KVPY – CLASS-XII

PART TEST – 2 (OLTS-1819-T2-PT-2-KVPY-XII)

PART – I

MATHEMATICS

1. The differential equation of 2

1 2 3x c y c y c= + + , where 1 2 3c ,c ,c are real arbitrary constants is

(A)

23 2

3 2

dy d y d y3

dx dx dx

=

(B)

23 2

3 2

d y dy d y3

dxdx dx

=

(C) 3

3

d y0

dx= (D) None of these

Ans. A

Sol. Differentiating we get 1 2

1

12c y c

y= +

22 1 1 12 3

1 1

y1y 2c y 2c

y y

−− = =

Differentiating again 2

1 3 2y y 3y=

2. Line y mx= bisects two distinct chords drawn from (4, 4) on 2y 4x= if

(A) 1

m2

−= (B)

3m

8= −

(C) 5

m8

= − (D) m 1=

Ans. D Sol. Any point on y mx= is (t, mt)

Equation of the chord of the parabola 2y 4x= having (t, mt) as the mid point is

( ) ( ) 2 2y 2mt 2 x t m t 4t− + = −

It passes through (4, 4)

( )2 2m t 2 4m 1 t 8 0− + + =

For two distinct chords, ( )D 0

( )2 24 4m 1 32m 0+ −

28m 8m 1 0+ +

21 1

m2 8

+

3. The sum of all real roots of the equation 6 4 3 2x 4x 10x 4x 1 0+ − + + = is

(A) 2 (B) 0 (C) 1 (D) 4 Ans. A

Sol. Clearly x 0= is not a solution.

Dividing by 3x

3

3

1 1x 4 x 10 0

xx

+ + + − =

Let 1

x zx

+ =

( ) ( )3 2z z 10 0 z 2 z 2z 5 0 + − = − + + =

For 2z 2z 5 0+ + = Discriminant < 0

Real roots of 1

x 2x

+ =

( )2

x 1 0 x 1,1 − = =

Sum = 2 4. Let w be a cube root of 1 other than 1. Then number of ordered pairs (a, b), where

a,b Z (set of all integers) such that ( ) ( )2a b a b 1 + + = is

(A) 4 (B) 5 (C) 6 (D) 8 Ans. C

Sol. On multiplying 2 2a b ab 1 0+ − − = ( )3 21, 1 0 = + + =

It is Quadratic in a discriminate 0

2b 3 b 1, 0, 1 −

If b 1 a 0, 1= − −

If b 0 a 1, 1= −

If b 1 a 0, 1=

5. Let 1 2 3 2016a ,a ,a ,..........a are the roots of the equation 1008x 7x 2016 0,− − = then the value of

2016 2016 2016

1 2 2016a a .................. a

2016

+ + + is

(A) 2016 (B) ( )2016

2016

(C) ( )2

2016 (D) ( )2016

Ans. A

Sol. Given 1 2 2016a ,a ..........a are roots of 1008x 7x 2016 0− − =

( )1008

i ia 7a 2016 = + i 1,2........2016=

( ) ( ) ( ) ( )( )2016 2 2

i i ia 7a 2016 14 a 2016 = + +

( ) ( ) ( ) ( )( )2016 2016 2016

22016 3

i i i

i 1 i 1 i 1

a 49 a 2016 14 2016 a= = =

= + +

but ( )2016 2016

2

i i

i 1 i 1

a 0 a= =

= =

Given expression ( )( )

32016

20162016

= =

6. Find the coefficient of 2009x in the expansion of ( ) ( )20072008 21 x 1 x x− + +

(A) –1 (B) –2 (C) 0 (D) 1 Ans. C

Sol. ( ) ( )20072008 21 x 1 x x− + +

( ) ( ) ( )2007

21 x 1 x 1 x x = − − + +

( ) ( )2007

31 x 1 x= − −

( ) ( )2007 2007

3 31 x x 1 x= − − −

7. The line passing through the extremity A of the major axis and extremity B of the minor axis

of the ellipse. 2 2x 9y 9+ = , meets the auxillary circle at the point M, then the area of the

triangle with vertices at A, M and the origin O is

(A) 31

10 (B)

29

10

(C) 21

10 (D)

27

10

Ans. D

Sol. Equation of the ellipse is 2 2x y

19 1

+ = and of the

auxillary circle is 2 2x y 9+ = .

Equation of AB is x y

13 1+ =

y – coordinate of M is given by ( )2 23 1 y y 9 − + =

210y 18y 0− =

y 0= or 9

y5

=

Area of ( ) ( )1

OAM OA MN2

=

1 9 27

32 5 10

= =

M

B

N 0 A (3, 0)

8. The range of a for which the expression ( )( )

( )

x 1 x 5y

x a

− −=

− is capable of taking all real

values for x R (the set of all real numbers), is

(A) ( ), 1− (B) ( )a 5,

(C) ( )a 1, 5 (D) a

Ans. C

Sol. ( )2x y 6 x 5 ay 0− + + + =

If x R, then ( ) ( ) ( )2 2y 6 4 5 ay 0 y 4 3 a y 16 0+ − + + − +

Disc. ( )2

0 16 3 a 64 0 1 a 5 − −

If a 1,5= then x cannot take values 1 and 5

( )a 1, 5

9. If equations 2x 3x 4 0− + = and ( )24x 2 3a b x b 0 a,b R− + + = have a common root

then the complete set of values of a, is (where [.] denotes the Greatest Integer function)

(A) 11 10

,3 3

− −

(B)

10, 3

3

− −

(C) 10

, 33

− −

(D)

103,

3

Ans. C

Sol. Since roots of 2x 3x 4 0− + = are imaginary, therefore both roots will be common

2 3a b4 b

b 16 3a b 6 3a 16 6 3a 101 3 4

+ = = = + = + = = −

10 10

10 3a 9 a 3 a , 33 3

− − − − −

10. A curve passes through the point 1,4

and its slope at any point is given by 2y y

cosx x

−

.

Then the curve has the equation

(A) 1 x

y x tan ne

− =

(B) ( )( )1y x tan n ex−=

(C) 1 e

y x tan nx

− =

(D) none

Ans. A

Sol. 2dy y ycos

dx x x

= −

2 2dy dy dv dxy vx v x v x v cos v sec v dv 0

dx dx dx x= = + + = − + =

tamv lnx c + =

y

tan lnx c c 1x

+ = =

y e

t v lnx lnx x

= − =

1y etan n

x x

− =

11. A circle with radius unity has its centre on the positive y – axis. If this circle touches the parabola y= 2x2 tangentially at the points P and Q then the sum of the ordinates of P and Q is

(A) 15

4 (B)

15

8

(C) 2 15 (D) 5

Ans. A

Sol. Let centre of circle be (0, k) i.e. circle is ( )22x y k 1+ − = . Solving with 2 y

x2

= , we get

( )2y

y k 12+ − =

i.e. 2 21y 2k y k 1 0

2

− − + − =

Since the two curves touch each other

Disc.

