KVPY_Stream (SB) Test Paper & Solution

KISHORE VAIGYANIK PROTSAHAN YOJANA - 2012 Date : 30-10-2011

Duration : 3 Hours

Max. Marks : 160

STREAM - SB / SX 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 Question No. 21 to 40 consist of ONE (1) mark for each correct response. CHEMISTRY Question No. 41 to 60 consist of ONE (1) mark for each correct response. BIOLOGY Question No. 61 to 80 consist of ONE (1) mark for each correct response.

PART-II : MATHEMATICS Question No. 81 to 90 consist of TWO (2) marks for each correct response. PHYSICS Question No. 91 to 100 consist of TWO (2) marks for each correct response. CHEMISTRY Question No. 101 to 110 consist of TWO (2) marks for each correct response. BIOLOGY Question No. 111 to 120 consist of TWO (2) marks for each correct response.

KVPY QUESTION PAPER - STREAM (SB / SX)

PART-I One Mark Questions MATHEMATICS 1.

Suppose logab + logb a= 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

1 , x

C=x+

then,

logba =

For a, b > 1

1 x logab > 1

By A.M.  G.M C= x 2.

Sol.

3. Sol.

4.

Sol.

1 2 x

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 (C) Hint n is multiple of = 4 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 2 (D) Given P = P For n = 2 (I + P)2 = I + P2 + 2PI = I + P + 2P = I + 3P = I + (22 – 1) P

(D) I + (2n–1)P

Consider the cubic equation x3 + ax2 + bx + c = 0 , where a,b,c are real numbers . Which of the following statements 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 roots (A) f ' (x) = 3x2 + 2ax + b D = 4a2 – 4(3b) = 4(a2 – 3b) If a2 – 2b < 0 then a2 – 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) =

1 1   1 x2 =    x 2 x   2x sin xy  1 or  –1  sin xy = 1 or –1 For x = –1 or 1 –sin y = 1 or sin y = –1

RESONANCE

PAGE - 2


  y =  2n   , n  I 2  (x, y) on pair of straight line.

6.

Sol.

Let A = (4,0) , B = (0,12) be two points in the plane . The locus of a point C such that the area of triangle ABC 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 x y 1 1 (C) 4 0 1  18 2 0 12 1

 7.

|–12x – y(4) + 48| = 36 (3x + y – 12)2 = 92 = 81

In a rectangle ABCD , the coordinates of A and B are (1,2) and (3,6) respectively and some diameter of the circumscribing circle of ABCD has equation 2x – y + 4 = 0. Then the area of the rectangle is (A) 16

Sol.

(B) 2 10

(C) 2 5

(D) 20

(A)

m1 m2 = –1  4t  6  6   6  2      = –1  2t  1  3   3  1   4t    (2) = –1  4t = –t + 2  2t  4 

 t=

2 5

 1 38  then C is   ,   5 5  2

 16  8      5  5

BC =

=

Sol.

=

320 25

8 5 8  5 5 8

 2 5 = 16 5 In the xy–plane , three distinct lines 1,2,3 concur at a point (,0). Further the lines 1,2,3 are normals to the parabola y2 = 6x at the points A = (x1 , y1) , B (x2 , y2) , C = (x3 , y3) respectively . Then we have (A)  < – 5 (B)  > 3 (C) – 5 <  < – 3 (D) 0 <  < 3 (B) h > 2a here 4a = 6, 2a = 3 h>3 Area = xy =

8.

2

RESONANCE

PAGE - 3


9.

Sol.

Let f(x) = cos5x + A cos 4x + Bcos3x + Ccos2x + Dcosx + E. and   2   3   8   9   – f  + ..............+ f   – f  Then T T = f(0) – f   + f  5 5 5 5          5  (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 (C) f(x) = cos5x + A cos 4x + Bcos3x + Ccos2x + Dcosx + E f( – x) = – cos5x + Acos4x – B cos3x + C cos2x – D cos x + E f (x) – f( – x) = 2 [A cos4x + C cos2x + E) for x = 0 f(0) – f() = 2(1 + B + D) ...... (i)   4   3   = 2 | –1 + B cos , f  – f + D cos | ...... (ii) 5 5 5 5  5 

at x =

at x =

 2   3  2  – f  =2 , f 5 5    5 

 2   6  1  B cos     D cos 5 5    

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

..... (iii)

...... (iv)

     9     2    8     3    7   T = f(0) – f() –  f  5   f  5   +  f  5   f  5   –  f  5   f  5   +               

  4   6     f    f     5   5 

 2   8   =f  from equation iind f   5   5 

  9   f  = f 3  3   3   7   4   6  = f  and f   = f  f  5   5   5   5 

     4     2    3   So T = f(0) – f() – 2  f  5  – f  5     f  5  – f  5             from equation 1, 2 & 3 T = 2(1 + B + D) – 2 [–1 + B cos

3  6 2 + Dcos | + 2 [ 1 + B cos + D cos ] 5 5 5 5

Will depend on B and D only 10.

