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1. Ans. C.

Based on the given sentences option ‘C’ is the correct

sentence which is in the comparative degree. Option ‘A’ and ‘B’ convey the wrong comparison and ‘D’ has double comparative and so they are wrong.

2. Ans. C.

Clutching means grasp (something) tightly, so it would be correct answer.

3. Ans. B.

Q and R are the son and Daughter of M, E is the mother of P and daughter-in-law of M Means Q and E are married couples in the family

P is the grandchild of M

4. Ans. C.

2 2 2324 18 ;441 21 ;64 8 but

297 x for any positive integer, i.e. 97 is odd man out

5. Ans. A.

1 1

2 2

1 2

120 10 12

150 15 10

| | 2

lengthSpeed length speed time

time

s s

s s

s s

6. Ans. D.

The odometer reading increases from starting point to end

Area of the given diagram = Odometer reading

Area of the velocity and time graph per second

1st sec triangle = 1/2 * 1 * 1 1/2

2nd sec square = 1 1 = 1

3rd sec square + triangle = 1 1 + 1/2 1 1 = 3/2

4th sec triangle =1/2 * 1 * 2 1

5th sec straight line = 0

6th sec triangle = 1/2 * 1 * 1 1/2

7th sec triangle = 1/2 * 1 * 1 1/2

Total Odometer reading at 7 seconds = 1/2 + 1 + 3/2 + 1 + 0 + 1/2 + 1/2 = 5

7. Ans. A.

According to passage, correct answer would be A.

8. Ans. C.

Manu age = sravan age + 2 months

Manu age = Trideep age 3 months

Pavan age = Sravan’s age + 1 month

From this Trideep age > Man> Pavan > Sravan

Trideep can occupy the extra space in the flat

9. Ans. B.

14,0

3

0,7

5,02

50,3

4,1

A

B

C

D

E

Required area is area of ∆OAB – area of ∆CEA

1 14 1 13

7 1 15.25 .2 3 2 6

sq units

10. Ans. A.

y a bx , where 1nx x and

0.1, 0.02

0.02( )

ab

b

a x

0.002a

0.002 0.02( )x

at x 5, y 0.002 0.02 1.609 0.03018

0.030

11. Ans. D. Product of eigen values = Det

12. Ans. A.

3 5

2 2

2

sin 1( ) .....

3! 5!

1.......

3! 5!

z z zf z z

z z

z z

z

Residue coefficient of 1

z =1

13. Ans. A. Required probability = TTTTH = (0.7)(0.7)(0.7)(0.7)(0.3) =0.07203

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14. Ans. B. 11 1 11 2 2 2

1 1

0 0

0

1120

(1 ) (1 )(1 ) | |

1 11

2 2

2(1 ) | 2

x xx dx

x

15. Ans. A.

16. Ans. C. Shifting doesn’t effects the sampling rate due to scaling

by a factor ‘2’ spectral components are doubled.

Thus, maximum frequency of | ( ) | 8Y f

∴ Nyquist rate 8 2 16Hz

17. Ans. A.

sin 1( )

( ) ( )2

F

tsa t

t

sa t rect

1( )

2

F

sa t rect

Convolution in time domain leads to multiplication in frequency domain

( ) ( )2 2

1 sin( )

rect rect X x t

tsa t

t

18. Ans. C.

3 5

3 5

3 5

[ ] [ 3] 2 [ 5]

( ) 2

( ) ( ) 2( 2)

( ) 2

[ ] [ 3] 2 [ 5]

[ ] [ ]

x n n n

x z z z

x z z

y z z z

y n n n

y n x n

19. Ans. B.

( ) 2cos200 4sin500 ,iV t t t

since there are 2 frequency term output will also have 2 frequency term. If we take 4sin500t first i.e. W = 500

then on the output section, this parallel LC combination

have LCZ so it is open circuit and V0 = Vi

So w.r.t. 4sin500t output must be 4sin500t without any

change in amplitude and phase, this is satisfied by only option B. 20. Ans. C. Answer C would be correct answer.

21. Ans. A. The semiconductor used in the MOSFET is n-type. At the surface, the intrinsic level is above EF as it is found at the distance of below EF, So, the surface is in inversion region. 22. Ans. D.