2

21 170 2k 4k 4 0 k

2 8

= − − + = =

Putting 17

k8

= in above equation, we get 2 15 225y y 0

4 4− + =

1 2

15y y

4 + =

12. AB is a double ordinate of hyperbola 2 2

2 2

x y1

a b− = such that AOB (where ‘O’ is the origin) is

an equilateral triangle, then the eccentricity of the hyperbola satisfies

(A) e 3 (B) 2

1 e3

(C) 2

e3

= (D) 2

e3

Ans. D

Sol. If ( )1 1A x ,y ,= then 2 2

1 1 1 1 1x y 2y x 3 y asec 3 b tan+ = =

( )2

2 2 2

2

b 1 sec 1 sec 1 4 1e 1 1 1 cot cot

a tan 3 3 3 3tan3

= = + = + + = +

2

e3

13. If all the real solutions of the equation ( ) ( )x x4 a 3 2 a 4 0− − + − = are non positive, then

(A) 4 < a 5 (B) 0 < a< 4 (C) a> 4 (D) a< 3 Ans. A

Sol. ( ) ( )x x4 a 3 2 a 4 0− − + − =

x 0, Let xy 2=

( ) ( )2y a 3 y a 4 0− − + − =

The roots of Quadratic must lie b/w (0, 1]

(a) ( ) ( )2

a 3 4 a 4 0− − −

2a 9 6a 4a 16 0+ − − +

2a 10a 25 0− +

a R

(b) a 3

0 12

−

0 a 3 2 − 0 a 5

(c) f(0) = a – 4 0 a 4

(d) f(1) = 1 – a + 3 + a – 4 0 a (4, 5] 14. If p(x) = ax² + bx and q(x) = lx² + mx + n with p(1) = q(1), p(2) – q(2) = 1 and p(3) – q(3) = 4,

then p(4) – q(4) is (A) 0 (B) 5 (C) 6 (D) 9 Ans. D

Sol. We have, p(1) – q(1) = 0 (a + b) – (l + m + n) = 0 …… (i)

p(2) – q(2) = 1 (4a + 2b) – (4l + 2m + n) = 1 …… (ii)

p(3) – q(3) = 4 (9a + 3b) – (9l + 3m + n) = 4 …… (iii) now use the operation 3 × (3) + 1 × (1) – 3 × (2),

we get (16a + 4b) – (16l + 4m + n) = 9 p(4) – q(4) = 9. 15. Circles are drawn on chords of the rectangular hyperbola xy=4 parallel to the line y=x as

diameters. All such circles pass through two fixed points whose coordinates are (A) (2,2) and (-2,-2) (B) (2,-2) and (-2,-2) (C) (-2,2) and (-2,-2) (D) (2,-2) and (2,2) Ans. A

Sol. Circle with points 11

2P 2t ,

t

and 22

2Q 2t ,

t

as diameter is given by

( ) ( )1 21 2

2 2x 2t x 2t y y 1

t t

− − + − − =

Also, slope of PQ is given by

1 21 2

11 t t 1

t t− = = −

hence, from (1), circle is

( ) ( )( )2 21 2x y 8 t t x y 0+ − − + − =

which is of the form S L 0+ = .

Hence, circles passé through the points of intersection of the circle 2 2x y 8 0+ − = and the

line x=y. The points of intersection are (2,2) and (-2,-2) 16. Sum of all the possible real values of x for which the sixth term of the expansion of

( ) ( ) x 2x 2

73

71/5 log 4 .3 9log 9E 3 7

−− −

= +

is 567, is

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

Ans. D

Sol. Put x 2

3x 2log 9

3 3y 3 log y log 9− −

= =

x 2 x 2

y 9 3− −

= =

Next, put ( ) ( ) x 2

71/5 log 4 .3 9

z 7− −

=

( ) x 2

7 7

1log z log 4 3 9

5

− = −

( )1/5

x 2

7log 4 3 9− = −

( )1/5

x 2z 4 3 9

− = −

Now, ( )7

E y z= + and 6th term is given by

( ) ( ) 2

x 2 x 27 7 5 5

6 5t C y z 21 3 4 3 9− −−= = −

( ) 2 x 2 x 2567 21 3 4 3 9

− −= −

( ) ( ) ( ) x 2 2 x 227 4 3 9 3

− −= −

( ) ( )2 x 23 x 2

27 4 3 9−−

= −

3 24u 9u 27 0− − = where x 2

u 3−

=

Now that u 3= satisfies this equation

x 2

3 3 x 2 1 x 2 1−

= − = − =

x 2 1 3= = or 1.

17. Let a,b,c are complex quantities and roots of 3 2z az bz c 0+ + + = are unimodular then

a b− equal to

(A) 0 (B) 1 (C) –1 (D) 2 Ans. A

Sol. a ++ =

b + + =

1 1 1

b + + =

b a + + = =

18. If y1 and y2 are two different solutions of the equation dy

P(x)y Q(x)dx

+ = and y = 3y1 + 2y2

is also the solution of the equation, (where - = 4), then which of the following is not

correct. (, ϵ R)

(A) 4 + = 5 (B) + 4 = 7

(C) 5 + 5 + 2 = 0 (D) 5 + 10 + 13 = 0 Ans. B

Sol. 11

dyP(x)y Q(x)

dx+ = , 2

2

dyP(x)y Q(x)

dx+ =

( )1 21 2

d(3 y y )P(x) 3 y 2 y Qx

dx

+ + + =

3Q(x) + 2(Q(x)) = Q(x) 3 + 2 = 1 and - = 4

9

5 = ,

11

5

− =

19. If , be the roots of ax2 + bx + c = 0 (a, b, c R), c

a<1 and 2b 4ac 0− , f(n) = ( )

nr r

r 1=

+ ,

then nLt→

f(n) is equal to

(A) 1

a1

c−

(B) 1

a1

c−

(C) c

a c− + (D)

2

a1

c−

Ans. D

Sol. 2 c c

za a

= = =

20. Let ( ) ( )n

2 2

k

k 0

f n n 2k C=

= − f find f (6).

(A) 3021 (B) 3022 (C) 3023 (D) 3024 Ans. D

Sol. We have ( ) ( ) ( )2 2 2n 2k n k k 2k n k− = − + − −

Thus, ( ) ( )n n

2 22 2

k n k

k 0 k 0

n 2k C n k C−

= =

− = −

( )n n

2 2 2

k k

k 0 k 0

k C 2 k n k C= =

+ − − k n kC C−

=

( )n n

2 2 2

k k

k 0 k 0

2 k C 2 k n k C= =

= − −

But ( ) ( ) ( )( )2

k k n kk n k C k.C n k C−

− = −

( ) ( )n 1 n 1

k 1 n k 1n C n C− −

− − −=

( ) ( )2 n 1 n 1

k 1 kn C C− −

−=

Thus, ( )n

2

k

k 0

k n k C=

−

( )( ) ( )n 1

2 n 1 n 1 2 2n 2

k 1 k n

k 1

n C C n C−

− − −

−

=

= =

Hence,

( ) ( ) ( )n

2 2 2 2n 2 2 2n 2

k n 1 n

k 0

n 2k C 2n C 2n C− −

−

=

− = −

( )( )

( )

( )

( )2

2

2n 2 ! 2n 2 !2n

n! n 2 !n 1 !