Sol.

In triangle ABC , we are given that 3sinA + 4 cosB = 6 and 4sinB + 3cosA = 1. Then the measured of the angle C is (A) 30º (B) 150º (C) 60º (D) 75º (B) 3sinA + 4cosB = 6 ...... (i) 3 cosA + 4sinB = 1 .......(ii) Squaring and adding , we get 9 + 16 + 24 sin(A+B) = 37 sin(A+B) = sin( – C) = sin C =

1 2 1 2

1 2

RESONANCE

PAGE - 4


11.

Sol.

12.

Which of the following intervals is possible domain of the function f(x) = log{x} [x] + log[x] {x} , where [x] is the greatest integer not exceeding x and {x} = x – [x] ? (A) (0,1) (B) (1,2) (C) (2,3) (D) (3,5) (C) [x] > 1  [x]  2 x2 But x cannot integers  Possible domain = (2,3) 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) +

f " (0 ) f (n –1)(0) 2! + ......+ (n – 1)! is (A) (2011)n Sol.

(B) (2012)n

(D) n(2011)n

f " (0 ) f (n –1)(0) + ......+ 2! (n – 1)!

(C) f(0) + f(0) +

(2011) +

(C) (2012)n – 1

n(2011)n –1 n(n – 1)(2011)n – 2  n(n – 1)...2! (2011)'  ,  ......    n – 1! 1! 2!  

= 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

1 (A)

Sol.

(B)

2

(C)

2

3

(D) 2 2

(B)

Point where slope of tangent is 1 of the curve y = ex is y = ex = 1 x=0  (0,1) and for y = lnx point is (1,0)  required minimum distance is

2 8

14.

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

 f ( x ) dx is equal to 2

(A) 17

(B) 22

8

Sol.

(B)

2

4

3

f ( x ) dx =

2

f ( x ) dx 

(C) 23

3 f ( x) dx

(D) 25

8

+ ...... +

7 f ( x ) dx

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

RESONANCE

PAGE - 5


15.

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

2012

e cos(  { x })



e cos( { x })  e – cos(  { x })

0

equal to (A) 0

(B) 1006

e cos x

1

Sol.

(B) I = 2012



0e

cos x

 e cos x

(D) 2012 

dx

e  cos x   cos x  e cos x 0 e 1

again I = 2012

(C) 2012

dx is

 dx  

1

 dx

2I = 2012

I = 1006



16.

 1  The value of lim   2 n  4n – 1 1 4

(A)

Sol.

0

n

1

=

 .......... .. 

(C)

 4

(D)

 6

  2 2  4n – n  1

1  r  2n 1–    2n 

2

dx

 0

 12

4n – r 2

n

1

4n 2 – 4

  is 2 2  4n – n  1

2

lim 

n  r  1

1

 .......... .. 

1

 = nlim  r  1

1 = 2

4n 2 – 4 (B)

 1  (D) lim   2 n  4n – 1

=

1

1–

x2 4

dx



4 – x2

0

1

 1 x   =  sin     2  0 

17.

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

1 3

RESONANCE

(B)

5 9

(C)

4 9

(D)

1 9

PAGE - 6


Sol.

(C) E1 : A wins when A takes out 3 & 6 and B takes out 8 & 9 1 1 1 × = 3 3 9 E2 : A wins when A takes out 5 & 6 and B takes out any two no.

P (E1) =

P (E2) =

1 1 ×1= 3 3

Required probability = 18.

Sol.

        Let a, b, c be three vectors in the xyz space such that a  b = b  c = c  a  0 . If A,B,C are points with  position vectors a, b, c respectively , then the number of possible positions of the centriod of triangle ABC

is (A) 1       (A) a  b  b  c  c  a      ab bc  0     ab  cb  0  a  c  b  0     a  c  b  a  c  b      (bc ) = ca   a  c   c = c  a      a  c    c  a     a  c   a  c   0   (1 + ) ( a  c ) = 0    From (ii) a  c  b    abc  0 

(C) 3

(D) 6

........(i)

........(ii)



 = –1

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 n

Sol.

(B) 2

   abc  0 centroid of ABC = 3

 19.

1 1 4 + = 9 3 9

2 (B)  r  r  1 r !



r 1



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)

RESONANCE

PAGE - 7


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 1, x  A f(x,A) =  .Then f(x , A B) equals 0, x  A (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.

