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GIFT-2016

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1

Answer Keys:

1 D 2 A 3 C 4 C 5 D 6 D 7 A

8 C 9 B 10 A 11 D 12 B 13 C 14 D

15 B 16 C 17 C 18 D 19 A 20 A 21 A

22 B 23 D 24 A 25 C 26 A 27 B 28 A

29 D 30 D 31 C 32 B 33 B 34 A 35 B

36 D 37 B 38 A 39 D 40 B 41 C 42 B

43 D 44 D 45 B 46 C 47 A 48 A 49 D

50 C

Explanations:

SECTION – I

3. (A + B) + (C) can do in 15 10

6 days15 10

Since A‟s days = (B+ C)‟s days

B B B C C can do in 15 12

60 days15 12

6. Let the total amount be x.

Sum of the ratio = 3 + 3 + 3 + 4 = 13

4 3

x 500013 13

x = 65000

The amount of three brothers 9

65000 4500013

7. Amount of the solutions x and y in ratio 2:5

Amount of acid solutions 2

x 21 6L7

Amount of acid solution 5

y 21 15L7

9. Cost price of the book = Rs x

Printed price = Rs y

y 90 x 114

100 100

x 90 45

y 114 57

x : y 45:57

95

2 5

90 97 Acid solution x Acid solution y

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10.

n G 60; n M 22, n P 18; n C 14;n M P 9; n M C 7; n P C 5;

n M P C 2

No. of students are not taking any of these courses.

60 8 7 2 5 6 3 4 60 35 25

11. The given sentence indicates a hypothetical case. So If-----were structure is required. The

structure of the conditional is: (if + past + would + present).

13. Let 7x, 7y be two numbers where x & y are co-primes.

LCM 7xy 196

xy 28

Possible pairs for x,y are (1, 28), (2, 14), (4, 7)

x, y are co-primes x, y 1,28 & 4,7

Numbers would be 7,196 , 28,49

28,49 is correct pair difference is 21

Larger number 49

Alternate Approach:

2 2

7x - 7y = 21

x - y = 3 (1)

(x + y) = 4xy + (x - y) = 112 + 9= 121

x + y = 11 (2)

⇒

∴

⇒

Solving we get

x = 7, y = 4

Larger Number is7x i.e., 49∴

G

M P

8 7

52

3

4

C

25

6

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15. Let zara = x

2x 5xzen x

3 3

5x 8xtotal x

3 3

Total Whole number

x multiple of '3' 3,6,9....

Total = 8 or 16 or 24 …..

17. The number of permutations of the letters of the word HINDUSTAN is 9!

n s2! (i.e., total

letters = 9, „N‟ repeat two times)

Let A, B, C be the number of permutations that the pattern „HIN‟, „DUS‟, „TAN‟ respectively

appears, then the required number of permutations is

n A B C n s n A B C

9! 7!7! 7! 5! 5! 5! 3!

2! 2!

169194

18. P[drawing a red ball from Box II after transfer)

= P[transfer of a red ball) × P(drawing of a red ball) + P[transfer of a black ball) × P(drawing of a

red ball)

2 6 4 5 1 6 2 5 16

2 4 6 7 2 4 5 8 3 13 3 13 39

19. In order to satisfy the given conditions, some Brazilians are not Venezuelans is the right

Statement as it is not clear whether Chileans drink coffee or not.

20. Distance covered by thief in 10 minutes 6 10

1km60

Relative speed = 8 – 6 = 2km/hr.

Time taken to cover 1 km distance move = 30 minutes

Hence the policeman will catch the thief after 1

2 hour chase.

Thus, distance covered in 1

2 hour by policeman = 4 km

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SECTION – II

21. Given that eigen values of A are 3 and 5

eigen values of 2A 5A are 9 15 , 25 25 i.e., – 6 and 0.

Determinant of 2A 5A

product of eigen values

6 0

0

22.

a b c a b 1

b c a b c 1

1 1 1 c a 1

(Since |AT|=|A| )

2 3

2

1 2

a 1 b

1 b 1 c c c

c 1 a

1 a b 1 a b

1 1 b c c c 1 b c D

1 c a 1 c a

23. Area of contact between tyre and road is more in tyre without projection, so grip will be more

which is usefull when applying brakes. Therefore, tyres having less projection are suitable for dry

road.