− −= −

−−

( ) ( )

( )( )

22n 2n 2 !n n 1

n! n 1 !

− = − − −

( ) ( )2n 2

n 12n C−

−=

PHYSICS

21. A transverse wave is described by the equation y = y0 sin 2 (ft – x/a). The maximum particle velocity is equal to four times the wave velocity if a is equal to

(A) y0 / 4 (B) y0/2

(C) y0 (D) 2y0 Ans. B

Sol. The maximum particle velocity of a SHM of amplitude Y0 and frequency f is 2fY0. The wave

velocity is f. For 2 fY0 to be equal to 4f, has to be Y0 / 2. (Here = a). 22. If the time of revolution of a satellite is T, then Kinetic energy is proportional to (A) 1/T (B) 1/T2

(C) 1/T3 (D) 1/T2/3 Ans. D

Sol. Orbital velocity 2 r

vT

=

Hence, KE 2 4/3

2 2 2/3

r T 1

T T T= =

23. A rubber rod of length 10 m density 1.3 103 kg/m3 and Young’s modulus 6 106 N/m2 hangs from the ceiling of a room. The elongation produced in the rod due to its own weight is

(A 2 cm (B) 2.3 cm (C) 6.6 cm (D) 10.6 cm Ans. D Sol. Mass of the rod = AL if A is its cross sectional area

Weight acts at the mid point Y = ( )L / 2mg

A L

if L is the original length

L = mgL

2AY =

2g L

2Y

=

9.8 1.3

120

= 10.6 cm

24. A stationary observer receives sound waves from two tuning forks, one of which

approaches and the other recedes with the same velocity. The observer hears beats of

frequency = 2 Hz. What is the velocity of each turning fork if their actual frequency 0 = 680 Hz, and velocity of sound in air is Vs = 340 m/s.

(A) 0.1 m/s (B) 0.2 m/s (C) 0.3 m/s (D) 0.5 m/s Ans. D

Sol. n = 0

s

C vn

C v

−

−

n = 340 0

340 v

−

−× 680 =

1340 680 v v

1 680 1340 340 340

−

− = +

340

n 680340 V

= +

= v

680 1340

−

n - n = 2

v v

680 1 1 2340 340

+ − + =

4v = 2

v = 1/2. Putting all the values v = 0.5 m/s 25. An open pipe is closed at one end. As a result the frequency of the third harmonic of the

closed pipe is found to be higher than the fundamental frequency of the open pipe by 100 Hz. What is the fundamental frequency of the open pipe?

(A) 50 Hz (B) 100 Hz (C) 200 Hz (D) 300 Hz Ans. C

Sol. 3 V

V 100 Hz4L 2L

= +

1 V

100 Hz4 L

=

V

200 Hz2L

=

26. An ideal gas ( = 1.5) is expanded adiabatically. How many times has the gas to be expanded to reduce the root mean square velocity of molecules 2.0 times

(A) 4 times (B) 16 times (C) 8 times (D) 2 times Ans. B

Sol. rms

3RTV

M=

rmsV T

Vrms is to reduce two times i.e, temperature of the gas will have to reduce four times or

T 1

T 4

=

During adiabatic process

1 1TV T V− − =

or, ( )

111

21.5 1

V T4 4 16

V T

−−

= = = =

V 16V =

27. A small ball of mass m and density is dropped in a long tube filled with a liquid of density

3

. The viscous force acting on the ball after a long time will be

(A)mg

3 (B)

2mg

3

(C) mg (D) cannot be determined Ans. B

Sol. F = 6rv

= ( )f6 r v after long 34r . m

3 =

= 6 ( ) 2.gr2

r.9

−

= 34. r .g

3 3

−

= mg.2

3

= 2

mg3

28. A particle of mass 2 kg is performing SHM, given by equation x = 10 sin 2t (x is in m and t in s). The work done on the particle in time 0.25 sec to 0.75 sec, will be

(A) 2002J (B) 1002J

(C) 50J (D) zero Ans. D

Sol. F = 2

2

2

d xm 2x (10 4 sin 2 t)

dt= −

= − 80 2 sin 2t

dx = 20 cos 2 t dt

W = 0.75 0.75

2

0.25 0.25

F dx 80 20 sin2 t cos2 t dt= −

= − 40 2 20 0.75

0.25

sin 4 t dt = 3

0.75

0.25

800cos 4 t 0

4

=

29. A mass M is oscillating with frequency fo if hung with a spring of

stiffness k. Now it is cut in two parts in the ratio of 2 : 1 and connected as shown. The new frequency is now

(A)1 k

2 M (B)

1 3k

2 M

(C)1 3k

2k 2M (D)

3 k

2 2M

M M

k1

k2

Ans. D

Sol. 1

3k k

2= ; 2k 3k=

keff = k1 + k2 = 1 9k 9

k2 2m 2

=

30. A simple pendulum oscillating with small amplitude has time period T. What will be the

percentage change in its time period if its amplitude is decreased by 5 %? (A) 6 % (B) 3 % (C) 1.5 % (D) 0 % Ans. D Sol. For small amplitude the time period is independent of the amplitude therefore there will be

no change in its time period.

31. A string of length 1 m and mass 10 gm is tightly clamped at its ends. The tension in the string is 4 N. Identical wave pulses are produced at one end at equal intervals of time t. What is the minimum value of t which allows a constructive interference between successive pulses?

(A) 0.05 s (B) 0.1 s (C) 0.15 s (D) 0.2 s Ans. B

Sol. v = 3

T 4 120 m / s

10 10−

= =

t = 2

0.1v

= sec

32. A liquid is kept in a vertical cylindrical vessel which is being rotated about its axis. The liquid

rises at the sides. If the radius of the vessel is 0.05 m and the speed of rotation is 2 rev/s, find the difference in the heights of the liquid at the centre of the vessel and at its sides.

(A) 10 cm (B) 12 cm (C) 6 cm (D) 2 cm Ans. D Sol. According to Bernoulli's theorem

P + 1

2v2 = constant

or P1 + 2

1

1v

2 = P2 + 2

2

1v

2

or P = 1

2v2

or hg = 1

2v2

or h = 1

2

2v

g

But v = r = 2 2 0.05 = 0.2 m/s

20.04

h2 9.8

=

= 0.02 m

h

33. A cubical block of wood of specific gravity 0.5 and a chunk of concrete of specific gravity

2.5 are fastened together. The ratio of the mass of wood to the mass of concrete which makes the combination to float with its entire volume submerged under water is

(A) 1

5 (B)

1

3

(C) 3

5 (D)

2

5

Ans. C Sol. let mass of wood = m1 And mass of concrete = m2

( )1 21 2

m m. .g m m g

0.5 2.5

+ = +

m1 = 2

1.5m

2.5

1

2

m 3

m 5=

34. Find work done on 2 moles of an ideal gas in the process

A→B shown in the P-V diagram. Path A to B is semicircular.