1, x  A (C) If f (x , A  B) =  0, x  A

Case (i) x  A, x  B then, f(x, A) + f(x, B) – f(x, A) f(x, B) = 1 + 0 – 0 = 1 Case (ii) x  A, x  B f(x, A) + f(x, B) – f(x, A) f(x, B) = 0 + 1 – 0 = 1 Case (iii) x  A, x  B f(x, A) + f(x, B) – f(x, A) f(x, B) = 0 + 0 – 0 = 0 Case (iv) x  A and x  B f(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 1, x  A  B If f (x , A) =  0, x  A  B

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

PHYSICS 21.

Ans. Sol.

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 the centre 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. (A) acceleration of cabin is g relative acceleration of A and B w.r.t. cabin is zero so both A and B continue to be exactly at rest relative to the cabin.

22.

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

Ans.

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

RESONANCE

PAGE - 8


Sol. 23.

Ans. Sol.

 W mg    k k  = mg  k

 W = 2mg

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

(A)

(B)

(C)

(D)

(A) I (A)

(B) II a = 2r

v =  A 2  r 2 ...(i) from (i) and (ii) v2 2 A 2



a2 4 A 2

(C) III

(D) IV

...(ii)

1

It is the equation of ellipse 24.

Ans. Sol.

25.

Ans. Sol.

A container with rigid walls is convered with perfectly insulating material. The container is divided into two parts by a partition. One part contains a gas while the other is fully evacuated (vacuum). The partition is suddenly removed. The gas rushes to fill the entire volume and comes to equilibrium after a little time. If the gas 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 (A) Work done in this process is zero so that (Ui = Uf) Two bulbs of identical volumes connected by a small capillary are initially filled with an ideal gas at temperature T. Bulb 2 is heated to maintain a temperature 2T while bulb 1 remains at temperature T. Assume throughout that the heat conduction by the capillary is negligible. Then the ratio of the final mass of the gas in 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 (B) Using mole conservation 2n = n1 + n2 ...(i) PV = n1RT ...(ii) PV = n2R(2T) ...(iii) n1 = 2n2 2n = 2n2 + n2  2n 

n2 =  3   

;

RESONANCE

M2 f 2n / 3 2   M2i n 3

PAGE - 9


26.

Ans. Sol.

Two rods, one made of copper and the other steel of the same length and cross sectional area are joined together. (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 copper end 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 (D) dQ dT  KA dt dx K1A( T  100 ) K 2 A  (0  T )  

K1(T – 100) = K2(–T) K1T – K1 × 100 = –K2T T(K1 + K2 ) = K1 × 100  385  100   = 88°C  435 

T=  27.

Ans. Sol.

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. (B) P1- T= constant    T1  P    1 T   P 2    2

 233K     T 

7/5

   

 1

233

T=

(7 / 5)1

 0.28     1 

(0.28 )2 / 5

 466 J

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

28.

Ans. Sol.

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 > f2 (C) f1 < f2 (D) f1 = f2 > f0 (C)  v 

f1 =  v  u  f0  

...(i)

 v u

f2 =  v f0  

...(ii)

f1  v 2   f2  v 2  u2     

v2

 

f1 = f2  2 2   v u  f2 > f1

RESONANCE

PAGE - 10


29.

Ans. Sol.

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’s axis is parallel to the plane of polarization of the incident light, the intensity of light at the exit of the fifth polarizer 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 (C)  = 0 (cos2)4 3 4

4

 = 0 ×    = 30% of 0 30.

Ans. Sol.

At 23C, a pipe open at both ends resonates at a frequency of 450 hertz. At what frequency does the same pipe 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 (C)  nv 

f =  2    f’ =

n  v' 2

f v  v     f ' v '  1.04 v 

f’ = (1.04)(450) = 468 Hz 31.

Ans. Sol.

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 frequency of 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 (D) × 1 1  2  2

f11 = f22 f1  2 f1    2 f2 1  2f1 1 1 2  2 1 1 1  2 2

 2 = 2 1 = 2.0 mm 32.

Ans. Sol.

For a domestic AC supply of 220 V at 50 cycles per second, the potential difference between the terminals of a two -pin electric outlet in a room is given by : (A) V(t) = 220 2 Cos(100t)

(B) V(t) = 220Cos(50t)

(C) V(t) = 220Cos(100t) (A) V = V0 cost

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

Vrms =

V0 2

V0 = 220 2 Volt  = 50 × 2 rad/sec = 100 rad/sec. v = 220 2 Cos(100t)

RESONANCE

PAGE - 11


33.

Ans. Sol. 34.

