol. (A)



PAGE - 15


Ag+ + 2CN – [Ag(CN)2] –

2[Ag(CN)2] – + Zn+2 [Zn(CN)

4] –2 + 2Ag+

43. Assuming ideal behaviour, the enthalpy and volume of mixing of two liquids, respectively, are

(A) zero and zero (B) +ve and zero

(C) –ve and zero (D) –ve and –ve

Sol. (A)

Ideal mixtureH = 0

V = 0

44. At 298 K, the ratio of osmotic pressures of two solutions of a substance with concentrations of 0.01 Mand 0.001 M, respectively, is :

(A) 1 (B) 100 (C) 10 (D) 1000

Sol. (C)

= CRT2

1

=001.0

01.0= 10

45. The rate of gas phase chemical reactions generally increases rapidly with rise in temperature. This is mainlybecause :(A) the collision frequency increases with temperature.(B) the fraction of molecules having energy in excess of the activation energy increases with temperature.(C) the activation energy decreases with temperature.(D) the average kinetic energy of molecules increases with temperature.

Sol. (B)Energy distribution at two different temperatures

46. Among i-iv :

the compound that does not undergo polymerization under radical initiation, is :(A) i (B) ii (C) iii (D) iv

Sol. (D)

The radical formed as intermediate or is unstable , where as in all other cases

resonance stabilised radicals are formed.

47. Two possible stereoisomers for

are :(A) enantiomers (B) diastereomers (C) conformers (D) rotamers

Sol. (A)



PAGE - 16


The given compound is chiral and the possible stereoisomers are enatiomer to each other.

48. For a process to occur spontaneously.(A) only the entropy of the system must increase.

(B) only the entropy of the surrounding must increase.(C) either the entropy of the system or that of the surroundings must increase.(D) the total entropy of the system and the surroundings must increase.

Sol. (D)According to second law of thermodynamics, for spontaniety : S

UNIV> 0

49. When the size of a spherical nanoparticle decreases from 30 nm to 10 nm, the ratio surface area/volumebecomes(A) 1/3 of the original (B) 3 times the original (C) 1/9 of the original (D) 9 times the original

Sol. (B)r

1= 30 nm r

2= 10 nm

volume

areasurface

=

3

r4

r43

2

= r

3

)r / 3(

)r / 3(

1

2=

10

30= 3

50. The major product of the following reaction is :

H

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

Sol. (C)The reaction is example of Pinacole – Pinacolone reaction with classical ring expanssion rearrangement .

51. For the transformation

the reagent used is :(A) LiAlH

4(B) H

3PO

2(C) H

3O+ (D) H

2 /Pt

Sol. (B)H

3PO

2a mild reducing agent is used to remove the diazonium group with H.



PAGE - 17


52. The values of the limiting molar conductivity (º) for NaCl, HCl and NaOAc are 126.4. 425.9 and 91.0 Sccm2.respectively. For HOAC. º in S cm2 mol –1 is :(A) 390.5 (B) 299.5 (C) 208.5 (D) 217.5

Sol. (A)

0NaOAc + 0

HCl – 0NaCl

53. To obtain a diffraction peak, for a crystalline solid with interplane distance equal to wavelength of incident X-ray radiation, the angle of incidence should be :

(A) 90º (B) 0º (C) 30º (D) 60º

Sol. (D)

BRAGG’S Law

2dsin = nd = , n = 1

sin =2

1 = 30°

angle of incidence = 60°

54. The standard Gibbs free energy change (Gº in kJ mol –1), in a Daniel cell (Eºcell

= 1.1 V), when 2 moles ofZn(s) is oxidized at 298 K, is closed to :(A) – 212.3 (B) – 106.2 (C) – 424.6 (D) – 53.1

Sol. (C)

Zn Zn+2 + 2e –

2mol 2mol 4mol

G° = –nFE°cell

= – 4 × 96500 × 1.1 = – 424.6 kJ/mol

55. All the products formed in the oxidation of NaBH4

by I2, are :