(A) ( ) ( ) ( )1 2 1 2 1 2 1

1P V V V V P P

4− + − −

(B) ( ) ( ) ( )1 2 1 2 1 2 1

1P V V V V P P

2+ + + +

(C) ( ) ( ) ( )1 2 1 2 1 2 1

1P V V V V P P

2− + − +

(D) ( ) ( ) ( )1 2 1 2 1 2 1

1P V V V V P P

2+ + + −

B A

V1 V2 V

P1

P2

P

• •

Ans. A Sol. work done = area enclosed from V1 to V2

= ( ) ( ) ( )1 2 1 2 1 2 1

1P V V V V P P

4− + − −

35. A black body is at a temperature of 2880K. The energy of radiation emitted by this object

with wave length between 499 nm and 500 nm is U1, between 999 nm and 1000 nm is U2 and between 1499 nm and 1500 nm is U3. Wien’s constant b=2.88 x 106 nmK. Then

(A) U1 = 0 (B) U3 = 0 (C) U1 = U2 (D) U2 > U1

Ans. D

Sol. From Wien’s Law, mT = constant, where T is temperature of black body and m is

wavelength corresponding to maximum energy of emission. Energy distribution of blackbody radiation is given below :

1. 1U and 3U are not zero because a blackbody emits nearly radiation’s of all wavelengths.

2. Since 1U corresponds lower wavelength and 3U corresponds higher wavelength and 2U

corresponds medium wavelength. Hence 2 1U U .

36. Infinite numbers of particles each of mass 2kg each are placed on the positive x-axis at a

distance 1m, 2m, 4 m and 8 m respectively from the origin. The net intensity of gravitational field at the origin will be

(A) 8G/3 (B) (C) zero (D) data insufficient Ans. A Sol. |E| = GM/r2

Enet = G 22 2 2

1 1 1 1. . .

1 2 4 8

+ + + +

(as a + ar + ar2 + . . . = a/1−r)

= G 21

1 1 4

−

= 8G/3

37. An open vessel containing water is moving with a constant acceleration a in the horizontal

direction. The free surface of water gets shaped with the horizontal at an angle

(A) = cos-1 g/a (B) = tan-1 a/g

(C) = sin-1 a/g (D) = tan-1 g/a Ans. B Sol. The free surface becomes perpendicular to the resultant force

tan = ma

mg

( )1tan a / g− =

mg

ma

water particle

38. An open pipe of length 47cm has fundamental frequency 340 Hz. What is the diameter of

the tube if velocity of sound in air is 340 m/sec? (A) 2 cm (B) 5 cm (C) 8 cm (D) 10 cm Ans. B Sol. Antinodes are formed at a distance 0.6r from the end of the tube, where r is radius of the

tube.

47 + 2 .6r = /2

= v/n = 100 cm 47 + .6D = 100/2 D = 5 cm 39. P –V diagram of an ideal gas is as shown, work done by the

gas in the process ABCD is

(A) 0 04 P V

(B) 0 02 P V

(C) 0 03 P V

(D) 0 0P V

A

P

2P0

P0

V0 2V0 3V0 V

B

C D

Ans. C

Sol. AB 0 0W P V= −

CD 0 0W 4P V= +

ABCD 0 0W 3P V= +

40. In the figure the intensity of waves arriving at D from two

coherent sources S1 and S2 is I0. The wave length of the

wave is = 4m. Resultant intensity at D will be (A) 4I0

(B) I0 (C) 2I0

(D) zero

S1

S2

4m

3m

D

Ans. C

Sol. Path difference = 5 – 4 = 1 m

Phase difference = = 2

. x

=

2

4

=

2

( )R 1 2 1 2I I I 2 I I cos= + +

R 0 0I I I 0= + +

R 0I 2I=

CHEMISTRY

41. Which monomer will give radical polymerisation most readily?

(A) H2C = CH2 (B) C6H5 – CH = CH2

(C) CH3 – CH = CH2 (D)

Ans. B

Sol. 6 5 3C H CHCH•

is most stable free radical.

42. Ziegler-Natta catalyst is: (A) R3Al (B) TiCl4 (C) R3Al/TiCl4 (D) R3B/TiCl2

Ans. C Sol. Fact based

43. ( ) ( )2 26 4 nNH CH NHCO CH CO

is a :

(A) homopolymer (B) copolymer (C) addition polymer (D) thermosetting polymer Ans. B Sol. Fact based 44. Which of the following is the first member of aldose monosaccharides?

(A) 2 2CH OH C CH OH− −

O

(B) CH2OH – CHOH - CHO

(C) CH2OH – CHOH – CHOH - CHO (D)

2 2CH OH CHOH C CH OH− − −

O

Ans. B

Sol. Glyceraldehyde is first member of monosaccharides.

45. Which of the following hexoses will form the same osazone when treated with excess phenyl hydrazine?

(A) D-glucose, D-fructose and D-galactose (B) D-glucose, D-fructose and D-mannose (C) D-glucose, D-mannose and D-galactose (D) D-fructose, D-mannose and D-galactose

Ans. B Sol. Fact based 46. Which of the following is NOT a reducing sugar?

(A)

(B)

(C)

(D)

Ans. C Sol. Acetal sugars are non-reducing sugar. 47.

HOH

C - OH

OHH

OHH

CH2OH

HC OH

, the given structural formula is enol form of:

(A) D-glucose (B) D-mannose (C) D-fructose (D) All of these

Ans. D

Sol. Due to Van Ekenstein transformation. 48. If Pd vs P(Where P denotes pressure in atom and d denotes density in gm/L) is plotted for

He gas(assume ideal) at a particular temperature, and ( )P 8.21 atm

dPd 5

dP =

=

, then the

temperature will be (A) 160 K (B) 320 K (C) 80 K (D) none of these

Ans. A

Sol. PM = dRT

( )

2 MPd p

2T

d Pd 2PM5

dP RT

=

= =

0.2 8.21 4

50.0821 T

=

T = 160 K 49. Two closed vessels A and B of equal volume containing air at pressure P1 and temperature

T1 are connected to each other through a narrow open tube. If the temperature of one is now maintained at T1 and other at T2 (Where T1 > T2) then that will be the final pressure?

(A) 1

1 2

T

2PT (B) 1 2

1 2

2PT

T T+

(C) 1 1

1 2

2PT

T T− (D) 1

1 2

2P

T T+

Ans. B Sol. Let T1 > T2: final pressure will be same, Let x mole transfer from A to B vessel

PAV = (n - x)RT and PAV = (n + x)RT2

( )1 2

1 2

n T Tx

T T

− =

+

( )( )

1

1

1 21A 1

1 1 2

1 2A

1 2

nRTV

P

n T TnRTP n RT

P T T

2PTP

T T

=

− = − +

=+

50. A given volume of ozonised oxygen (containing 60% oxygen by volume) required 220 sec to

effuse while an equal volume of oxygen took 200 sec only under identical conditions. If density of O2 is 1.6 g/L then find density of O3.

