KVPY – CLASS-XII FULL TEST – 1

(OLTS-1819-T5-FT-1-KVPY-XII)

PART – I

MATHEMATICS

1. If one vertex of an equilateral triangle of side 2 is the origin and another vertex lies on the

line x 3y= , then the third vertex can be

(A) (0, 2) (B) ( )3, 1− −

(C) (–2, –2) (D) ( )3, 1

Ans. A Sol.

(0, 2)

60o 30o

2 2

2. The minimum distance of the surface 2xyz 2= from the origin is equal to

(A) 0 (B) 1 (C) 2 (D) 5 Ans. C

Sol. Let 2 2 2d x y z= + + .

( )22 2 2 2 2

d x y x y 2xy 2xy 4xy xy xy

= + + = − + + +

3. If the equation sec cosec c + = has solutions in , then the least integer value that 2c

can take is (A) 4 (B) 6 (C) 8 (D) 10

Ans. C

Sol. 2

2 2

1 1 2c

sin coscos sin= + +

24cosec 2 4 cosec2= +

4 4 +

2c 8

4. If ( )90

1

2 sin 2 ,=

= r

x r r then the value of x is equal to

(A) 90cot1 cos 1ec (B) 90sec1

(C) 90cot1 (D) none of these

Ans. C

Sol. S 1sin2 2sin4 3sin6 ... 89sin178= + + + +

S 89sin178 88sin176 87sin174 ... 1sin2= + + + +

Adding the two, we get

( )sin89

S 90 sin2 sin4 ... sin178 90 sin90 90cot1sin1

= + + + = =

5. If ( ) ( ) ( )2

2 2f x x 2x 3 2 x 2x 3 3= + + + + + + , then which of the following statements is

correct?

(A) The equation ( )f x 0= has four real roots.

(B) The equation ( )f x 0= has two real roots and two imaginary roots.

(C) The minimum value of ( )f x equals 11.

(D) The minimum value of ( )f x equals 12.

Ans. C

Sol. Let 2x 2x 3 t+ + = .

( ) ( )( )2

22f x t 2t 3 x 1 3 2= + + = + + +

6. If 1 2z , z lie on z r= and 1

2

zarg

z 3

=

, then

( )2

1 2

1 2

z z

z z

+ is equal to

(A) 1 (B) 2 (C) 3 (D) none of these Ans. C

Sol. ( )

2

1 2 21 2

2 1 1 2

z zz z2cos 4cos 3

z z 3 z z 6

+ + = = =

7. The graph of a function ( )y f x= is symmetric about the lines x a= and ( )x b b a= . The

period of the function is equal to (A) a b+ (B) a b−

(C) ( )2 a b+ (D) ( )2 a b−

Ans. D

Sol. ( ) ( )f a x f a x− = + and ( ) ( )f b x f b x− = +

( ) ( )f 2a x f 2b x − = −

( ) ( )( )f x f 2 a b x = − +

8. There are 3 bags. Bag 1 contains 2 red and 2a 4a 8− + black balls, bag 2 contains 1 red and 2a 4a 9− + black balls, and bag 3 contains 3 red and 2a 4a 7− + black balls. A ball is drawn

at random from a bag. The maximum value of the probability that it is a red ball is

(A) 1

3 (B)

1

2

(C) 2

9 (D)

4

9

Ans. A

Sol. The required probability is ( ) 2

1 6P A

3 a 4a 10=

− +

9. Let a, b, c be positive real numbers forming an A.P. If the equation 2 0ax bx c+ + = has real

roots, then

(A) 2 3a c

c a− (B) 2 3

a c

c a−

(C) 2 3a c

c a+ (D) 2 3

a c

c a+

Ans. A

Sol. 20 4 0D b ac −

( )2

16 0a c ac+ −

2 12 2 3a c a c

c a c a+ − −

10. The smallest natural number n that satisfies 200 30012 n is (A) 5 (B) 6 (C) 7 (D) 8 Ans. B

Sol. ( )100

200 300 100 3 312 n 144 n 144 n

The smallest number is 6.

11. The equation x 3 4 x 1 x 8 6 x 1 1+ − − + + − − = has

(A) no solution (B) only one solution (C) only two solutions (D) more than two solutions Ans. D

Sol. Put x 1 t− = or 2x t 1= + so that the given equation becomes

2 2t 4 4t t 9 6t 1+ − + + − = or ( ) ( )2 2

t 2 t 3 1− + − =

or t 2 t 3 1− + − =

This equation is satisfied for all values of t lying between 2 and 3 i.e. 2 t 3 .

Thus, the given equation is satisfied for all values of x lying between 5 and 10.