(A) B2H

6and NaI(B) B

2H

6, H

2and NaI (C) BI

3and NaH (D) NaBI

4and HI

Sol. (B)

2NaBH4

+ 2 2Na + B

2H

6+ H

2

56. The spin-only magnetic moments of [Mn(CN)6]4 – and [MnBr

4]2 – in Bohr Magnetons, respectively, are :

(A) 5.92 and 5.92 (B) 4.89 and 1.73 (C) 1.73 and 5.92 (D) 1.73 and 1.73Sol. (C)

[Mn(CN)6] –4 & [(Mn/Br)

4] –2

Mn+2 in both Mn+2 : [Ar]3d5 45°

With CN –

:

With Br – :

Magnetic moments : )21(1 BM & )25(5 BM

57. In a zero-order reaction, if the initial concentration of the reactant is doubled, the time required for half thereactant to be consumed :(A) increases two-fold (B) increases four-fold (C) decreases by half (D) does not change

Sol. (A)Half life of zero order reactions initial concentration.



PAGE - 18


58. The adsorption isothermal for a gas is given by the relation x = ap/(1 + bp) where x is moles of gas adsorbedper gram of the adsorbent, p is the pressure of the gas, and a and b are constants. Then x :(A) increases with p(B) remains unchanged with p(C) decreases with p(D) increases with p at low pressure and the same at high pressure.

Sol. (D)

x =)bp1(

ap

59. The reaction

+ CHCl3

H

Heat / NaOH

is known as :

(A) Perkin reaction (B) Sandmeyer reaction(C) Reimer-Tiemann reaction (D) Cannizzaro reaction

Sol. (C)It is fact.

60. Among i-iii

the boiling point follows the order(A) ii < i < iii (B) iii < ii < i (C) i < ii < iii (D) ii < iii < i

Sol. (C)Due to intramolecular H-bonding boiling point of (ii) is less than (iii) but greater than (i).

BIOLOGY

61. The major constituents of neurofilaments are

(A) microtubules (B) intermediate filaments

(C) actin filaments (D) protofilamentsAns. (B)

62. In which phase of the cell cycle are sister chromatids available as template for repair ?

(A) G1 phase (B) G2 phase

(C) S phase (D) M phase

Ans. (C)

63. A person has difficulty in breathing at higher altitudes because

(A) oxygen is likely to diffuse from lungs to blood.

(B) oxygen is likely to diffuse from blood to lungs

(C) partial pressure of O2is lower than partial pressure of CO

2

(D) overall intake of O2 by the blood becomes low.



PAGE - 19


Ans. (D)

64. In humans , the composition of a zygote that will develop into a female is

(A) 44A +XX (B) 44A + XY

(C) 22 + X (D) 23 A

Ans. (A)

65. If you fractionate all the organelles from the cytoplasm of a plant cell. In which one of the following sets

of fractions will you find nucleic acids ?

(A) nucleus, mitochondria, chloroplast, cytoplasm

(B) nucleus, mitochondria, chloroplast, glyoxysome

(C) nucleus, chloroplast , cytoplasm and peroxisome

(D) nucleus, mitochondria, chloroplast, Golgi bodies

Ans. (A)

66. A protein with 100 amino acid residues has been translated based on triplet genetic code. Had the

genetic code been quadruplet. the gene that codes for the protein would have been.

(A) same in size (B) longer in size by 25%

(C) longer in size by 100% (D) shorter in size

Ans. (B)

67. If the sequence of base in DNA is 5'- ATGTATCTCAAT- 3', than the sequence of bases in its transcript

will be :

(A) 5' - TACATAGAGTTA - 3'

(B) 5' - UACAUAGAGUUA - 3'

(C) 5' - AUGUAUCUCAAU - 3'

(D) 5' - AUUGAGAUACAU - 3'

Ans. (D)

68. The Na+ /K+ pump is present in the plasma membrane of mammalian cells where it

(A) expels potassium from the cell

(B) expels sodium and potassium from the cell.

(C) pumps sodium into the cell.

(D) expels sodium from the cell.