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

(A) R1 (B) R2 (C) R3 (A) Current through R1 is maximum so power dissipates at R1 is maximum.

(D) R4

An electron collides with a free molecules initially in its ground state. The collision leaves the molecule in a excited 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 

Ans.



(A) K = K’, P  P' (B)





(B) K’ < K, P  P' 





(C) K = K’, P  P'





(D) K’ < K, P  P'



Sol.

Momentum is conserved during collision so P  P' . 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 oscilloscope?

Ans. Sol.

(I)

(II)

(III)

(IV)

(A) I (B) II (C) III (D) Potential difference across inductor during current growth is v = v0e–t/

RESONANCE

(D) IV

PAGE - 12


36.

Ans. Sol. 37.

Ans. Sol.

Given below are three schematic graphs of potential energy V(r) versus distance r for three atomic particless electron (e), poton (p+) and neutron (n), in the presence of a nucleus at the origin O. The radius of the nucleus is r0. 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–) (A) Neutron is chargeless so graph is (1) Proton is positive charge so graph is (2) and for e– graph is (3) Due to transitions among its first three energy levels, hydrogenic atom emits radiation at three discrete wavelength 1,2 and 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 (D) 1 8  R  1 9 1 1   3R    R 1      2 4  4   1  1 1   5R  5R  R      3  4 9   9  4  36

1 1 1   1 2 3

38.

Ans.

Sol.

The total radiative power emitted by spherical blackbody with radius R and temperature T is P. If the radius is doubled and the temperature is halved then the radiative power will be : (A) P/4 (B) P/2 (C) 2P (D) 4P (A) P  P'

AT 4 T 4 A  2

4

P 4 P'

P’ = 39.

P 4

Ans.

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 (B)

Sol.

RH =

V i

V. i



2

i

RESONANCE



P

 h     2 i e  2

PAGE - 13


40.

Ans. Sol.

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

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

CHEMISTRY 41.

Sol.

Ther hydridizations of Ni (CO)4 and Cr(H2O)62+, respectively. (A) sp3 and d3sp2 (B) dsp2 and d2sp3 3 2 3 (C) sp and d sp (D) dsp2 and sp3 d2 (C) 2,2,2 2,2 Ni(CO)4 Ni : [Ar] 3d8 4s2 = t 2g e g

CO is a strong filed ligand

3d 4s Cr (H2O)6+2 Cr+2 : [Ar] 3d4 45°

3d 4s 2 octahedral complex : d sp3

42.

Sol.

4p

4p

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 (A) Ag+ + 2CN–  [Ag(CN)2]– 2[Ag(CN)2]– + Zn+2  [Zn(CN)4]–2 + 2Ag+

RESONANCE

PAGE - 14


43.

Sol.

44.

Sol.

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 (A) Ideal mixture H = 0 V = 0 At 298 K, the ratio of osmotic pressures of two solutions of a substance with concentrations of 0.01 M and 0.001 M, respectively, is : (A) 1 (B) 100 (C) 10 (D) 1000 (C)  = CRT

45.

Sol.

1 0.01  2 = 0.001 = 10

The rate of gas phase chemical reactions generally increases rapidly with rise in temperature. This is mainly because : (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. (B) Energy distribution at two different temperatures

46.

Among i-iv :

Sol.

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

Sol.

are : (A) enantiomers (A)

(B) diastereomers

(C) conformers

(D) rotamers

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

RESONANCE

PAGE - 15


48.

Sol.

49.

Sol.

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. (D) According to second law of thermodynamics, for spontaniety : SUNIV > 0 When the size of a spherical nanoparticle decreases from 30 nm to 10 nm, the ratio surface area/volume becomes (A) 1/3 of the original (B) 3 times the original (C) 1/9 of the original (D) 9 times the original (B) r1 = 30 nm r2 = 10 nm 2  surface area  3   = 4r = 3  volume  r 4r 3

(3 / r2 ) 30 (3 / r1 ) = 10 = 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



Sol.

52.

Sol.

the reagent used is : (A) LiAlH4 (B) H3PO2 (C) H3O+ (D) H2/Pt (B) H3PO2 a mild reducing agent is used to remove the diazonium group with H. 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 (A)

0NaOAc + 0HCl – 0NaCl

RESONANCE

PAGE - 16


53.

Sol.

To obtain a diffraction peak, for a crystalline solid with interplane distance equal to wavelength of incident Xray radiation, the angle of incidence should be : (A) 90º (B) 0º (C) 30º (D) 60º (D)

BRAGG’S Law 2dsin = n d = , n = 1

1   = 30° 2 angle of incidence = 60° sin =

54.