(A) 1.936 g/L (B) 2.16 g/L (C) 3.28 g/L (D) 2.24 g/L Ans. D Sol. Let V mL of gas effused

( )22O

mix

mix

dV / 220d 1.6 1.1 1.936g / L

V / 200 d= =

Let density of ozone is d; In 100 volume ozonised oxygen, 60% O2 and 40% by volume O3 is present.

Mass of mixture = mass of ozone + mass of oxygen

100 1.936 = 40 d + 60 1.6 Density of O3 is 2.44 g/L 51. Fe2O3(s) may be converted to Fe by the reaction

( ) ( ) ( ) ( )2 3 2 2Fe O s 3H g 2Fe s 3H O g+ +

for which KC = 8 at temp 720oC. What percentage of the H2 remains unreacted after the reaction has come to equilibrium? (A) ~ 22% (B) ~ 34% (C) ~ 66% (D) ~ 78% Ans. B

Sol. Let initial moles of H2(g) is 1

( ) ( ) ( ) ( )2 3 2 2Fe O s 3H g 2Fe s 3H O g+ +

At equ 1 – 3x 3x

3

c 3

3x

VK

1 3x

V

3x8

1 3x

=−

=

−

x = 0.22 % of H2 unreacted

1 3 0.22

100 34%1

− = =

52. For the reaction (1) and (2)

( ) ( ) ( ) ( )

( ) ( ) ( )

A g B g C g ..... 1

X g 2Y g ..... 2

+

1 2P PGiven, K :K 9 :1=

If the degree of dissociation of A(g) and X(g) be same then the total pressure at equilibrium (1) and (2) are in the ratio

(A) 3 : 1 (B) 36 : 1 (C) 1 : 1 (D) 0.5 : 1 Ans. B

Sol. ( ) ( ) ( )A g B g C g+

t = 0 a mole

teq (a - a) (a) (a) if PTotal = P1 Then

1

1 1

p

1

P P1 1

K1

.P1

+ + =

−

+

Also

( ) ( )X g 2 Y g

t = 0 b mole

teq (b - b) (2b) if PTotal = P2 Then

2

1

2

2

2

p 1

2

P

P

2.P

1K

1.P

1

K 1Given :

K 9

+ =

−

+

=

53. Le-Chatelier principle is NOT applicable to

(A) ( ) ( ) ( )2 2H g I g 2HI g+ (B) ( ) ( ) ( )Fe s S s FeS s+

(C) ( ) ( ) ( )2 2 3N g 3H g 2NH g+ (D) ( ) ( ) ( )2 2N g O g 2NO g+

Ans. B Sol. Fact based. 54. Two solid compounds X and Y dissociates at a certain temperature as follows

( ) ( ) ( )

( ) ( ) ( )

1

2

3 3

P

3 3

P

X s A g 2B g ;K 9 10 atm

Y s 2B g C g ;K 4.5 10 atm

−

−

+ =

+ =

The total pressure of gases over a mixture of X and Y is (A) 4.5 atm (B) 0.45 atm (C) 0.6 atm (D) None of these Ans. B Sol. Let x is partial pressure of A and y is partial pressure of C when both equilibrium

simultaneously established in a vessel.

( ) ( ) ( )( )

( )

( ) ( ) ( )( )

( )

1

2

2

P A B totalx 2x 2y

2

P C B totaly 2y 2x

P1

P2

X s A g 2B g K P .P

Y s C g 2B g K P .P

K xx 2y

K y

+

+

+ =

+ =

= =

KP1 = x(2x + 2y)2

x = 0.1 atm

y = 0.05 atm Total pressure of gases = PA + PB + PC = 3(x + y) = 0.45 atm 55. In a saturated solution of AgCl, NaCl is added gradually. The concentration of Ag+ is plotted

against the concentration of Cl–. The graph appears as:

(A)

(B)

(C)

(D)

Ans. C

Sol. For AgCl Ksp = [Ag+][Cl–] y x

xy = constant i.e., Rectangular hyperbolic. 56. Which of the following is most soluble in water?

(A) Ba3(PO4)2 (Ksp = 6 10–39) (B) ZnS (Ksp = 7 10–19)

(C) Fe(OH)3(Ksp = 6 10–38) (D) Ag3PO4(Ksp = 1.8 10–18) Ans. D Sol. For most soluble salt, solubility should be maximum.

57. At 25oC, Ksp for PbBr2 is equal to 8 10–5. If the salt is 80% dissociated, what is the solubility of PbBr2 in mol/litre?

(A)

1/3410

1.6 1.6

−

(B)

1/3510

1.6 1.6

−

(C)

1/3410

0.8 0.8

−

(D)

1/2510

1.6 1.6

−

Ans. A

Sol. ( ) ( ) ( )2

20.8 S 2 0.8 S

PbBr s Pb aq 2Br aq+ −

+

Ksp = [Pb2+][Br–]2

8 10–5 = (0.8 S) (1.6 S)2

S =

1/3410

1.6 1.6

−

58. The simultaneous solubility of AgCN(Ksp = 2.5 10–16) and AgCl(Ksp = 1.6 10–10) in 1.0 M

NH3(aq) are respectively: ( ) 7

f 3 2Given :K Ag NH 10

+ =

(A) 0.037, 5.78 10–8 (B) 5.78 10–8, 0.037

(C) 0.04, 6.25 10–8 (D) 1.58 10–3, 1.26 10–5 Ans. A

Sol. ( ) ( ) ( ) ( ) ( )3 3 2AgCl s 2NH aq Ag NH aq Cl aq

+ −+ +

K1 = 1.6 10–10 107 = 1.6 10–3

( )

3 2

2

3

Ag NH Cl

NH

+ −

=

( ) ( ) ( ) ( ) ( )3 3 2AgCN s 2NH aq Ag NH aq CN aq

+ −+ +

K2 = 2.5 10–16 107 = 2.5 10–9

( )

3 2

2

3

Ag NH CN

NH

+ −

=

( )

35

9

2

1 2

8

5

Cl 1.6 106.4 10

2.5 10CN

x xK 0.04

1 2x1 2x

x 0.037

0.037CN 5.78 10

6.4 10

− −

−−

− −

= =

= =−−

=

= =

59. In the closest packing of atoms (A) the size of tetrahedral void is greater than that of octahedral void. (B) the size of tetrahedral void is smaller than that of octahedral void. (C) the size of tetrahedral void is equal to that of octahedral void. (D) the size of tetrahedral void may be greater or smaller or equal to that of octahedral void

depending upon the size of atoms. Ans. B

Sol. c

a

r

r= 0.225 (for tetrahedral void)

c

a

r

r= 0.414 (for octahedral void)

60. Which of the following figures represents the cross-section of an octahedral site?

(A)

(B)

(C)

(D)

Ans. D Sol. Fact based

PART – II

MATHEMATICS

61. A particle P starts from the point 0z 1 2i,= + where i 1= − . It moves first horizontally away

from the origin by 5 units and then vertically away from the origin by 3 units to reach a point

1z . From 1z the particle moves 2 units in the direction of ˆ ˆi j+ and then it moves through

an angle 2

in the anticlockwise direction on a circle with centre at origin, to reach point 2z .