Ans. (D)

69. The CO2

in the blood is mostly carried

(A) by haemoglobin in RBCs

(B) in the cytoplasm of WBCs

(C) in the plasma as bicrbonate ions

(D) by plasma proteins

Ans. (C)

70. Patients who have undergone organ transplants are given anti-rejection medications to(A) minimize infection

(B) stimulate B- macrophage cell interaction

(C) prevent T- lymphocyte proliferation

(D) adopt the HLA of donor

Ans. (C)

71. Saline drip is given to a Cholera patient because

(A) NACl kills Vibrio cholera

(B) NACl generates ATP

(C) Na+ ions stops nerve impulse and hence sensation of pain

(D) Na+ ions help in retention of water in body tissue



PAGE - 20


Ans. (D)

72. A water molecule can form a maximum of hydrogen bonds.

(A) 1 (B) 2

(C) 3 (D) 4

Ans. (D)

73. Circadian Rhythm is an endogenously driven cycle for biochemical, physiological and behavioral

processes. In humans, the approximate duration of this 'biological clock' is :

(A) 1 Hours (B) 6 Hours

(B) 12 Hours (D) 24 Hours

Ans. (D)

74. Modern evolutionary theory consists of the concepts of Darwin modified by knowledge concerning :

(A) population statistics

(B) Mendel's laws

(C) the idea of the survival of the fittest

(D) competition

Ans. (A)

75. Soon after the three germ layers are formed in a developing embryo, the process of organogenesis

starts. The human brain is formed from the

(A) ectoderm

(B) endoderm

(C) mesoderm

(D) partly endoderm and partly mesoderm

Ans. (A)

76. Puffs in the polytene chromosomes of drosophila melanogaster salivary glands represent

(A) transcriptionally active genes

(B) transcriptionally inactive genes

(C) heterochromatin

(D) housekeeping gense

Ans. (A)

77. The process of cell death involving DNA cleavage in cells is known as

(A) necrosis (B) apoptosis

(C) cytokinesis (D) endocytosis

Ans. (B)

78. According to the original model of DNA. as proposed by Watson & Crick 1953, DNA is a

(A) left handed helix

(B) helix that makes a full turn every 70 nm.(C) helix where one turn of DNA contains 20 basepairs

(D) two stranded helix where each strand has opposite polarity.

Ans. (D)

79. At which stage of meiosis I does crossing over occur ?

(A) lepoptene (B) zygotene

(C) pachytene (D) diplotene

Ans. (C)

80. An electrode is placed in the axioplasm of a mammalian axon and another electrode is placed just

outside the axon. The potential difference measured will be

(A) 0 (B) – 70mV



PAGE - 21


(B) –70V (D) +70 V

Ans. (B)

PART-IITwo Mark Questions

MATHEMATICS

81. Let A and B be any two n × n matrices such that the following conditions hold : AB = BA and there exist

positive integers k and l such that Ak = I (the identity matrix) and B = 0 (the zero matrix), Then(A) A + B = l

(B) det (AB) = 0(C) det (A + B) 0

(D) (A + B)m = 0 for some integer mSol. (B)

AK = I |A|K = 1B = 0 |B| = 0

|AB| = 0

82. The minimum value of n for which 2222

2222

)1 –n2(.......531

)n2(.......642

< 1.01

(A) is 101 (B) is 121 (C) is 151 (D) does not exist

Sol. (C) Let 12 + 22 + .......+ (2n –1)2 = x and c =

)1r4 –r4( 2

=3

n(4n2 – 1)

x

x –)n2(......21 222 < 1.01

x6

)1n4()1n2(n2 < 2.01

)1 –n4(3

n3

)1n4()1n2(n

2

< 2.01

)1 –n2(

)1n4(

< 2.01

4n + 1 < (2.01) (2n – 1)

4n + 1 < 4.02 n – 2.01

3.01 < 0.02 n

2

301< n

150.5 < n

83. The locus of the point P = (a , b) where a , b are real numbers such that the roots of x3 + ax2 + bx + a = 0are in arithmetic progression is(A) an ellipse (B) a circle

(C) a parabola whose vertex is on the y –

axis (D) a parabola whose vertex is on the x –

axisSol. (C)

A - d , A, A + d



PAGE - 22


3A = –a

A =3

a –

3

ab –

9

a

27

a –33

+ a = 0

3

ab –a

27

a2 3 = 0

a = 0 or

27

a2 2

+ 1 =3

b

2x2 = 9 (y - 3) 27

h2 2

=

1 –

3

k

parabola whose vertex on y axis.