In rainy season, there is a very thin water film between tyre and road. To apply brake, it is

necessary to push more water back that can be only achieved by projecting surfaces. Therefore in

rainy season, tyres with more projection are suitable.

While walking on ice, one should put small steps to avoid slipping because friction is very less.

Limiting frication = μ normal reaction

Since μ is very very low, therefore limiting friction will be very low to avoid slipping

Action F Sin Fcos

tan

friction

Act

ion

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25. Energy conservation between released point and at bottom of inclined plane

2 2 2 2 21 1 1 1 2mgh mV I mV mR

2 2 2 2 5

10V gh

7

Horizontal component of velocity (XV ) = V cos30

Velocity of ball “A” at max height/position(XV )= V cos30

Since, collision is elastic, so velocity of “B” becomes XV and velocity of “A” becomes zero.

Therefore ball “A” fall down vertically to X. Energy conversation of ball “B” at its lowest

position and its stream of swing.

2 2

x x

2 2

1 1mV mg h V g h

2 2

1 V cos 30 15h h 5.4m

2 g 28

26. We need to first find unit vector 1 2( , )u u u

giving the direction from (3,1) to (1,3)

2 2

(1 3,3 1) ( 2,2)

( 2) (2) 2 2

v

v

v 1 1

u , unit vectorv 2 2

DD of f at (3, 1) in the direction of at 3,1u f .u

x y

1 1f 3,1 f 3,1

2 2

1 17 4sinh 3 18 12sinh 3

2 2

111 8sinh 3

2

= -48.89 (up to two decimal places)

27. From expression pV = nRT we write P1/P2 = T1/T2

1

1 1

273 t c k1atm1.1 273 t c

1.1atm 273 t c 10 k

283 1.1 273t c 1.73 or t 173 C

0.1

12gn 0.1mol and T 273 173 k 100k

120g mol

0.1mol 0.082 Latm k mol 100knRTv 0.82L

P 1atm

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28. Since on electron has a spin of ½, „n‟ unpaired electrons will have a spin of n/2

S = n/2

Hence, s

4n n1

2 2

2nn 2 n n 2

2

29. 3 2 3 2 2 3 2 3u x, y x 3x y y x 2y ; v x, y x 3xy y 4xy 3y

2 2 2

2 2 2

u v3x 6xy 2x 3x 3y 4y

x x

u v3x 3y 4y 6xy 3y 4x 3

y y

u v

V z Cy x

Then u v

x y

iff

2 23x 2x 3y 4x 3

i.e.,2 2x y 2x 1

i.e., 2 2x 1 y 2

i.e., z 1 2

C-R equations are satisfied on the circle z 1 2.

f(z) is differentiable on the circle z 1 2.

30. Let us consider a small element at the ring at point A having charge dQ

The field at P due to this element

2 20

2 20

x

x 2 2 2 20

2

x3

0 0 2 2 2

x 32 2 2

0

1 dQdE

4 R x

QRd

1 2 RdE

4 R x

dE dEcos

Qd

1 x2dE4 R x R x

Q 1 xE d

4 2x R

QxE

4 x R

A

P

dESin

dE

dECos

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yE = 0 (By symmetry)

Force on charge particle

x x

x 32 20 2

F q E

Q xF q

4x R

Newton‟s second law

2

x 2

2

3 22 2 2

0

32 20 2

d xF m

dt

qQx d xm

dt4 x R

1 qQ xx 0

4 mx R

Since x < R, so x2 + R

22R

3

o

n 3

0

3

0

qQ xx 0

4 m R

qQ

4 mR

4 mRT 2

qQ

31. 23 23

11 10 1Na Na e

The balancing of charge number gives +1

33. Since D>>d

Apply continuity Equation between (1) and (2)

2 2

1 2D V d V4 4

2

2 1

DV V

d

––––––– (1)

Apply Bernoulli‟s Equation between (1) and (2)

2 2

1 1 2 21 2

1 2 atm

2 2

1 2

2

2 1

P V P VZ Z

eg 2g eg 2g

P P P

V VH h D

2g 2g

V V 2g H h

D

1

2

2V

h

1V

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Since, V1 is the velocity at topmost layer of water in the tank. So it is very small as compared to

2g (H – h).Therefore it is neglected

2V 2g H h is in horizontal direction.