The point 2z is given by

(A) 6 7i+ (B) 7 6i− +

(C) 7 6i+ (D) 6 7i− +

Ans. D

Sol. Affix of B is ( )1 2i 5 6 2i+ + = +

Affix of C is ( )6 2i 3i 6 5i+ + = +

Affix of D is ( )1 i

6 5i 2 7 6i2

+ + + = +

Affix of E is ( ) ( )i/27 6i e 7 6i i 6 7i+ = + = − +

E

Z2

2

C Z1

D

B A Z0 (1+ 2i)

5 3

62. If , , are such that 2 2 2 3 3 32, 6, 8,++ = + + = + + = then 4 4 4 + + is

(A) 5 (B) 18 (C) 12 (D) 36 Ans. B

Sol. We have ( ) ( )2 2 2 2 2 + + = + + + + +

( )4 6 2= + + +

1 + + = −

Also, 3 3 3 3 + + −

( ) ( )2 2 2= + + + + − − −

( )8 3 2 6 1− = +

3 8 14 6 = − = − or 2 = −

Now, ( )2

2 2 2 4 2 22 + + = +

( ) ( )22 2 2 = + −

( ) ( ) ( )24 36 2 1 2 2 2 18 = − − − − =

63. Through any point (x, y) of a curve which passes through the origin, lines are drawn parallel

to the coordinate axes. The curve, given that it divides the rectangle formed by the two lines and the axes into two areas, one of which is twice the other, represents a family of

(A) circles (B) parabolas (C) hyperbolas (D) straight lines Ans. B

Sol. Let ( )P x,y be the point on the curve passing through

the origin O (0, 0), and let PN and PM be the lines parallel to the x – and y – axes, respectively (Figure).

If the equation of the curve is ( )y y x= , the area

POM equals x

0y dx and the area PON equals

x

0

xy y dx− . Assuming that ( )2 POM PON,= we

therefore have x x x

0 0 02 y dx xy y dx 3 y dx xy= − = .

Differentiating both sides of this gives

dy dy dy dx3y x y 2y x 2

dx dx y x= + = =

log y 2log x C= + 2y Cx ,= with C

being a constant. This solution represents a parabola. We will get a

similar result if we had started instead with 2 (PON) = POM.

O

y

P(x,y)

M x

N

64. The orthogonal trajectories for 2y Cx ,x 0= where C is arbitrary constant represent

(A) a family of circles (B) a family of straight lines (C) a family of ellipses (D) a family of hyperbola Ans. C

Sol. We have ( ) 2f x,y,c y cx 0= − =

2

yc

x= . Differentiating

2

4

x y ' 2xy0

x

−=

2y

y 'x

=

Replacing dy

dxby

dx,

dy− we have

xy '

2y

−=

2y dy x dx+ = constant

2 21y x

2+ = constant which represent a family of ellipses

65. Equation of the curve obtained by reflecting the parabola P; 2y 4x= in the tangent to the

parabola P at the point (1, 2) is

(A) 2x 4y= (B) ( ) ( )2

x 1 4 y 2− = −

(C) ( ) ( )2

x 1 4 y 1+ = − (D) ( ) ( )2

x 1 4 y 1− = +

Ans. C

Sol. Equation of tangent to the parabola at (1, 2) is x y 1 0− + =

Any point P on the parabola is ( )2t , 2t whose reflection Q (h, k) in the line x y 1 0− + = is

given by ( )22 2 t 2t 1h t k 2t

1 1 2

− − +− −= =

−

h 2t 1 = − and 2k t 1= +

Now eliminate t.

66. A monic quadratic trinomial P(x) is such that ( )P x 0= and ( )( )( )P P P x 0= have a

common root, then (monic polynomial has its leading coefficient equal to 1)

(A) ( ) ( )P 0 .P 1 0 (B) ( ) ( )P 0 .P 1 0

(C) ( ) ( )P 0 .P 1 0= (D) none

Ans. C

Sol. Let x = be root of both ( )P x 0= as well as of ( )( )( )P P P x 0=

( )P 0 =

( )( )( ) ( )( )P P P 0 P P 0 0 = =

If ( ) 2P x x ax b,= + + then ( ) ( )( ) ( ) ( )P 0 b P P 0 P b P b 0= = =

( ) ( ) ( )2b ab b 0 b 1 a b 0 P 0 P 1 O + + = + + = =

67. Let 1 2 3z ,z ,z be three distinct complex numbers lying on a circle with centre at the origin such

that 1 2 3 2 3 1z z z ,z z z+ + and

3 1 2z z z+ are real numbers, then 1 2 3z z z equals

(A) – 1 (B) 0 (C) 1 (D) none of these Ans. C

Sol. As 1 2 3z ,z ,z lie on a circle with centre at the origin,

1 2 3z z z r= = = (say)

As 1 2 3z z z R+ . 1 2 31 2 3z z z z z z+ = +

2 4

1 2 3

r r

z z z= +

( )2 2

2 3 1

1 2 3

r z z z z

z z z

+=

2

1 2 3

2

1 2 3 2 3 1

z z zr

z z z z z r z

+=

+ (1)

Similarly, 2

1 2 3

2

1 2 3 3 1 3

z z zr

z z z z z r z

+ +=

+ (2)

and 2

3 1 2

2

1 2 3 1 2 3

z z zr

z z z z z r z

+=

+ (3)

From (1), (2) and (3), we get

2

1 2 3 2 3 1 3 1 2

2 2 2

1 2 3 2 3 1 3 1 2 1 2 3

z z z z z z z z zr

z z z z z r z z z z z z z r z

+ + += = =

+ + +

( )

( )1 3 3 2 3 1

2 2

2 3 1 3 1 2

z z z z z z

z z r z z z r z

+ − +=

+ − +

( ) ( )

( ) ( )1 2 3 3

2231 2 3

z z z 1 z 1

z rz z z r

− − −= =

−− +

2

3 2 1

2 2 2

1 2 3 3 2 1

z 1 z 1 z 1r

z z z z r z r z r

− − −= = =

− − −

( ) ( )