12. If 12 24a log 18, b log 54= = , then the value of ( )ab 5 a b+ − is equal to

(A) 0 (B) 4 (C) 1 (D) none of these Ans. C

Sol. We have 2 212

2 2

log 18 1 2log 3a log 18

log 12 2 log 3

+= = =

+ and 2 2

24

2 2

log 54 1 3log 3b log 54

log 24 3 log 3

+= = =

+

Putting 2x log 3,= we have

( )1 2x 1 3x 1 2x 1 3x

ab 5 a b . 52 x 3 x 2 x 3 x

+ + + + + − = + −

+ + + +

( )

( ) ( ) ( ) ( )

2 2 26x 5x 1 5 x 1 x 5x 61

x 2 x 3 x 2 x 3

+ + + − + + += = =

+ + + +

13. If the (n + 1) numbers a, b, c, d, … are all different and each of them is a prime number, then

the number of different factors (other than 1) of ( )ma b c d... m N is equal to

(A) nm 2− (B) ( ) nm 1 2+

(C) ( ) nm 1 2 1+ − (D) none of these

Ans. C

Sol. Number of different factors of ma b c d...

( ) ( ) ( ) ( ) m 1 1 1 1 1 1 1 ... n factors 1= + + + + −

( ) nm 1 2 1= + −

14. If ( )n

n r

r

r 0

1 x a x=

+ = , rr

r 1

ab 1

a−

= + and ( )

100n

r

r 1

101b ,

100!=

= then n is equal to

(A) 99 (B) 100 (C) 101 (D) 102 Ans. B

Sol. ( )n n

n n r r

r r

r 0 r 0

1 x C x a x= =

+ = =

n

r ra C=

Also, n

r rr n

r 1 r 1

a C n 1b 1 1

a rC− −

+= + = + =

( )

nn n

r

r 1 r 1

n 1n 1b

r n!= =

++ = =

( )100

101

100!=

n = 100

15. If x, y, z are integers in A.P. lying between 1 and 9 and x51, y41 and z31 are three digit

numbers, then the value of

5 4 3

x51 y 41 z31

x y z

is equal to

(A) x y z+ + (B) x y z− +

(C) 0 (D) x 2y z+ +

Ans. C

Sol. Let

5 4 3

x51 y41 z31

x y z

=

5 4 3

100x 50 1 100y 40 1 100z 30 1

x y z

= + + + + + +

Apply ( )2 2 3 1R R 100R 10R .→ − +

5 4 3

1 1 1

x y z

=

Applying ( )2 2 1 3

1C C C C ,

2→ − + we get

( )

5 0 3

1 0 1

1x y x z z

2

=

− +

5 0 3

1 0 1 0

x 0 z

= =

16. If ( )5,12 and ( )24, 7 are the foci of an ellipse passing through the origin, then the

eccentricity of the ellipse is

(A) 386 / 13 (B) 386 /38

(C) 386 / 25 (D) 1/ 2

Ans. B

Sol. If the foci be ( )S 5, 12 and ( )S' 24, 7 and the ellipse passes through origin O, then SO

13; S 'O= = 25 and SS’ = 386 .

Now, SO S'O 2a+ = and SS’ = 2ae

SS' 386

eS'O SO 38

= =+

17. The consecutive odd integers whose sum is 2 245 21− are

(A) 43, 45, ..., 75 (B) 43, 45, ..., 79

(C) 43, 45, ..., 85 (D) 43, 45, ..., 89

Ans. D Sol. Let n consecutive odd integers be 2m 1, 2m 3, 2m 5, ..., 2m 2n 1+ + + + − .

Given that ( ) ( ) ( ) ( ) 2 22m 1 2m 3 2m 5 ... 2m 2n 1 45 21+ + + + + + + + − = −

( ) 2 22mn 1 3 5 ... 2n 1 45 21+ + + + + − = −

2 2 22mn n 45 21+ = −

( )2 2 2 2n m m 45 21+ − = −

On comparing, n m 45, m 21+ = =

18. The values of x and y satisfying the equation

7 3 2 3 2 2 4sin y x x 9x 9 x x 4x 4 sec 2y cos y= − − + + − − + + + are

(A) x 1, y n , n I= = (B) x 1, y 2n , n I2

= = +

(C) x 1, y 2n , n I= = (D) 3

x 1, y 2n , n I2

= = +

Ans. B

Sol. 7 3 2 3 2 2 4sin y x x 9x 9 x x 4x 4 sec 2y cos y= − − + + − − + + +

( )( ) ( )( )2 2 2 4x 1 x 9 x 1 x 4 sec 2y cos y 1= − − + − − + + and 7sin y 1

19.

1 1 1a a

a a a

n

a 1nk 1

k n k

limn

− −

+→=

+

is equal to

(A) 1 (B) 2 (C) 3 (D) none of these Ans. A

Sol.

1 1a a

1/a a a

n

a 1nk 1

k n k

limn

− −

+→=

+

1an a

nk 1

1 k klim

n n n→=

= +

( )1

1/a a

0x x dx= +

20. Let 1 2F, F be the foci of the hyperbola

2 2x y1

16 9− = and

3 4F ,F the foci of its conjugate

hyperbola. If He and

Ce are their eccentricities respectively, then pick the correct statement.

(A) The equations of their asymptotes are different.