The water molecules follow parabolic trajectory

2

y y

2

x x

2

2

1h U t g t

2

2ht

g

1x U t g t

2

1 2hx V t 0 t 2g H h

2 g

x 2 h H h

x f h

For maxima & minima of „x‟ 0dx

dh

H – 2h = 0

h = H/2

h = 10/2 = 5m

34. We have

2

1

12

1

2phOH phOH

m 1 m / 2

75.2g / 94gmolnm 0.8mol,kg

n 1kg

Molality of solution = m(1 – α ) + 2

m

= m (1 – 2

)

Since f f fT k molality k m 12

1 1 17k 14k.kg mol 0.8molkg 1 2 1 0.75

2 2 0.8

35. x

x 0Let y lim x

ye

x 0 x 0 x 0

2

1

log x xlog lim x log x lim lim 01 1

x x

(By L‟Hospital Rule)

0y e 1

2V

h

x

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36. (A) 3

3 6Co NH

Co3+

in ground state electronic configuration EC = 27 – 3 – 18 = 6

d6 configuration.

During hybridization inner paring of electrons takes place due to the influence of powerful

cyanide ligands.

Inner d-orbital complex

(B) 4

6Mn CN

Ec of Mn+2

= 25 – 2 – 18 = 5

d 5 configuration

3 α 5 4s° 4p°

↑↓ ↑ ↑ ↑ ↑

3 α 4s 4p

↑↓ ↑↓ ↑↓ XX XX XX XX XX XX

Inner α-orbital complex

(C) 4

6Fe CN

EC = 26 – 2 – 18 = 6

d6 configuration

3d 6 4s° 4p°

↑↓ ↑ ↑ ↑ ↑

Fe+2

ion in 4

6Fe CN

complex ion

3 α 6 4s 4p


Inner d-orbital complex

3 d 6 4s° 4p°

↑↓ ↑ ↑ ↑ ↑

3d 6 4s 4p


2 3d sp

2 3d sp

2 3d sp

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(D) 2

3 6Ni NH

EC of Ni2+

= 28 – 2 – 18 = 8

d6 configuration

3 α 8 4s 4p

↑↓ ↑↓ ↑↓ ↑ ↑ XX XX XX XX XX XX

Outer α-orbital complex

Hence 2

3 6Ni NH

is the outer d-orbital complex.

37. We have 3 3 4 2CH OCH g CH g H g CO g

Total pressure of the system at time t = Po + 2p

The rate expression for first order kinetics is

0

A klog t

A 2.303

In the present case we get

0.5o

o

0.693 tP Plog t

P 2.303

o

o

0.693 14.5 P P12 0.249or 0.564

2.303 P

P = Po – 0.564 Po = Po (1 – 0.564) = (0.40 atm) (0.436) = 0.1744 atm

Total pressure of system is

Po + 2P = (0.40 + 2 × 0.1744) atm = 0.7488 atm

38.

1 2 1

dTQ k A

dx

r r r rslope

0 L o

x

1 2

1

2

2 1 21

r rr r

L

r rA r r

L

x

x

2

2

1 21

TL

2

o T11 21

r r dTQ k r

L dx

dQ k dT

r rr

L

x

x

x

3d 8 4s° 4p°

↑↓ ↑ ↑ ↑ ↑

2 3d sp Hybridisation

x

r

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1 2

2 1 1 2

1 21 2

1 2 1 2

1 22 21 2

1 1 LQ k T T

r r r r

r r LQ k T T

r r r r

QL 500 6.6T T 400 1400K

22r r k5 10 10 10 210

7

39. 0

2

20 2

0 2

I 4 cos x dx cos x is periodic with period ' '

4 cos x dx cos dx 4 sin x | sin x | 8

40. Let the amounts of F and cl atoms in mixture be x and y respectively.

(x6.0231023

)(27.9110–22

) + (y6.0231023

)(20.7710–22

) = 284 kJ

x(1681) + y (1251) = 284 mol –––––––– (1)

(x6.0231023

)(5.5310–22

) + (y6.0231023

)(5.7810–22

) = 68.8 kJ

x(33.1) + y (348.1) = 68.8 mol ––––––– (2)

Solving (1) and (2),

x = 0.076 mol and y = 0.125 mol

0.076%of F 37.81

0.076 0.125

0.125%of cl 62.19

0.076 0.125

42. Area bounded by the parabola 2y 4ax and y mx is

2

3

8 a

3 m

Required area 8

3

Alternate Approach:

Curves: 2y 4x (parabola)

2y

x = = g(y) ...(1)4

⇒

and

y x(line) x y f (y) …(2)