( ) ( )2 1

2 2

2 1

z 1 z 1

z r z r

− − −=

− − −

2 1

2 1

z z1

z z

−= =

−

2

3 3z 1 z r− = − 2r 1=

Hence, 1 2 3z z z 1=

68. If coefficient of 20x in ( )20

21 x x− + and in ( )20

21 x x+ − are respectively a and b, then

(A) a b= (B) a b

(C) a b (D) a b 0+ =

Ans. B

Sol. Let ( )20

2 40

0 1 401 x x a a x ..... a x− + = + + +

( )20

2 2 40

0 1 2 401 x x a a x a x ..... a x+ + = − + − +

Thus, coefficient of 20x in ( )20

21 x x− +

= coefficient of 20x in ( )20

21 x x+ +

Therefore, we consider the expansion

( )20

2 2 40

0 1 2 401 x px b b x b x .... b x+ + = + + + +

Dividing both the sides by 20x , we get

20b = coefficient of constant term in

201

1 pxx

+ +

But

20 2 20

20 20 20

1 2 20

1 1 1 11 px 1 C px C px ....... C px

x x x x

+ + = + + + + + + +

( ) ( ) ( ) ( ) ( ) ( )20 2 20 4 2 20 20 10

20 2 1 4 2 20 10b 1 C C p C C p .... C C p= + + + +

Thus, 20 p 1a b

== and 20 p 1

b b=−

=

Hence, a b

69. If P is any point on the ellipse with foci, S1 and S2 and eccentricity 1

2 such

that 1 2 2 1 1 2PS S , PS S and S PS = = = , then cot ,cot ,cot2 2 2

are in

(A) A.P. (B) G.P. (C) H.P. (D) none of these

Ans. A

Sol. 1

e2

=

we know for a triangle cot cot cot cot .cot .cot2 2 2 2 2 2

+ + =

____(1)

( )

( ) ( )

s s bcot

2 s a s c

− =

− −

( )

( ) ( )

s s ccot

2 s a s b

− =

− −

1 1 2 2

1 2 1 2

PS S S PSscot .cot

2 2 s a PS PS S S

+ + = =

− + −

cot .cot 32 2

=

put in equation (1)

1 2PS PS 2AA '+ = and 1 2S S 2AA 'e=

70. Sum of the series ( )r 4 n n 4 rn 3

r r

r 4 n 4r 0 r 04 4

3 C C 3S

C C

+ +

+ += =

= + is

(A) n 44 + (B) ( )n 4 2n

44 C+

(C) n 4

n 4

4

4

C

+

+ (D)

n 4 n 4

n 4

4

3 2

C

+ +

+

+

Ans. C

Sol. ( ) ( )

n

r

r 4

4

C n! 4!r !

r ! n r ! r 4 !C+=

− +

( ) ( ) ( ) ( )

( )n 4

r 4

4!C

n 1 n 2 n 3 n 4

+

+=

+ + + +

n 4

r 4

n 4

4

C

C

+

+

+=

Thus, ( )n 3

r 4 n 4 n 4 r

r 4 rn 4r 0 r 04

1S 3 C C 3

C

+ + +

++= =

= +

n 4n 4

n 4 k

kn 4 n 4k 04 4

1 4C 3

C C

+++

+ +=

= =

PHYSICS

71. At t = 0, source starts accelerating with an acceleration a and observer starts moving with constant velocity v0 as shown in the figure simultaneously. Source emits a frequency f0 and velocity of sound in the air is v. The frequency detected by the observer initially is

source

a0

observer

v0

(A) 2

o(v v )f

(2vf a)

−

− (B)

2

o2(v v )f

(2vf a)

−

−

(C) 2

o(v v )f

2(2vf a)

−

− (D)

2

o2(v v )f

(vf a)

−

−

Ans. B

Sol. Let first pulse be released at t = 0.

Time when first pulse reaches O = t1 = ov v−

Time when second pulse reaches O =

t2 = T +

2

o

o

1v T aT

2v v

+ −

−

T = t2 – t1 = 2

o o

vT aT

v v 2(v v )−

− −

O

a v0

u = 0

2

o2f (v v )f

2fv a

− =

−

72. A planet of mass m moves along an ellipse around the sun so that its maximum and minimum distances from the sun are equal to R and r respectively. The angular momentum of this planet relative to the centre of the sun is

(A) GMmrR

2r R+

(B) GMmR

2Rr R+

(C) GMrR

2mr R+

(D) 2GMrR

mr R+

Ans. D Sol. According to Kepler’s Second Law the angular momentum of the planet is constant, we have

mv1R = mv2r, v1R = v2r If the mass of the Sun is M conserving total mechanical energy of the system at two given

positions we have,

– 2 2

1 2

GMm 1 G M m 1mv m v

R 2 r 2+ = − +

2 2

2 1v v1 1GM

R r 2 2

− − = −

R S r

v1

v2

Or 2 2 2

1 1

2

v R vr RGM

Rr 22r

− − = −

Or 2 2

1

2

v Rr RGM 1

Rr 2 r

− − = −

2 2

1

2

v R r

2 r

−=

( )

( ) ( )

2

2

1 2 2

2 GM R r r 2 GM rv

R R rRr R r

− = =

+−

Or ( )

2 GM r

R R r+

Now angular momentum = mv1R = m( )

2 GM rR

R r+.

73. Three solid spheres each of mass m and radius R are released from

the position shown in figure. The speed of any one sphere at the time of collision would be

(A)1 3

Gmd R

−

(B)

3 1Gm

d R

−

(C)2 1

GmR d

−

(D)

1 2Gm

R d

−

d

d d

• •

•

Ans. D Sol. From conservation of mechanical energy

2 2

21 Gm Gm3 mv 3

2 2R d

= −

2 1 2v Gm

R d

= −

1 2

v GmR d

= −

2R

2R

2R

74. In a room where temperature is 30oC a body cools from 61oC to 59oC is 4 minutes. The time taken by the body to cool from 51oC to 49oC will be:

(A) 4 minutes (B) 6 minutes (C) 5 minutes (D) 8 minutes Ans. B Sol. Rate of cooling difference in temperature

−dT

dt

dT

Kdt

=−

dT = − K.dt In First Case

dT = 61−59=2

= 60 − 30 = 30 dt = 4 minutes

dT 2 1

Kdt 30x 4 60

= − = − = −

For second case dT 2=

50 30 20 = − =

160

dT 2dt 6 min.

K x 20= = =

75. Two rods of length 1l and 2l are made of materials whose coefficients of linear expansion

are 1 and 2 . If the difference between two lengths is independent of temperature then,

(A) 1 1

2 2

l

l

=

(B) 1 2

2 1

l

l

=

(C) 2 2

2 1 1 2l l = (D)2 2

1 2

1 2l l

=

Ans. B Sol. If change in length is l

Then I1 = L01 1T

I2 = L02 2T

difference in length is ( ) ( )2 1 02 2 01 1l l L l L l− = + − +

=(L02 − L01) + (I2 − I1)

L02 − L01 is independent of temperature for I2 − I1 to be independent of temperature.

I2 − I1 must be equal to zero

i.e. L1 1T = L2 2T

i.e. L1 1 = L2 2

i.e. 1 2

2 1

I

I

=

.