(B) H Ce e

(C) Area of the quadrilateral formed by their foci is 50 sq. units. (D) Their auxiliary circles have the same equation. Ans. C

Sol. ( ) ( )H 1 2

9 5e 1 F 5, 0 ; F 5, 0

16 4= + = −

2 2

H C

1 11

e e+ =

C

5e

3 =

( ) ( )3 4F 0, 5 ;F 0, 5 −

PHYSICS

21. A stone thrown down with a speed u takes a time t1 to reach the ground, while another stone, thrown upwards from the same point with the same speed, takes time t2. The maximum height the second stone reaches from the ground is

(A) ½ gt1 t2 (B) g/8(t1 + t2)2 (C) g/8(t1 – t2)2 (D) ½ gt22 Ans. B

Sol. –h = 21 1

1Ut g t

2

−− +

21 1

1h Ut g t

2= + …(1)

22 2

1h Ut g t

2− = − …(2)

h

U Point from where particle is thrown downward

2 21 2 2 1

1 1Ut Ut g t g t

2 2+ = − (from 1 & 2)

1 2 2 1 1 2

gU(t t ) (t t ) (t t )

2+ = − +

2 1

gU (t t )

2= −

h

U Point from where particle is thrown upward

22. The river ‘A’ and river ‘B’ merge to form a river ‘C’ with their speeds in the ratio 1 : 1.5. The

cross-sectional areas of the river ‘A’, river ‘B’ and the river ‘C’ are in the ratio 1 : 2 : 3. Assuming streamline flow, the ratio of the speed of river ‘C’ to that of the river ‘B’.

(A) 7 : 9 (B) 4 : 3 (C) 8 : 9 (D) 5 : 3 Ans. C Sol. Let

Area of river ‘A’ A

Area of river ‘B’ 2A

Area of river ‘C’ 3A

VA : VB : VC 1

3V : V : V

2

By equation of continuity

3

AV A 2V2

+ 3A V1

river 'C '

4V V

3=

river 'B '

river 'C'

V

V

3V

924 8

V3

=

river 'C' river 'B'V : V = 8 : 9.

23. A block of mass m is placed on the top of a wedge having undefined smooth curved surface. If the wedge is now accelerated horizontally with acceleration a, then the speed of block with respect to wedge when it reaches the bottom of wedge is

(A) 2gh (B) 2(a g)h+

(C) 2(a gh)− (D) 2(a gh)+

m

a

h

Ans. D

Sol. ma + mgh = 21mv

2

v 2(a gh)= +

24. A block of mass 2.0 kg is sliding on a frictionless horizontal floor with a velocity of 5.0 m/s

due east. It is subjected to a variable net force for 10 seconds resulting in an impulse of 6 N-s in the North and 18.0 N-s in the west. The total work done by this force is

(A) 0.0 J (B) 90 J (C) 100 J (D) 200 J Ans. A

Sol. 2 2final North Westv v v= + North

6V

2= = 3 m/s

= 2 2(3) (4)+ West

18 10V

2

−= = 4 m/s

= 5

2 2f i

1W m (v v )

2= − = 0.

25. A ball dropped from a high altitude acquires a terminal velocity before hitting the ground, where it bounces off elastically. If air resistance depends on the speed of the ball, what will it acceleration be just after the first bounce?

(A) zero (B) g downwards (C) 2g downwards (D) 3g downwards

Ans. C

Sol. Let Fair = kv then mg = kv

SQ mg

vk

=

But just after the bounce Fnet = mg + kv

a = kv

gm

+ = 2g

26. A sphere of mass m has to purely roll on a rough inclined plane of

coefficient of friction ‘’. The friction force acting on the sphere is

(A) mg cos

(B) 7

sinmg2 downward

(C) 7

sinmg2 upward

(D) 7

sinmg5 downward

m

Ans. C

Sol.

2

2

K

R1

sinmgf

+

=

27. Two particles move parallel to x axis about the origin with same amplitude a and frequency

. At a certain instant they are found at a distance a/3 from the origin on opposite sides but their velocities are in the same direction. What is the phase difference between the two.

(A) 1 7cos

9

−

(B) 1 5cos

9

−

(C) 1 4cos

9

−

(D) 1 1cos

9

−

Ans. A

Sol.

21 7

cos 1 23 9

= − =

28. A hollow tilted cylindrical vessel of negligible mass

rest on horizontal plane as shown in figure. The diameter of the base is a and the side of the

cylinder makes an angle with the horizontal. Water is then slowly poured into the cylinder. The cylinder topple’s over when the water reaches a certain height h, given by

(A) h = 2a tan

(B) h = 2a tan2

(C) h = a tan

(D) a

h tan2

=

h

a

Ans. C

Sol. h

ACsin

=

h

OD2sin

=

; But OD = a

2cos

h a

2sin 2cos=

h = a tan

h

a

2

O

a

2 A B

C

D

cm

29. The sun orbits the centre of the galaxy (Milky-way) in almost circular path of radius ‘R’ in a

period ‘T’ and earth also orbits the sun in an almost circular path of radius ‘r’ in a period ‘t’. Assume whole mass of the galaxy concentrated at its centre and find an expression for the ratio of mass of galaxy to that of sun.