Sol.

The standard Gibbs free energy change (Gº in kJ mol–1), in a Daniel cell (Eºcell = 1.1 V), when 2 moles of Zn(s) is oxidized at 298 K, is closed to : (A) – 212.3 (B) – 106.2 (C) – 424.6 (D) – 53.1 (C) Zn  Zn+2 + 2e– 2mol 2mol 4mol G° = –nFE°cell = – 4 × 96500 × 1.1 = – 424.6 kJ/mol

55. Sol.

All the products formed in the oxidation of NaBH4 by I2, are : (A) B2H6 and NaI(B) B2H6, H2 and NaI (C) BI3 and NaH (B)

(D) NaBI4 and HI

2NaBH4 + 2  2Na + B2H6 + H2  56. Sol.

The spin-only magnetic moments of [Mn(CN)6]4– and [MnBr4]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 1.73 (C) [Mn(CN)6]–4 & [(Mn/Br)4]–2 Mn+2 in both Mn+2 : [Ar]3d5 45° With CN– : With Br– : Magnetic moments :

57.

Sol.

58.

1(1  2) BM &

5 (5  2) BM

In a zero-order reaction, if the initial concentration of the reactant is doubled, the time required for half the reactant to be consumed : (A) increases two-fold (B) increases four-fold (C) decreases by half (D) does not change (A) Half life of zero order reactions   initial concentration. The adsorption isothermal for a gas is given by the relation x = ap/(1 + bp) where x is moles of gas adsorbed per 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.

RESONANCE

PAGE - 17


Sol.

(D)

x=

59.

ap (1  bp)

The reaction

NaOH / Heat

+ CHCl3      H

is known as : (A) Perkin reaction (C) Reimer-Tiemann reaction (C) It is fact.

Sol.

60.

(B) Sandmeyer reaction (D) Cannizzaro reaction

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

Sol.

BIOLOGY 61.

Ans. 62.

Ans. 63.

Ans. 64.

Ans.

The major constituents of neurofilaments are (A) microtubules (B) intermediate filaments (C) actin filaments (D) protofilaments (B) 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 (C) 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 O2 is lower than partial pressure of CO2 (D) overall intake of O2 by the blood becomes low. (D) 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 (A)

RESONANCE

PAGE - 18


65.

Ans. 66.

Ans. 67.

Ans. 68.

Ans. 69.

Ans. 70.

Ans. 71.

Ans. 72.

Ans.

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 (A) 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 (B) 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' (D) 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. (D) 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 (C) 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 (C) 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 (D) A water molecule can form a maximum of hydrogen bonds. (A) 1 (B) 2 (C) 3 (D) 4 (D)

RESONANCE

PAGE - 19


73.

Ans. 74.

Ans. 75.

Ans. 76.

Ans. 77.

Ans. 78.

Ans. 79.

Ans. 80.

Ans.

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 (D) 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 (A) 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 (A) Puffs in the polytene chromosomes of drosophila melanogaster salivary glands represent (A) transcriptionally active genes (B) transcriptionally inactive genes (C) heterochromatin (D) housekeeping gense (A) The process of cell death involving DNA cleavage in cells is known as (A) necrosis (B) apoptosis (C) cytokinesis (D) endocytosis (B) 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. (D) At which stage of meiosis I does crossing over occur ? (A) lepoptene (B) zygotene (C) pachytene (D) diplotene (C) 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 (B) –70V (D) +70 V (B)

RESONANCE

PAGE - 20


PART-II Two Mark Questions MATHEMATICS 81.

Sol.

82.

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 m (B) AK = I |A|K = 1 B = 0 |B| = 0 |AB| = 0

The minimum value of n for which (A) is 101

Sol.

2 2  4 2  6 2  .......  ( 2n)2 1 2  3 2  5 2  .......  ( 2n – 1)2

< 1.01

(B) is 121

(C) is 151

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

 (4r 2 – 4r  1) = 3 (4n

n

(D) does not exist 2

– 1)

12  2 2  ......  (2n)2 – x < 1.01 x 2n(2n  1) ( 4n  1) < 2.01 6x n (2n  1) ( 4n  1) n 3 ( 4n 2 – 1) < 2.01 3 

( 4n  1) < 2.01 (2n – 1)

 4n + 1 < (2.01) (2n – 1)  4n + 1 < 4.02 n – 2.01 3.01 < 0.02 n

301
Sol.

The locus of the point P = (a , b) where a , b are real numbers such that the roots of x3 + ax2 + bx + a = 0 are 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 – axis (C) A - d , A, A + d 3A = –a A =



–a 3

– a3 a3 ab  – +a=0 27 9 3

RESONANCE

PAGE - 21




2a3 ab a– =0 27 3

a = 0 or b 2a2 +1= 3 27

k  2h2 =  – 1 3   27 parabola whose vertex on y axis.