84. The smallest possible positive slope of a line whose y –intercept is 5 and which has a common point with

the ellipse 9x2 + 16y2 = 144 is

(A)43 (B) 1 (C)

34 (D)

169

Sol. (B) Let line

y = mx + 5

ellipse 19

y

16

x 22

tangent condition c2 = a2m2 + b2

25 = 16m2 + 9

m2 = 1

m = ±1For positive slope m = 1

85. Let A = { R | cos2(sin) + sin2(cos) = 1} and B = { R | cos(sin) sin(cos) = 0 }. Then A B(A) is the empty set(B) has exactly one element(C) has more than one but finitely many elements(D) has infinitely many elements

Sol. (A)A ={R|cos2(sin) = cos2 (cos)}A = {R| sin = n ± cos}A = {R| sin ± cos = n}

A = {R| sin ± cos = 0}sin ± cos[ 2,2 ]

B {R| cos (sin) sin(cos) = 0} by AB {R| cos (± cos) sin(cos) = 0}

B {R|2

1sin(2cos) = 0}

2cos = p, p

cos =2

p



PAGE - 23


p = 0 only

cos = 0

O = 2k±2

so AB

86. Let f(x) = x3 + ax2 + bx + c , where a,b,c are real numbers. If f(x) has local minimum at x = 1 and a local

maximum at x =31 – and f(2) = 0 , then

1

1 –

dx)x(f equals

(A)3

14(B)

3

14 –(C)

3

7(D)

3

7 –

Sol. (B) f'(x) = 3x2 + 2ax + b f'(1) = 0 3 + 2a + b = 0 .........(i)

3

1'f = 0 b

3

a2

9

3 = 0

1 – 2a + 3b = 0 .........(ii)by (i) + (ii) we get

4 + 4b = 0 b = –1

f(2) = 8 + 4a + 2b + c = 0

8 – 4 – 2 + c = 0

c = –2

f(x) = x3 – x2 – x – 2

dx)2x(2dx)x(f1

0

21

1

=

1

0

3

x23

x2

= –

3

14

87. Let f(x) = x12 – x9 + x4 – x + 1. Which of the following is true ?

(A) f is one –one

(B) f has a real root(C) f never vanishes(D) f takes only positive values

Sol. (D) f(x) = x12 – x9 + x4 – x + 1

for x = 1 , f(1) = 1x = 0 , f(0) = 1for x > 1 f(x) = x9(x3 –1) + x(x3 –1) + 1 = positive

for 0 < x < 1 f(x) = x12 – x4 (x5 –1) + (1 – x) = positive

for x < 0 f(x) = x12 + x4 – x (1 + x8) + 1 = positve

f(x) is always positive

88. For each positive integers n, define fn(x) = minimum

!n

)x –1(,

!n

x nn

, for 0 x 1. Let In

= 1n,dx)x(f

1

0

n .

Then

1nnI is equal to

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

Sol. (A) In = dx!n

x2

1

0

n

+dx

!n

)x1(1

2

1

n



PAGE - 24


=!1n

x 1n

1

2

1

1n2

1

0!1n

)x1(

In

=!1n

2

1

.2

1n

1n I

n=

.................!3

2

1

!2

2

1

.2

32

= 2

2

11e2

1

= 2 3e

89. The maximum possible value of x2 + y2 – 4x – 6y , x,y real subject to the condition | x + y | + | x – y | = 4

(A) is 12 (B) is 28 (C) is 72 (D) does not exist

Sol. (B) |x + y| + |x –

y| = 4

Now, x2 + y2 – 4x – 6y

= (x – 2)2 + (y – 3)2 – 13

A point on square ABCD which is maximum distance from (2, 3) is C( –2, –2)So, required maximum value is 28.