At the point of intersection, we have

2y

y y(y 4) 0 y 0,44

Required area is 2

4

y 0

yy dy

4

42 3

0

y y 16 88

2 12 3 3

O

y

A

y x

2y 4x

O (0,0)

A (4,4)

x

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43. Given D. E can be written as

2

2

1 2y 1dy 2x 1 x dx V.S.F

2 21 y

Integrating we get

3 2

2 211 y 1 x c

3

44. Each photon at 2.5 Hz carries energy of E = h

34 9

24

2

6.63 10 2.5 10

1.6575 10 joule

power 7.5 10 J / sec

Number of photons emitted per second 2

26

24

7.5 104.53 10

1.6575 10

45

2

2

z 2 z 4log 4

z 1

Simplifies to

2z 2 z 4 4 z 1

2

z 1 1 z 2 as z 0 not possible.

Hence, the locus of „z‟ is exterior of the circle.

46. OP is rotating clockwise and the direction of magnetic field is into the plane, therefore positive

charge will be at P and negative charge at the center

Electromotive force reduced due to rotation of rod OP

2

2

1emf E Ba

2

ECurrent i

R

1 Bai

2 R

Force on rod OP due to current (i)

T = i B a [left of rod OP]

2 31 B a

T2 R

Since angular speed is constant, therefore torque on rod (OP) is zero

T = F

2 3

2 31 B a 10F 0.5 0.5 0.1 125 N

2 R 10

i

R 10

PA

EO

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47. Let the roots be ,

Sum of the roots

2

3a 12

a 5a 3

2

3a 13

a 5a 3

Product of roots 2

2.2

a 5a 3

2

2

2

2 22

2 2

22

a 5a 3

3a 1 1

a 5a 39 a 5a 3

3a 1 9 a 5a 3

45a 6a 27 1

26 2a

39 3

48. Since process is ready state process therefore time derivative of variable will be zero

Apply continuity equation between inlet and outlet.

2 2

in out

in out

0.08 V 0.08 V 0.84 4

V V 159.15m / sec

Apply momentum equation in vertical direction

in

out

V 159.15m / sec

V 0m / sec

Force on fluid:

2

y in

2 23

y

y

F eV dA

F 10 159.15 0.084

F 127323.95 127324N

Apply Newton‟s third law

Force on tank yF = + 127324 N

New force on tank = Weight + force due to flow = 1500 + 127324 = 128824 N

49. If α is the degree of dissociation of pcl5(g)

5 3 2pcl (g) pcl g cl g

t = 0 3mol 0 0

t = teq 3mol(1 – α) (3mol) α (3mol) α

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Total amount of gases in the vessel

2 5 3 2n n N n pcl n pcl n cl

1mol 3mol 1 3mol 3mol

4mol 3mol 1

From the given volume of equilibrium pressure

Total amount of gases,

1 1

2.05atm 100litpvn

RT 0.082atm LK mol 500k

n 5.0mol (2)

Equating (1) and (2)

4mol 3mol 5mol

1 / 3 0.333

5

3

2

n Pcl 3mol 1 0.333 2mol

n Pcl 3mol 0.333 1mol

n cl 3mol 0.333 1mol

5

5 eq

3 2

3 2

5

n Pcl 2P Pcl P 20.5 0.82atm

n 5

1P Pcl P cl 2.05 0.41atm

5

P Pcl P cl 0.41 0.41kp 0.205atm

P Pcl 0.82

50. Let radius of circle =R

1 1

1

2 2

2

2 2

1

2

1

2

2 2

1

2

PQ R sin

OQ R cos

S Q OQ Os

S Q R cos

S Q S O OQ

S Q R cos

S P S Q PQ

S P S Q

Path difference x S P S P

PQ

2 2 2 2

R cos R sin R cos R sin

2S O 1S Q

P

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1 12 2 2 22 2

2

2 2

2 2 2 2 2 2

R 2 R cos R 2 R cos

1 1 1 11 1

1 2 R Cos 2 R cos 2 2 R cos2 2 2 2R 1 1

2 R 2 R 2 R 2

2 2

2 2 2

2 R cosx

R

R R R

x 2 cos ...(1)

For constructive interference

x n ...(ii)

Where n= 1, 2, 3, 4,..

From (i) & (ii)

0

0

n 2 cos

ncos

2

1for n

2

60

for n 2

0

for n 3 not possible

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