2x2 = 9 (y - 3) 

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)

Sol.

3 4

(B) 1

(C)

4 3

(D)

9 16

(B) Let line y = mx + 5 x2 y2  1 16 9

ellipse tangent condition

c2 = a2m2 + b2 25 = 16m2 + 9 m2 = 1 m = ±1

For positive slope m = 1 85.

Sol.

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 (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 A B {R| cos (± cos) sin(cos) = 0} B {R|

1 sin(2cos) = 0} 2

2cos = p, p  cos =

p 2

p = 0 only cos = 0 O = 2k±

 2

so AB 

RESONANCE

PAGE - 22


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 = –

1 1 and f(2) = 0 , then f ( x ) dx equals 3



–1

14 3 (B) f'(x) = 3x2 + 2ax + b f'(1) = 0  3 + 2a + b = 0

(A) Sol.

(B)

– 14 3

(C)

7 3

(D)

–7 3

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

 1 3 2a f '    = 0   b = 0 3 9 3  

 1 – 2a + 3b = 0 by (i) + (ii) we get 4 + 4b = 0  f(2) = 8 + 4a + 2b + c = 0  8 – 4 – 2 + c = 0 c = –2  f(x) = x3 – x2 – x – 2 1

1

1

0

.........(ii) b = –1

2  f ( x ) dx  2  (  x  2) dx 1

 x3    2  2x  = – 14 =  3   0 3

87.

Sol.

88.

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 (D) f(x) = x12 – x9 + x4 – x + 1 for x = 1 , f(1) = 1 x = 0 , f(0) = 1 for 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 1  x n (1 – x )n    For each positive integers n, define fn(x) = minimum  n! , n!  , for 0  x  1. Let In = fn ( x ) dx, n  1 .   0





Then  I n is equal to n1

(A) 2 e – 3 Sol.

(A) In

(B) 2 e – 2

(C) 2 e – 1

(D) 2 e

1 2

1 (1  x )n xn dx =  dx +  n! 1 0 n! 2

x n1 = n  1!

1 2

1

 (1  x )n1     n  1 !  1   0 2

RESONANCE

PAGE - 23

KVPY QUESTION PAPER - STREAM (SB / SX) n1

 1   2 In = 2 . n  1!

  1  2  1 3         2 2  In = 2.       .......... .......   2!  3! n1      1 = 2 e 2  1   

89. Sol.

 1 = 2 e 3 2 

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 (B) |x + y| + |x – y| = 4

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. Now,

90.

Sol.

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 + y equals (A) 82 (B) 116 (C) 130 (D) 148 (C) Let AM and GM be ab and ba respectively AM = 10a + b GM = 10b + a

xy = 10a + b, xy = (10b + a)2 2 (x – y)2 = (x + y)2 – 4xy = 4[10b + a]2 – 4[10b + a]2 2 (x – y) = 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 when a + b = 11 a–b=1 a = 6, b = 5 x + y = 2 (10 × 6 + 5) = 130

RESONANCE

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PHYSICS 91.

Ans.

92.

Ans.

An isolated sphere of radius R contains uniform volume distribution of positive charge. Which of the curves shown 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 (B) Electric field due to uniformly charged isolated sphere (In volume) Case (a) r < R

E=

Case (b) r = R

E=

Case (c) r > R

E=

KQ R3

(D) IV

r

KQ R2 KQ r2

The surface of a planet is found to be uniformly charged. When a particle of mass m and no charge is thrown at an angle from the surface of the planet, it has a parabolic trajectory as in projectile motion with horizontal range L. A particle of mass m and charge q, with the same initial conditions has a range L/2. The range of particle of mass m and charge 2q with the same initial conditions is : (A) L (B) L/2 (C) L/3 (D) L/4 (C) Let gravitational acceleration due to mass of planet is g and due to uniformly charge on surface of planet electric field is E and which is constant because path of projectile is parabolic. All projetile have same condition I. Range for uncharged particle Range = L = II.

u2 sin 2 g

...(1)

Range for a particle of mass m and charge q L

u2 sin 2

Range = 2  qE g

...(2)

m

RESONANCE

PAGE - 25


From eq. (1) and (2) L u2 sin 2  2  qE   g 1  mg  



L L  2 1  qE mg

 1

qE =2 mg

qE

 mg  1 III.