90. The arithemetic mean and the geometric mean of two distinct 2 – digit numbers x and y are two integers

one of which can be obtained by reserving the digits of the other (in base 10 representation). Then x + yequals(A) 82 (B) 116 (C) 130 (D) 148

Sol. (C) Let AM and GM be ab and ba respectivelyAM = 10a + bGM = 10b + a

2

yx = 10a + b, xy = (10b + a)2

(x – y)2 = (x + y)2 – 4xy= 4[10b + a]2 – 4[10b + a]2

(x – y)2 = 22 × 18 (a + b) (a – b)

(x – y)2 = 11 × 9 × 4 (a + b) (a – b)

this should be a perfect square of an integer, only possible case whena + b = 11a – b = 1



PAGE - 25


a = 6, b = 5

x + y = 2 (10 × 6 + 5) = 130

PHYSICS

91. An isolated sphere of radius R contains uniform volume distribution of positive charge. Which of the curvesshown below correctly illustrates the dependence of the magnitude of the electric field of the sphere as a

function of the distance r from its centre ?

(A) I (B) II (C) III (D) IVAns. (B)

Electric field due to uniformly charged isolated sphere (In volume)

Case (a) r < R E = rR

KQ3

Case (b) r = R E = 2R

KQ

Case (c) r > R E = 2r

KQ

92. The surface of a planet is found to be uniformly charged. When a particle of mass m and no charge is thrownat an angle from the surface of the planet, it has a parabolic trajectory as in projectile motion with horizontalrange L. A particle of mass m and charge q, with the same initial conditions has a range L/2. The range ofparticle of mass m and charge 2q with the same initial conditions is :(A) L (B) L/2 (C) L/3 (D) L/4

Ans. (C)Let gravitational acceleration due to mass of planet is g and due to uniformly charge on surface of planetelectric field is E and which is constant because path of projectile is parabolic. All projetile have sameconditionI. Range for uncharged particle

Range = L =g

2sinu2 ...(1)



PAGE - 26


II. Range for a particle of mass m and charge q

Range =

m

qEg

2sinu

2

L 2

...(2)

From eq. (1) and (2)

mg

qE1g

2sinu

2

L 2

mgqE1

L

2

L

mg

qE

1 = 2

1mg

qE

III. Range for a particle of mass m and charge 2q

Range = 3

L

21

L

mg

qE21g

2sinu

m

qE2g

2sinu 22

93. Figure below shows a small mass connected to a string, which is attached to a vertical post.If the ball is

released when the string is horizontal as shown , the magnitude of the total acceleration (including radialand tangential) of the mass as a function of the angle is

(A) g Sin (B) g 1Cos3 2 (C) g Cos (D) g 1Sin3 2

Ans. (D)

Tangential acceleration at angle isa

t= g cos

From energy conservation between horizontal and position.PE

i+ KE

i= PE

f+ KE

f

0 + 0 = –mg sin +2

1mv2

v2 = 2gsinCentripetal acceleration at angle is



PAGE - 27


ac

=

2v= 2gsin

Total acceleration at angle is

a = 2222c

2t

)sing2(cosgaa = 1sin3g 2

94. One mole of an ideal gas at intial temprature T, undergoes a quasi –static process during which the volume V

is doubled .During the process the internal energy U obeys the equation U = aV3

, where a is a constant .Thework done during this process is(A) 3RT / 2 (B) 5RT / 2 (C) 5RT / 3 (D) 7RT / 3

Ans. (D)n = 1 moleinitial temperature = TInitial volume = VFinal volume = 2VInternal energy V = aV3

nCVT = aV3 n RT

2

3= aV3

PV = nRT From ideal gas equation

2

3PV = aV3 P =

3

aV2 2

Work done by gas during this process is

W = dv3

av2Pdv

v2

v

2v2

v

= v2

v

2dvv3

a2

= PV3

7v.

3

av2.