(A)

2 2R t

r T

(B)

3 2R T

r t

(C)

2 3R t

r T

(D)

2 3R T

r t

Ans. A

Sol. = 3R

T 2GM

2 3

2

4 RM

GT

=

30. Two rubber balloons filled with same ideal gas when held at the bottom of the lake in thermal

equilibrium with surrounding water occupy equal volumes. Rubber of the first balloon is a good conductor of heat while that of the second balloon is a good insulator. Both the balloons are set free simultaneously. If temperature of water in the lake is uniform, which balloon will occupy more volume when it comes near surface of lake?

(A) The first balloon. (B) The second balloon (C) Both the balloons will occupy equal volumes. (D) Decision depends on the adiabatic exponent of the gas. Ans. A Sol. For 1st balloon

0 1 o 1(P h g) V P V+ =

For 2nd balloon

0 2 o 1(P h g) V P V + =

31. A point source is emitting sound in all directions. The ratio of distance of two points from the

point source where the difference in loudness levels is 3 dB is: (loge 2 = 0.3)

(A) 1

2 (B)

1

2

(C) 1

4 (D)

2

3

Ans. B

Sol.

2

2

1

r3 10 log

r

=

32. A certain sample of monoatomic ideal gas is subject to a thermodynamic process in which V

and T are related as 2V kT= (k is constant). The molar specific heat of the gas in this process is

(A) 3R

2 (B) 2R

(C) 5R

2 (D)

7R

2

Ans. B

Sol. 2v kT= or PV–1 = constant

C = CV + R

2 = 2R

33. The velocities are in ground frame and the cylinder is

performing pure rolling on the plank. Velocity of point ‘A’ would be

(A) 2Vc (B) 2Vc + Vp

(C) 2Vc − Vp (D) 2(Vc+ Vp)

Vp

VC C

A

Ans. C Sol. VAp = 2Vc

VAg = VAp + VPg = 2Vc − Vp 34. A superconducting rigid planer loop of area ‘A’ and self inductance ‘L’ carrying a current is

held motionless in a region of free space. Now a uniform magnetic field of induction ‘B’

pointing everywhere parallel to the magnetic moment 'm' of the loop s switched on. The current in the loop after magnetic field is switched on is given by

(A) AB

L (B)

m

A

(C) m AB

A L− (D)

m AB

A L+

Ans. C

Sol. Initial current m

iA

= ( m = iA)

Induced current AB

iL

= ( = Li)

Net current = m AB

A L−

35. The equivalent capacitance between points A and B of the

circuit will be (A) 12C (B) 6 C (C) 3 C (D) 24 C

4C 6C

4C

4C

4C

2C

A B

Ans. C Sol. Equivalent circuit of the above figure can be drawn as CAB = 3C

6C

2C

A B

2C

2C

6C 6C

A B

36. An inductance L is split into three equal parts AB, BC and CD and a capacitor C is connected across its two centre terminals (figure). The outer terminals are short circuited. The frequency of oscillation is

L A B C D

C

(A) 3 1

2 2 LC´

p (B)

1 1

2 LC´

p

(C) 3 1

2 LC´

p (D)

3 1

2 LC´

p

Ans. A

Sol. ( ) ( )

( ) ( ) ( )

2 / 3 L. 1/ 3 L 2 1L ' L,

2 / 3 L 1/ 3 L 9 2 / 9 L C= = w =

+ ´

37. Graph shown is between deviation () and angle of incidence I then angle of prism is

(A) –x + y + z (B) x + y – z (C) x + z – y (D) x – y + z

Z

O i

X Y Ans. B

Sol. = x + y – z 38. A hollow lens is made of thin glass and in the shape of a double concave lens. It can be filled

with air, water of refractive index 1.33 or CS2 of refractive index 1.6. It will act as a diverging lens if it is

(A) filled with air and immersed in water. (B) filled with water and immersed in CS2. (C) filled with air and immersed in CS2. (D) filled with CS2 and immersed in water. Ans. D

Sol. air = I.

water = 1.33

2CS = 1.6

2

1 1 2

n1 1 11

f n R R

= − − −

2

1 1 2

n1 1 11

f n R R

= − − +

n2

n1

For diverging lens f must be –ve.

for this 2

1

n1

n

n2 > n2

Answer (D) is the correct option as

2CS water

39. When the electron in a hydrogen atom jumps from the second orbit to the first orbit, the

wavelength of the radiation emitted is . When the electron jumps from the third to the first orbit, the wavelength of the radiation emitted as

(A) 9

4 (B)

4

9

(C) 27

32 (D)

32

27

Ans. C

Sol. 2 2

hc 1 1 3R R

41 2

= − =

…(i)

2 2

hc 1 1 8R R

' 91 3

= − =

…(ii)

' 3 9

4 8

=

,

27'

32 =

40. If photoelectron emitted by a metal on irradiating it by a light of wavelength cannot move away farther than a distance ‘d’ in presence of a retarding electric field ‘E’. The threshold wavelength of the metal is