Range for a particle of mass m and charge 2q u2 sin 2 u2 sin 2 L L    Range = 2qE   1  2 3 2qE g  g1  m mg  

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 radial and tangential) of the mass as a function of the angle  is

(A) g Sin

(B) g

3 Cos 2   1

(C) g Cos 

(D) g

3Sin 2   1

Ans. (D)

Tangential acceleration at  angle is at = g cos From energy conservation between horizontal and  position. PEi + KEi = PEf + KEf 0 + 0 = –mg sin +

1 mv2 2

v2 = 2gsin Centripetal acceleration at  angle is ac =

v2 = 2gsin 

Total acceleration at  angle is a=

a2t  ac2  g2 cos 2   ( 2g sin )2

RESONANCE

= g 3 sin2   1

PAGE - 26


94.

Ans.

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 .The work done during this process is (A) 3RT / 2 (B) 5RT / 2 (C) 5RT / 3 (D) 7RT / 2 (D) n = 1 mole initial temperature = T Initial volume = V Final volume = 2V Internal energy V = aV3 3 2

nCVT = aV3

 n RT = aV3

PV = nRT

From ideal gas equation

3 PV = aV3 2

P=

2aV 2 3

Work done by gas during this process is 2v

W=



2v

Pdv 

v

2a = 3



2av 2 dv 3

v

2v 2

 v dv v

2v

2a  v 3  14av 3 7 2av 2 7 = 3  3   9  3 . 3 . v  3 PV   v

W= 95.

7 RT 3

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 ending at A is nearly (A) 600 KJ (B) 300 KJ (C) – 300 KJ (D) – 600 KJ Ans. (C) VA = VC = 2m3, PA = PB = 200K Pa, PC = 500 KPa BC is isothermal process PCVC = PBVB VB =

PC VC 500  2 3  PB 200 = 5m

Total workdone by gas during one complete cycle. W = W AB + W BC + W CA = 200 × 103 (5 – 2) + 200 × 103 × 5n = 600 × 103 – 1000 × n

RESONANCE

200 +0 500

5 = –300 kJ 2

PAGE - 27


96.

Ans.

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 the material is

(A) 60.0 (B) 53.1 (B) For total internal reflection 1.5 sinC = 1.2 sin90°

(C) 43.5

(D) 32.3

1 .2 4  1 .5 5

sinC =

C = 53.1 maximum incident angle is 53.1 Ans. 97.

At a distance  form a uniformly charged long wire , a charged particle is thrown radially outward with a velocity u in the direction perpendicular to the wire . When the particle reaches a distance 2 from the wire its speed is found to be

2u . The magnitude of the velocity , when it is a distance 4l away from the wire ,

is (ignore gravity)

Ans.

(A) 3u (B) 2 u (C) 2 2 u (D) 4 u (A) Let linear charge density on wire is  and mass of the particle is m & charge on particle is q. From energy conservation between  and 2 distance. Loss in potential energy = gain in kinetic energy –2kq n

1  2 1 2 = m( 2u)  mu 2 2 2

–2kqn2 =

1 1 m2u2  mu2 2 2

1 mu 2 2

2kqn2 =

...(i)

conservation between  and 4 distance. Loss in potential energy = gain in kinetic energy –2kq n

 1 2 1 2 = mv  mu 2 2 4e

4kqn2 =

1 1 mv 2  mu2 2 2

From eq (i) mu2 = v=

1 1 mv 2  mu2 2 2 3u

RESONANCE

PAGE - 28


98.

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

Ans.

The loop is pulled to the right as indicated .What are the directions of the induced current in the loop and the magnetic 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 (B)

Sol. As loop in pulled to right, there free downward magenetic field is decrease. So, induce emf induced as to oppose 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 left and force on right side of loop is towards right 99.

Two batteries V1 and V2 are connected to three resistors as shown below

Ans.

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 V and V2 = 10V , the current I will be (A) 7 mA (B) 15 mA (C) 20 mA (D) 25 mA (D)

Sol.

V1r1  V2r2 I = r r  (r  r )R 12 2 2 Case – I

V1 = 2V

Case – II V2 = 4V from case (I and Case II)

Case – III

V1 = 10V

V2 = 0V

I = 3mA

V1 = 0

I = 4mA

 3 r1      2 r2  V2 = 10V

................. (i) I=?

10(r1  r2 ) I = r r  (r  r )R 12 1 2

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

from (i) and (ii) I = 25 mA

RESONANCE

PAGE - 29


100.

Ans. Sol.

y2 + = 1. At point (0, b), the x-component of a2 b2 velocity is u. the y-component of acceleration at this point is : (A) – bu2 / a2 (B) – u2 / a2 (C) – au2 / b2 (D) – u2 / a (A) A particle moves in a plane along an eliptic path given by

x2 a2



y2 b2

=1

Different w.r.t.