3

7

9

av14

3

v

3

a2 23v2

v

3

W = RT3

7

95. A constant amount of an ideal gas undergoes the cyclic process ABCA in the PV diagram shown below.

The path BC is an isothermal .The work done by the gas during one complete cycle , beginning and endingat A is nearly(A) 600 KJ (B) 300 KJ (C) – 300 KJ (D) – 600 KJ

Ans. (C)V

A= V

C= 2m3, P

A= P

B= 200K P

a, P

C= 500 KP

a

BC is isothermal processP

C

VC

= PB

VB

VB

= 200

2500

P

VP

B

CC = 5m3

Total workdone by gas during one complete cycle.



PAGE - 28


W = WAB

+ WBC

+ WCA

= 200 × 103 (5 – 2) + 200 × 103 × 5n

500

200+ 0

= 600 × 103 – 1000 × n2

5= –300 kJ

96. A material is embedded between two glass plates . Refractive index n of the material varies with thickness as

shown below .The maximum incident angle (in degrees) on the material for which beam will pass through thematerial is

(A) 60.0 (B) 53.1 (C) 43.5 (D) 32.3Ans. (B)

For total internal reflection1.5 sinC = 1.2 sin90°

sinC =5

4

5.1

2.1

C = 53.1maximum incident angle is 53.1 Ans.

97. At a distance form a uniformly charged long wire , a charged particle is thrown radially outward with avelocity u in the direction perpendicular to the wire . When the particle reaches a distance 2 from the wire

its speed is found to be u2 . The magnitude of the velocity , when it is a distance 4l away from the wire ,

is (ignore gravity)

(A) u3 (B) 2 u (C) u22 (D) 4 u

Ans. (A)Let linear charge density on wire is and mass of the particle is m & charge on particle is q. From energyconservation between and 2 distance.Loss in potential energy = gain in kinetic energy

–2kq n

2=

22 mu2

1)u2(m

2

1

–2kqn2 =22 mu

2

1u2m

2

1

2kqn2 =

2mu2

1

...(i)conservation between and 4 distance.Loss in potential energy = gain in kinetic energy

–2kq ne4

=

22 mu2

1mv

2

1



PAGE - 29


4kqn2 =22 mu

2

1mv

2

1

From eq (i)

mu2 =22 mu

2

1mv

2

1

v = 3 u

98. A rectangular loop of wire shown below is coplanar with a long wire carrying current I .

The loop is pulled to the right as indicated .What are the directions of the induced current in the loop and themagnetic forces on the left and the right sides of the loop ?

Inducded Current Force on left side Force on right side(A). Counterclockwise To the left To the right

(B). Clockwise To the left To the right(C). Counter clockwise To the right To the left(D). Clockwise To the right To the left

Ans. (B)

Sol.

As loop in pulled to right, there free downward magenetic field is decrease. So, induce emf induced as tooppose the decrement of magnetic field.

Therefore directions of the induced current in the loop is clock wise and force on left side of loop is toward leftand force on right side of loop is towards right

99. Two batteries V1and V

2are connected to three resistors as shown below

If V1

= 2V and V2

= 0V , the current I = 3mA. If V1

= 0V and V2

= 4V , the current I = 4 mA. Now if V1

= 10 Vand V

2= 10V , the current I will be

(A) 7 mA (B) 15 mA (C) 20 mA (D) 25 mAAns. (D)

Sol.

I = R)rr(rr

rVrV

2221

2211

Case – I V1

= 2V V2

= 0V I = 3mA

Case – II V2

= 4V V1

= 0 I = 4mAfrom case (I and Case II)



PAGE - 30


2

1

r

r

2

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

Case – III V1

= 10V V2

= 10V I = ?

I =R)rr(rr

)rr(10

2121

21

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

from (i) and (ii)

I = 25 mA

100. A particle moves in a plane along an eliptic path given by2

2

a

x+

2

2

b

y= 1. At point (0, b), the x-component of

velocity is u. the y-component of acceleration at this point is :(A) – bu2 / a2 (B) – u2 / a2 (C) – au2 / b2 (D) – u2 / a

Ans. (A)

Sol.2

2

2

2

b

y

a

x = 1 Different w.r.t.

dt

dy

b

y2

dt

dx

a

x222 = 0 ................. (i)

ux= u, at (o, b), u

y= 0 (from eq (i))

Different w.r.t.