(A) hc

eEd (B)

hc

eEd −

(C) hc

eEd + (D)

11 eEd

hc

−

−

Ans. D

Sol. o

hc hcEed= +

CHEMISTRY

41. Which of the following two substances can be distinguished by heating? (A) CaCO3 and MgCO3 (B) LiNO3 and NaNO3 (C) Ba(NO3)2 and Ca(NO3)2 (D) NaHCO3 and MgCO3 Ans. B

Sol. LiNO3 will evolve a coloured gas, NaNO3 → Colourless gas. 42. Which of the following can react with hydrogen in presence of light? (A) N2 (B) Cl2 (C) O2 (D) S8 Ans. B

Sol. Light

2 2H Cl 2HCl+ ⎯⎯⎯→

43. The energy of first orbit of hydrogen atom is -13.6 eV. How much energy is required to excite

the electron of hydrogen from ground state to first excited state? (A) 3.4 eV (B) 10.2 eV (C) 13.6 eV (D) 12.1 eV

Ans. B

Sol. E = E2 – E1 = - 3.4-(-13.6) = 10.2 eV 44. In which of the following molecules/ions both the central atoms have same type of

hybridization? (A) CO2 and SO2 (B) NF3 and BF3 (C) H2O and OF2 (D) BeCl2 and SCl2

Ans. C Sol. ‘O’ undergoes sp3 hybridization in both compounds. 45.

0 C

once

ntr

atio

n (

in M

)

Time (in sec)

0.2

0.4

0.6

0.8

Above graph is given for the following reversible reaction.

( ) ( )X g Y g

The equilibrium constant KC of above reaction is:

(A) 1

4 (B)

3

2

(C) 2

3 (D) 4

Ans. C

Sol. C

Y 0.4 2K

X 0.6 3= = =

46. 2CN H O HCN OH− −+ +

What will be the ionization constant (Ka) of HCN if the hydrolysis constant (Kh) of above reaction is 10–8?

(A) 10-4 (B) 10–6 (C) 106 (D) 10–10

Ans. B

Sol. 14

6wh 8

a

K 10K 10

K 10

−−

−= = =

47. Which of the following solution does NOT display common ion effect? (A) H2S + NaHS (B) HCN + KH (C) NaHS + HCl (D) H2CO3 + KHCO3

Ans. B Sol. No common ion present between HCN and KH.

48. What is the order of the reaction for which the above graph is given?

(A) Zero (B) First (C) Second (D) Third

t1/2

Initial concentration Ans. A

Sol. t½ a for zero-order reaction.

49. The solubility product of PbI2 is 32 10–9 mol3 L–3 at a certain temperature. What is the molarity of the saturated solution of PbI2?

(A) 0.0002 M (B) 0.002 M (C) 0.2 M (D) 0.02 M Ans. B

Sol. 4s3 = 32 10–9 s = 2 10–3 M. 50. The most acidic compound out of the following is:

(A) CH3CHCH2COOH

F

(B) CH3CH2CHCOOH

F (C) CH3CH2CHCOOH

Cl

(D) CH2CH2CH2COOH

Cl Ans. B Sol. Electron withdrawing atom increases acidic strength. 51. Which isomer of CH3CH2CH2CH2NH2 is most basic in nature? (A) Chain isomer (B) Position isomer (C) Functional isomer (D) Metamer Ans. C Sol. 2o and 3o amines are the functional isomers of 2o-amine. 52. X X

CH3

3

3

CH Cl

Anhy.AlCl⎯⎯⎯⎯⎯⎯⎯→

In the above reaction (X) should be (A) OH (B) Cl (C) CH2OH (D) CHO Ans. D Sol. X(CHO)is a meta directing group.

53. Which of the following reaction releases maximum heat energy?

(A)

2

NiH ⎯⎯⎯→+

(B)

2

NiH ⎯⎯⎯→+

(C)

2

NiH ⎯⎯⎯→+

(D)

2

NiH ⎯⎯⎯→+

Ans. B Sol. Unsubstituted or less substituted alkenes are more reactive towards H2/Ni.

54. O - CH 2CH2( )HI excess

Pr oducts⎯⎯⎯⎯⎯⎯⎯→

Which of the following is NOT a product of above reaction?

(A)

OH

(B)

CH2CH2I

(C)

CHI CH3

(D)

I

Ans. D

Sol.

OH

will not react with HI to form

I

55. Which of the following contains (3c-2e) bond? (A) H3BO3 (B) B2H6 (C) NaBH4 (D) B2O3 Ans. B Sol. B2H6 contains 2c – 3e bond. 56. Which of the following contains metal ions? (A) Silica gel (B) Silicones (C) Silanes (D) Silicates

Ans. D

Sol. Silicates are metal salts.