76. A particle executes SHM between x = -A and x = +A. The time taken for it to go from 0 to

A/2 is T1 and to go from A/2 to A is T2. Then (A) T1 < T2 (B) T1 > T2 (C) T1 = T2 (D) T1 = 2T2 Ans. A

Sol. Let x = A sin t From 0 – A/2

A/2 = A sin T1

or sin T1 = 1

2 = sin

6

1

2.T

T 6

= or T1 =

T

12

From 0 – A

A = A sin T’1

or, sin T’1 =1 = sin ( / 2)

or, 1

2.T

T 2

= or T’1 =

T

4

From A/2 – A :

T2 = T’1 − T1 = T T

4 12− =

T

6

T1 = T

12 and T2 =

T

6

T1 < T2 77. The work done in the process 1 – 2– 3–4 shown on

P-V diagram is (A) 300 J (B) 600 J (C) 900 J (D) 1200J

3 2

1 4

P(N/m2)

2 8 V (litre)

2 × 105

8 ×105

6 ×105

Ans. C

Sol. 4 1V V 6litre− =

From geometry 2 3V V 3litre− =

5 3

104

1W 4 10 6 10

2

−=

1200J=

5 3

230

1W 2 10 3 10

2

−= −

300J= −

W = 900 J

P(N/m2)

V(litre)

2 8

2105

6105

8105

O

4 1

3 2

78. The speed of sound through oxygen at T K is (300 m/s). When the temperature is increased

to 3T, the molecule dissociates into oxygen atom, now the speed of sound will be (A) 520 m/s (B) 801 m/s (C) 600 m/s (D) 580 m/s

Ans. B

Sol. vT = 3RT 7RT 32 10

5M

−= = 300 m/s

v3T = 35R3T 16 10

3

−

3

3T

3

T

v 5 R3T 32 10 5 50

v 3 7RT 716 10

−

−

= =

= 2.67

v3T = 300 2.67 = 801 m/s

79. A bullet of mass of 10 gm moving with a speed of 400 m/s hits an ice block of mass 990 gm kept on a frictionless floor and gets stuck in it. The mass of ice melted, if 50% of the lost Kinetic energy goes to ice will be (Temperature of ice block = 0oC).

(A) 1.2 gm (B) 2.4 gm (C) 0.6 gm (D) 3 gm Ans. A

Sol. Velocity of bullet + ice block, V = ( ) ( )10 x 400

1000m/s

V =4 m/s

Loss of K.E. = ( )2 21 1mv m M V

2 2− +

= 1

2[0.01(400)2−1(4)2] =

1

2[1600 − 16]= (1584/2) J = 792 J

Heat generated = 1

2( )1584 / 2 4.2 = 95 Cal

Mass of ice melted = 95 cal

80 Cal / gm = 1.2 gm.

80. A cubical block of wood of specific gravity 0.5 and a chunk of concrete of specific gravity 2.5 are fastened together. The ratio of the mass of wood to the mass of concrete which makes the combination to float with its entire volume submerged under water is

(A) 1

5 (B)

1

3

(C) 3

5 (D)

2

5

Ans. C

Sol. let mass of wood = m1 And mass of concrete = m2

( )1 21 2

m m. .g m m g

0.5 2.5

+ = +

m1 = 2

1.5m

2.5

1

2

m 3

m 5=

CHEMISTRY

81. How many effective Na+ and Cl– ions are present respectively in a unit cell of NaCl solid (Rock salt structure) if all ions along line connecting opposite face centres are absent?

(A) 3, 3 (B) 7

,42

(C) 7 7

,2 2

(D) 7

4,2

Ans. A Sol. No. of Na+ = 4 – 1 = 3

No. of Cl– = 1

4 2 32

− =

82. The pH of which of the following salt solution does depend on concentration of solution? (A) CH3COONH4 (B) NaCl (C) NH4CN (D) FeCl3 Ans. D Sol. CH3COONH4 and NH4CN are salts of weak acids and weak bases so their pH only depends

on their a bK Kp and p values. NaCl is a neutral salts of strong acid and strong base. It’s pH = 7

at room temperature. Fe3+ undergoes hydrolysis. Hence its pH depends on concentration. 83. Which of the following compound(s) form(s) primary amine(s) when treated with Br2 and

KOH? (A)

CH3CNH2

O

(B)

CH3 - C - NHCH 3

O

(C)

CH3 - C - NHC2H5

O

(D) CH3NHCH2CH3

Ans. A Sol. Unsubstituted amides responds to this reaction(Hoffmann’s bromamide reaction) 84. Which of the following substance(s) do NOT react with NaOH to form sodium salts?

(A)

OH

NO2

(B)

CH2COOH

(C)

CH2OH

(D)

OH

NO2 Ans. C Sol. Phenols and acids react with NaOH whereas alcohols do not react with it. 85. OC2H5

CH

OC2H5

( ) ( )2H O/HP Q

+

⎯⎯⎯⎯→ +

In the above reaction, the products (P) and (Q) can be distinguished by using (A) NaOH (B) Tollen’s reagent (C) Fehling’s solution (D) FeCl3 Ans. B Sol. OC2H5

CH

OC2H5

2H O/H+

⎯⎯⎯⎯→2 5C H OH+CHO

Benzaldehdye does not react with Fehling’s solution. So, Tollen’s reagent is used.

86. Which of the following reagent converts CH3CH2COOH to a chiral compound? (A) SOCl2 (B) PCl5 (C) Cl2/red P (D) PCl3 Ans. C Sol. ( ) ( )2

3 2 3

Cl /Pachiral CH CH COOH CH CHCOOH Chiral⎯⎯⎯⎯→

Cl 87. Which of the following can absorb the highest number of CH3I molecules? (in presence of

moist Ag2O) (A)

2 2 2 3NH CH CH CH CH− − − −

NH2

(B) CH3 – CH2 – NH – CH2 – CH2 – NH2

(C) 3 2 2 2CH NH CH N CH NH− − − − −

CH3

(D) CH3 – NH – CH2 – NH – CH2 – CH2 NH2

Ans. D Sol. No. of CH3I molecules can be absorbed by

1o – amine → 3, 2o – amine → 2, 3o – amine → 1 88.

( ) HI/red P

4CHOH Products⎯⎯⎯⎯→

CH2OH

CHO

The product of above reaction confirms that (A) glucose contains an aldehyde group (B) glucose contains five OH-groups (C) glucose contains a straight carbon chain (D) glucose contains four secondary OH groups Ans. C Sol.

( ) HI/red P

4CHOH 2 iodohexane n hexane⎯⎯⎯⎯→ − + −

CH2OH

CHO

89. ( )aK

2 2HOOC CH CH CH CH COOH p 2− − − − =

3NH+ ( )aKp 8=H

( )aKp 4=

What is the isoelectric point of the above amino acid? (A) 4 (B) 6 (C) 3 (D) 8 Ans. C

Sol. a a1 2

K Kp p 4 2

I.P 32 2

+ += = =

90. Which of the following on heating does NOT form cyclic ester?

(A) ( )2 4HO C CH OH− −

O

(B) ( )2 3HO C CH OH− − −

O

(C) 2 2HO C CH CH OH− −

O

(D) 2HO C CH OH− −

O

Ans. C

Sol. -hydroxy carboxylic acid gives H2O on heating.