2

22

2

2 dt

dx

a

2

dt

xd

a

x2

+

dt

dy

b

2

dt

yd

b

y222

2

2 = 0

accelerat at (o, b) is

22

y222

2

2o

b

2

b

ba2u

a

2

dt

xd

a

x2

(0) = 0

ay= – 2a

b u2

CHEMISTRY

101. XeF6hydrolyses to give an oxide. The structure of XeF

6and the oxide, respectively are

(A) octahedral and tetrahedral(B) distorted octahedral pyramidal(C) octahedral and pyramidal(D) distorted octahedral and tetrahedral

Sol. (B)

XeF6

+ 3H2O XeO

3+ 6HF

distorted pyramidaloctahedral

102. MnO4 – oxidizes (i) oxalate ion in acidic medium at 333 K and (ii) HCl. For balanced chemical equations, the

rations [MnO4 – : C

2O

42 –] in (i) and [MnO

4 – : HCl] in (ii), respectively, are :

(A) 1 : 5 and 2 : 5 (B) 2 : 5 and 1 : 8 (C) 2 : 5 and 1 : 5 (D) 5 : 2 and 1 : 8Sol. (C)

5e – + MnO4 – + 8H+ 4H

2O

C2O

4 –2 2CO

2+ 2e –

HCl 2

1Cl

2+ H+ + e –



PAGE - 31


103. If Fe / 2FeºE = – 0.440 V and 2Fe / 3Fe

ºE = 0.770 V, then Fe / 3FeºE is :

(A) 0.330 V (B) – 0.037 V (C) – 0.330 V (D) – 1.210 VSol. (B)

2e – + Fe+2 Fe – 0.44 V

Fe+3 + 3e – Fe 0.77 V

Fe+3 + 3e – Fe xV

then x =3

77.0)44.0(2 = – 0.037 V

104. The electron in hydrogen atom is in the first Bohr orbit (n = 1). The ratio of transition energies, E(n = 1 n= 3) to E (n = 1 n = 2), is(A) 32/27 (B) 16/27 (C) 32/9 (D) 8/9

Sol. (A)

21

31

E

E

=

22

22

2

1

1

1

3

1

1

1

=4 / 3

9 / 8=

27

32

105. In the following conversion,

O3H)ii(

MeMgBr)i(X

O3H

2 / NaOH IY

the major products X and Y, respectively, are :

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

106. The reaction sequence,

2HNOX Y

the major products X and Y, respectively, are :



PAGE - 32


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

Due to intramolecular hydrogen bonding reaction is favored at ortho position.

107. Optically active (S)--methoxyacetaldehyde on reaction with MeMgX gave a mixture of alcohols. The majordiastereomer ‘P’ on treatment with MeI/K

2CO

3gave an optically inactive compound. P is :

(A) i (B) ii (C) iii (D) ivAns. Incorrect question.Sol. The statement optically active (S) – – methoxy acetaldehyde is incorrect.

108. At 300 K the vapour pressure of two pure liquids, A and B are 100 and 500 mn Hg, respectively. If in a mixtureof A and B, the vapour pressure is 300 mm Hg, the mole fractions of A in the liquid and in the vapour phase,respectively, are :(A) 1/2 and 1/10 (B) 1/4 and 1/6 (C) 1/4 and 1/10 (D) 1/2 and 1/6

Sol. (D)P

total= P°

A+ (P0

B – P0

A) X

B

300 = 100 + (500 – 100) (1 – XA)

1 – XA

=2

1 X

A=

2

1

YA

=300

21100

=6

1

109. The crystal field stabilization energies (CFSE) of high spin and low spin d6 metal complexes in terms of 0,

respectively, are :(A) – 0.4 and – 2.4

(B) – 2.4 and – 0.4

(C) – 0.4 and 0.0



PAGE - 33


(D) – 2.4 and 0.0

Sol. (B)Low spin d6

CFSE = – 4 ×5

2

0+

5

3

0 × 2

= – 0.4 0

High Spin d6

CFSE = –6 ×52

0= – 2.4

0(ignoring pairing)