57. In a primitive cubic unit cell of a metal, the metal atoms are present at the (A) corners (B) corners and face centres (C) corners and body centre (D) face centre and body centre Ans. A Sol. In primitive cubic unit cell, only corners are occupied. 58. Which transition metal ions contains a half-filled electron configuration? (A) Ti2+ (B) Mn2+ (C) Cr2+ (D) Fe2+

Ans. B Sol. Mn2+ is [Ar]183d5. 59. Which of the following electrolyte(molten salt) on electrolysis deposits the maximum mass of

metal by passing same amount of electricity? (A) NaCl (B) MgCl2 (C) AlCl3 (D) CaCl2 Ans. A Sol. Maximum amount of Na is formed.

60. H = Enthalpy change of a thermodynamic process

E = Internal energy change of a thermodynamic process

The quantity (H – E) represents (A) entropy change (B) work done (C) temperature change (D) none of these Ans. B

Sol. H – E = PV(Work done).

PART – II

MATHEMATICS

61. If ( )

n

n

k 1

1a ,

k n 1 k=

=+ −

then for n 2 ,

(A) n 1 na a+ (B) n 1 na a

+

(C) n 1 na a+= (D) n 1 n

1a a

n+− =

Ans. B

Sol. We have n

n

k 1

1 1 1a

n 1 k n 1 k=

= +

+ + −

n

k 1

2 1

n 1 k=

=+

For n 2 ,

( )n n 1

n n 1

k 1 k 1

1 1 1 1 1a a

2 n 1 k n 2 k

+

+

= =

− = −+ +

( ) ( )

n

k 1

1 1 1 1

n 1 n 2 k n 1 n 2=

= − −

+ + + +

( ) ( )

n

k 2

1 10

n 1 n 2 k=

= + +

n n 1a a+

62. For real x, the expression ( ) ( )2 22 k x x x k− + + cannot exceed

(A) 2k (B) 22k

(C) 23k (D) none of these

Ans. B

Sol. Let ( ) ( )2 2y 2 k x x x k= − + + .

Put 1 x

tan , ,k 2 2

− = −

.

( )( )( )( )

( )( )

2

2

2

2

2

2

2k cos sin 1 siny 2k 1 tan tan sec

cos

2k 1 sin 1 sin2k

cos

− + = − + =

− + =

63. If 1 2 3z ,z ,z are 3 distinct complex numbers such that

2 3 3 1 1 2

3 4 5

z z z z z z= =

− − −, then

the value of 2 3 3 1 1 2

9 16 25

z z z z z z+ +

− − − equals

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

Sol. Let 2 3 3 1 1 2

3 4 5k

z z z z z z= = =

− − −

( )( ) ( )2 22 3 2 32 3 2 32 2

2 3

9 9 9z z z z z z k z z

z zk k − = − − = = −

−

Similarly, ( ) ( )2 23 1 1 2

3 1 1 2

16 25k z z ; k z z

z z z z= − = −

− −

64. Let ( )1

2

1

lim 2012sin 2013cos | | .

n

nn

k n x x x dx→

−

= + The value of 2012k − is equal to

(A) 1− (B) 0 (C) 1 (D) 2012

Ans. C

Sol. ( )1

2

1

1 1 1 1 1 12012sin 2013cos | | 4026 sin cos 1 4026 sin 2sin

2−

+ = + − = −

n

n

x x x dxn n n n n n

Hence,

2

2

1 1sin sin

24026 lim 20131 1

22

→

= − =

n

n nk

n n

65. Let ( )2

2

x ax 4f x

x bx 16

+ +=

+ + be defined for all real x. The number of possible ordered pairs

( )a, b (where a, b I ) is equal to

(A) 89 (B) 85 (C) 135 (D) 150 Ans. C

Sol. 2x bx 16+ + should be positive for all real x and 2x ax 4+ + should be non-negative for all

real x.

66. ( )2

2

ln x 1 xdx

1 x

+ +=

+

(A) ( )21ln x 1 x C

2+ + + (B) ( )( )

221

ln x 1 x C2

+ + +

(C) ( )2ln x 1 x C+ + + (D) ( )2x ln x 1 x C+ + +

Ans. B

Sol. Integrating by parts taking 2

1

1 x+ as 2nd function, we get

( )( )2

2I ln x 1 x= + +( )2

2

ln x 1 xdx

1 x

+ +−

+

( )( )2

22I ln x 1 x 2C = + + +

67. The number of functions f from set A = 0, 1, 2 to set B 0,1,2,3,4,5,6,7= such that

( ) ( )f i f j for i j and, i, j A is

(A) 8

3C (B) ( )8 8

3 2C 2 C+

(C) 10

3C (D) none of these

Ans. C

Sol. A function f : A B→ such that ( ) ( ) ( )f 0 f 1 f 2 falls in one of the following four categories:

Case 1 ( ) ( ) ( )f 0 f 1 f 2

There are 8

3C functions in this category

Case 2 ( ) ( ) ( )f 0 f 1 f 2=

There are 8

2C functions in this category

Case 3 ( ) ( ) ( )f 0 f 1 f 2 =

There are 8

2C functions in this category

Case 4 ( ) ( ) ( )f 0 f 1 f 2= =

There are 8

1C functions in this category

Thus, the number of desired functions is

8 8 8 8 9 9 10

3 2 2 1 3 2 3C C C C C C C+ + + = + =

68. If the equation 3 2x bx cx 1 0+ + + = ( )b c has only one real root , then the value of

( ) ( )1 1 22tan cosec tan 2sin sec− − + is equal to

(A) − (B) 2

−

(C) 2

(D)

Ans. A

Sol. ( )1,0 − . Hence, ( )1 1

2

2sintan 2tan sin

1 sin

− − =

− and

( )1 11tan cot sin

sin

− − = − +

.