2x dx 2y dy  =0 ................. (i) a 2 dt b 2 dt ux = u, at (o, b), uy = 0 Different w.r.t. 2x d2 x 2

2



2  dx    a 2  dt 

x2

2

a dt accelerat at (o, b) is

+

2 y d2 y 2

b dt

2



(from eq (i))

2  dy    =0 b 2  dt 

2xo  d2 x  2 2 2bay 2   u  (0) = 0 2  2 2 2 a  dt  a b b2

ay = –

b a2

u2

CHEMISTRY 101.

Sol.

102. Sol.

XeF6 hydrolyses to give an oxide. The structure of XeF6 and the oxide, respectively are (A) octahedral and tetrahedral (B) distorted octahedral pyramidal (C) octahedral and pyramidal (D) distorted octahedral and tetrahedral (B) XeF6 + 3H2O  XeO3 + 6HF distorted pyramidal octahedral MnO4– oxidizes (i) oxalate ion in acidic medium at 333 K and (ii) HCl. For balanced chemical equations, the rations [MnO4– : C2O42–] in (i) and [MnO4– : 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 : 8 (C) 5e– + MnO4– + 8H+  4H2O C2O4–2  2CO2 + 2e– HCl 

103. Sol.

1 Cl2 + H+ + e– 2

If E º Fe 2 / Fe = – 0.440 V and Eº Fe3  / Fe 2  = 0.770 V, then E º Fe3  / Fe is : (A) 0.330 V (B) – 0.037 V (C) – 0.330 V (D) – 1.210 V (B) 2e– + Fe+2  Fe – 0.44 V Fe+3 + 3e–  Fe

0.77 V

Fe+3 + 3e–  Fe

xV

then

x=

2  ( 0.44 )  0.77 = – 0.037 V 3

RESONANCE

PAGE - 30


104. Sol.

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 (A)

E13 E12 105.

1   1  2  2 3  1 8/9 32 =  1 = 1  = 3/4 27  2  2 2  1

In the following conversion, (i) MeMgBr

NaOH / I

(ii) H3O 

H3O

2      X    Y

the major products X and Y, respectively, are :

Sol.

(A) i (C)

(B) ii

106.

The reaction sequence,

HNO2    X

(C) iii

(D) iv

(C) iii

(D) iv

Y

the major products X and Y, respectively, are :

(A) i

RESONANCE

(B) ii

PAGE - 31


Sol.

(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 major diastereomer ‘P’ on treatment with MeI/K2CO3 gave an optically inactive compound. P is :

Ans. Sol.

108.

Sol.

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

At 300 K the vapour pressure of two pure liquids, A and B are 100 and 500 mn Hg, respectively. If in a mixture of 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 (D) Ptotal = P°A + (P0B – P0A) XB 300 = 100 + (500 – 100) (1 – XA) 

1 – XA =

YA =

109.

Sol.

1 2



XA =

1 2

1 1 2 = 6 300

100 

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 (D) – 2.4 and 0.0 (B) Low spin d6 CFSE = – 4 ×

2 3 0 + 0 × 2 5 5

= – 0.4 0 High Spin d6 CFSE = –6 × 110.

Sol.

2  = – 2.4 0 (ignoring pairing) 5 0

Emulsification of 10 ml of oil in water produces 2.4 × 1018 droplets. If the surfaces tension at the oil-water interface is 0.03 Jm–2 and the area of each droplet is 12.5 × 10–16 m2, the energy spent in the formation of oil droplets is : (A) 90 J (B) 30 J (C) 900 J (D) 10 J (A) Total area of droplets = 2.4 × 1018 × 12.5 × 10–16 = 3000 m2 Energy liberated = 3000 × 0.03 = 90 J

RESONANCE

PAGE - 32


BIOLOGY 111.

Ans. 112.

Ans. 113.

Ans. 114.

Ans. 115.

Ans. 116.

Ans. 117.

Ans.

Which sequence of events givens rise to flaccid guard cells and stomatal closure at night ? (A) low [Glucose ]  low osmotic pressure  low pH  high pCO2 (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 (D) 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 (C) 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 (D) 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 (C) 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 (B) 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 (A) 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 (C)

RESONANCE

PAGE - 33


118.

Ans. 119.

For a human male what is the probability that all the maternal chromosomes will end up in the same gamete ? (A) 1/23 (B) 223 (C) 2th (D) (½)23 (D)

Ans.

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 (A)

120.

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

Ans.

 (A) 34 A

(B) 34 nm

 (C) 3.4 A (A)

(D) 3.4 m

RESONANCE

PAGE - 34

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