110. Emulsification of 10 ml of oil in water produces 2.4 × 1018 droplets. If the surfaces tension at the oil-waterinterface is 0.03 Jm –2 and the area of each droplet is 12.5 × 10 –16 m2, the energy spent in the formation of oildroplets is :(A) 90 J (B) 30 J (C) 900 J (D) 10 J

Sol. (A)Total area of droplets = 2.4 × 1018 × 12.5 × 10 –16 = 3000 m2

Energy liberated = 3000 × 0.03 = 90 J

BIOLOGY

111. Which sequence of events givens rise to flaccid guard cells and stomatal closure at night ?(A) low [Glucose ] low osmotic pressure low pH high pCO

2

(B) low pH high pCO2 low [Glucose ] low osmotic pressure

(C) low osmotic pressure high pCO2

low pH low osmotic pressure

(D) high pCO2

low pH low [Glucose ] low osmotic pressure

Ans. (D)

112. Rice has a diploid genome with 2n = 24. If crossing over is stopped in a rice plant and then selfed

seeds are collected, will all the offsprings be genetically identical to the parent plant ?

(A) yes, because crossing-over is the only source of genetic variation

(B) no, because stopping of crossing over automatically increases rate of point mutation

(C) yes, only if the parent plant was a completely inbred line(D) yes, only if the parent plant was a hybrid between two prue-bred lines

Ans. (C)

113. Rodents can distinguish between many different types of odours. The basis for odour discrimination is

that

(A) they have a small number of odorant receptors that bind to many different odorant molecules

(B) the mechanoreceptors in the nasal cavity are activated by different odorant molecules found in the

air passing through the nostrils

(C) the part of the brain that processes the sense of smell has many different receptors for odorant

molecules

(D) a large number of different chemoreceptors are present in the nasal cavity that binds a variety of

odorant molecules

Ans. (D)

114. Although blood flows through large arteries at high pressure, when the blood reaches small capillaries

the pressure decreases because

(A) the values in the arteries regulate at he rate of blood flow into the capillaries

(B) the volume of blood in the capillaries is much lesser than the that in the arteries

(C) the total cross-sectional area of capillaries arising form an artery is much greater than that of the

artery

(D) elastin fibers in the capillaries help to reduce the arterial pressure

Ans. (C)

115. E.coli about to replicate was pulsed with tritiated thymidine for 5 min and then transferred to normal 1

medium . After one cell division which one of the following observations would be correct ?




(A) both the strands of DNA will be radioactive

(B) one strand of DNA will be radioactive

(C) none of the strands will be radioactive

(D) half of one strand of DNA will be radioactive

Ans. (B)

116. Selection of lysine auxotroph (bacteria which requires lysine for growth) from a mixed population of

bacteria can be done growing bacterial population in the presence of

(A) lysine (B) penicillin

(B) lysine and penicillin (D) glucose

Ans. (A)

117. Increasing the number of measurements of and experimental variable will

(A) increase the standard error of the sample

(B) increase the mean of the sample

(C) decrease the standard error of the sample

(D) result in all of the above

Ans. (C)

118. For a human male what is the probability that all the maternal chromosomes will end up in the samegamete ?

(A) 1/23 (B) 223

(C) 2th (D) (½)23

Ans. (D)

119. Nocturnal animal have retinas that contain

(A) a high percentage of rods to increase sensitivity to low light conditions

(B) a high percentage of cones so that nocturnal color vision can be improved in low light conditions

(C) an equal number of rods and cones so that vision can be optimized

(D) retinas with the photoreceptor layer present in the front of the eye to increase light sensitivity

Ans. (A)

120. The length of one complete true of a DNA double helix is

(A) 34 A (B) 34 nm

(C) 3.4 A (D) 3.4 m

Ans. (A)

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