69. Let ( )2010

2 r

r

r 0

2x 3x 4 a x=

+ + = . The value of 7

13

a

a is equal to

(A) 1

8 (B)

1

2

(C) 2 (D) 8 Ans. D

Sol. Replace x by 2

x.

70. The number of triplets of integers for which 2 22a b 8c 7+ − = is equal to

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

Sol. 2 22a b 8c 7+ = + b must be odd.

Let b 2m 1= + .

2 2 a 2m 2m 4c 3 + + = + a must be odd.

Let a 2n 1= + .

( )2 2n 2n m m 1 2c 1 + + + = + , which is not possible.

PHYSICS

71. A particle is projected with initial speed u 1500 m/s= such that the radius of the path at

highest point is equal to maximum height of the projectile. The horizontal range of the projectile is

(A) 150 m (B) 100 2 m

(C) 100 3 m (D) 150 2 m

Ans. B

Sol. Radius at highest point 2 2u cos

r Hg

= =

2 2 2 2u cos u sin

g 2g

= tan = 2

i.e. sin =2 1

and cos3 3

=

Range = 2u

2sin cosg =

1500 2 2 1

10 3 3

= 100 2 m

72. A particle moves along x-axis. The position of the particle at time t is given as

x = t3 − 9t2 + 24t + 1 The distance traveled in first 5 seconds is (A) 20 m (B) 10 m (C) 18 m (D) 28 m

Ans. D

Sol. Distance Travelled = 5 5

2

0 0

v dt 3t 18t 24 dt= − + = 28

73. A small object of mass m is suspended by a thread of length and released when thread is horizontal. Find out the velocity of the object when its vertical component is maximum

(A) g (B) 2

g3

(C) 3

g2

(D) 2

g3

Ans. B

Sol. Vertical velocity will be maximum when vertical acceleration is zero. mg Tcos=

2mv

mgcos cosr

= +

2

2 gr sinv

cos

=

also mgr cos = 21mv

2

2v 2grcos=

2tan 2 =

2gr

V3

=

74. Two identical shaped spheres of specific gravity 0.8 and 1.6 are connected through string and placed in liquid container as shown in figure. If spheres are in equilibrium with respect to container then tension in the string will (take volume of each sphere v0 = 250 CC)

(A) 1N (B) 2N (C) 3N (D) none of these

P0

a = 2g

Ans. A

Sol. b 1 1F m g T 2m g+ + =

m1g Fb

T

2m1g

b 1F T m g+ =

b 2 2F m g T 2m g+ = +

b 2F T m g− =

m2g

2m2g

Fb

T

b 1 1

b 2 2

F T m g 1.6

F T m g 0.8

+ = = =

−

b bF T 2F 2T+ = −

b3T F=

bF V gT

3 3

= =

1 2 1.6 0.81.2

2 2

+ + = = =

T = 1N

75. Air is blowing and is providing a constant horizontal acceleration ax = g to the particle as shown in the figure. Particle is projected from point P with a velocity u in upward direction. Let Q be the highest point of particle. Speed of the particle at highest point Q is

(A) 2 u (B) u

(C) u / 2 (D) None u

Q

a =gx

P

Ans. B

Sol. v u a t= +x x x

v 0 gt= +x

…..(1)

y y yv u a t= +

0 u gt= −

u

tg

= …..(2)

By (1) and (2) we get xv u= and yv 0=

Hence net velocity = u 76. A balloon filled with of 1.0 g of hydrogen is in thermal equilibrium with air in a room. Root

mean square speed of air molecules at the room temperature is 500 m/s, the average molecular mass of air molecules is 29 g/mol. If hydrogen in the balloon is considered as an ideal gas its total internal energy is near to

(A) 1813 J (B) 3021 J (C) 4223 J (D) 6042 J

Ans. B

Sol. rms

3RTv

M=

2rmsmv

RT3

= …(i)

2rmsnMvnRT

U1 3( 1)

= = − −

= 3021 J

77. A light bulb of resistance R 8( 5 1)= + is attached in series with an infinite resistor network

with identical resistances ‘r’ as shown in figure. A 20V battery drives current in the circuit. What should be the value of r such that the bulb dissipates maximum power?

r

20V

R

r

r r r

r

(A) 16 (B) 8

(C) 4 (D) 8( 5 1)+

Ans. A

Sol. Let ‘x’ be the equivalent resistance of infinite network.

AB

rxR r x

r x= + =

+

rx + r2 + rx = rx + x2 x2 – rx – r2 = 0

2 2r r 4r 5 1

x r2 2

+ + += =

A

r X

r

B

According to maximum power transfer theorem, for maximum power in R. x = R

( )5 1r 8 5 1

2

+= +

r = 16

78. A parallel plate capacitor consists of square plates of edge length ‘a’ separated by a distance d << a. It is charged to a potential difference ‘V’ and made to move with a constant velocity ‘v’ directed along one of its edges. The magnetic field exists inside the capacitor will be

(A) 0 0vV

d

(B) 0 0vV

a

(C) 0 0

2

dvV

a

(D) none of these

Ans. A

Sol. 0

qdV

A=

20 0AV Va

qd d

= =

0 02 Va 2 Vv adq dai

dt d dt d

= = =

0 00

vV1 iB

2 a d

= =

(For infinite sheet)

79. What amount of heat will be generated in a coil of resistance R due to a charge q passing

through it if the current in the coil decreases down to zero uniformly during a time interval t?

(A) 24 q R

3 t (B)

22 q R

3 t

(C) 23 q R

4 t (D)

23 q R

2 t

Ans. A Sol. Suppose initial current is i0, then

i(t) = i0t

1t

−

q = t

0

0

ti 1 dt

t

−

So, i0 = 2q

t

H =

2t

0

2q t1 Rdt

t t

−

80. An object moves with velocity ˆV ( j)− towards the inclined plane

mirror as shown in figure. The velocity of the image is

(A) ˆV j (B) 3 Vˆ ˆV ( i) j2 2

− +

(C) 3 Vˆ ˆV i j2 2

+ (D) V 3ˆ ˆ( i) V ( j)2 2

− +

V

ˆY( j)

ˆX(i) 30°

Ans. B

Sol. v = 2V cos 30° sin 30° ˆ( i)− + V(cos2 30° – sin2 30°) j

CHEMISTRY

81. 100 mL of 0.2 M CaCl2 solution is isotonic with [Assume complete dissociation of salts] (A) 100 mL of 0.2 M C6H12O6 solution (B) 200 mL of 0.3 M NaCl solution (C) 100 mL of 0.3 M KCl solution (D) 200 mL of 0.1 M Al2(SO4)3 solution Ans. C Sol. Isotonic solution have same osmotic pressure. 82. Which of the following complex displays ionization isomerism? (A) [Cr(H2O)3(NH3)3]3+

(B) [Cu(NH3)3Cl]Br (C) Na3[FeF6] (D) [Cr(NH3)6]ClBr Ans. B Sol. Ionization isomers occur in [Cu(NH3)3Cl]Br. 83. Which of the following forms a hydrocarbon, when reacts with C2H5MgBr? (A) HCHO (B) CH3COC2H5 (C) CH3COOH (D) CH3COOC2H5

Ans. C

Sol. ( )( )2 5 3 2 6 3C H MgBr CH COOH C H Mg Br CH COO+ ⎯⎯→ +

84. Glucose is converted to a black substance when reacts with (A) C6H5NHNH2 (B) Br2/H2O (C) Conc.H2SO4 (D) NH2OH Ans. C Sol. Conc. H2SO4 is a dehydrating agent. Hence, it removes water from glucose. 85. A 10 L vessel contains equal mass of the following gases at constant temperature. Which

will produce the maximum partial pressure? (A) CH4 (B) CO (C) CO2 (D) SO2 Ans. A Sol. Number of moles of CH4 is the highest, So, it forms maximum pressure. 86. The strongest reducing agent out of the following is: (A) NH3 (B) PH3 (C) BiH3 (D) AsH3

Ans. C

Sol. Due to less bond energy.

87. Which of the following reaction produces a substance that is used to purify water? (as a disinfectant) (A) Ca(OH)2 + HCl (B) Ca(OH)2 + Cl2 (C) Ca(OH)2 + Cl2O7 (D) Ca(OH)2 + HClO2

Ans. B

Sol. ( ) ( ) ( )2 22Ca OH Cl Ca OCl Cl Bleaching powder H O+ ⎯⎯→ +

88. The product of which of the following reaction does NOT undergo oxidation reaction with

acidified KMnO4 solution?

(A) 2

4

H3 2 Pd BaSO /quinoline

CH CH COCl−

⎯⎯⎯⎯⎯⎯⎯⎯→ (B)

O

3 3CH MgBr/H O+

⎯⎯⎯⎯⎯⎯⎯→

(C) 4LiAlH3 3CH COCH ⎯⎯⎯⎯→ (D)

2H O⎯⎯⎯→

CHCl2

Ans. B Sol. CH3OH

(3o-alcohol) will not be oxidized by 4MnO−

89. Which of the following substance will produce the heaviest product with NH3? (A) HCHO (B) HCOOH (C) HCOCl (D) HCOOCOH Ans. A Sol. HCHO will form urotropene with NH3. 90. Which of the following substance on heating undergoes decarboxylation?

(A)

COOH

COOH

(B)

COOH

O

COOH

(C)

COOHHOOC

COOH

(D)

COOH

COOH

O

Ans. B

Sol. -keto acids undergo decarboxylation on heating.