. 0.7 0.5 180Efficiency 100% 21%

(100 3)

oc sc

in

FFV I

P

23. Ans. C.

0 0V volts

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24. Ans. B.

2 2.5 1 1.8E BEV V V

2 20.7 1.1B EV V V

2

50.110

10BI A

k

2

550 10CI A

2

4 35 10 10 0.5CV V

25. Ans. C.

6

1.44

( 2 )

1.445.64

[2200 2 4700] 0.022 10

A B

fR R C

kHz

26. Ans. D. Accumulator

RLC Rotate left accumulator content without carry

27. Ans. D.

All the transistor are nmos. we know when input to gate is 0 n nmos behave as open circuit. Input to Gate is 1

is short circuit. we can redraw the circuit as

to get the functionality get find its truth table 28. Ans. A.

29. Ans. B.

1 2

1 2 1 11

GGYG

x GG GH

30. Ans. A.

For unit step input ess=1

1 pk

0 0

2lim ( ) lim

( 1)

1, 0

1

ps s

ss

k G ss s

so e

31. Ans. B.

0 0

0

?

( )

Im

image

s L si L

si

L

f

f f f f

f age freq

f Signal freq

0 0

0

0 1

0 1 0 0 1

2

2 3500 15 3485

s L Si L

Si L s

s L F

Si L F L L F

f f f f

f f f

f f f

f f f f f f mHz

32. Ans. A.

33. Ans. C.

( ) ( ) ( )

( ) ( )S

p t h t y t

h t P t t

34. Ans. B.

If incident wave is right handed polarized then the reflected wave is left handed polarized.

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35. Ans. C.

Differential from of Faraday’s law in given by

E’ B/t or

E’ H’/ t

36. Ans. A.

2

6

1 2

6

1 2

1

6 6

2 1 2

0

2 2

6

12 36 0

6, 6

. ( )

( )

(0) 3

3

( ) ( 6)

| 36

36 (3 0).1( 6) 18

(3 18 )

x

x

x x

x

x

D D

D

C F C C x e

y C C x e

y

C

dyC e C C x e

dx

dy

dx

C C

y x e

37. Ans. D.

1 2 3

2 113

3 5

2 11(1,0,2) 3(0,1,0) ( 2,0,1)

5 5

(4,3, 3)

u e e e

38. Ans. B.

2

3 3

0 0

(6 )

R

x

x y

Volume zdxdy

x y dydx

23 2 3

0 0

3

2

3

2 2

33

2 3

0

(6 ) .

2 1 4(6 )

3 2 9

2 24

3 9

8 82 2(9) (3 ) 10

9 3 27

x

x y

yx y dx

z

xx x dx

x x x dx

xx cubic units

39. Ans. B.

40. Ans. B.

0

22cos 2cos |

3 3output t t

In E.F.S 3rd harmonic

333

32cosj tj t

t e e

The coefficient of 3

3j t

e

is 1 So 3 1C

41. Ans. D.

| |[ ] (2) ,

[ ] 2 [ ] 2 [ 1]

2 [ ]; :| | 2

2 [ 1]; ;| | 0.5

n

n n

n

n

x n n

x n u n u n

u n ROC z

u n ROC z

Thus, combined ROC does not exist for x[n] 42. Ans. B.

At 0t the circuit is on steady state i.e. the capacitor is

open circuited so the circuit will be

3 2 3 3

2 3

[ ] [ ] 4

6

F

F

V V V V V

V V V

at t = 0+ when is open circuited, the capacitors will have

an ideal voltage source of values 4V and 6V so the circuit will be

So the current through 2 resistor at t = 0* should be 4/2+2 = 1A

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43. Ans. B. Nodal equation at V

8 80

1 1 1 1

4 16

4

V V V V

V

V V

By using KCL at node ‘a’.

1 1

8 41 0 5

1i i A

KCL at b

- 1

40 4 5 0

1

1

i i i

i A

44. Ans. A. Since the given network is symmetric and reciprocal

11 22

12 21

z z

z z

2

2 0

111 0

1

221

1

3 6| 2

3 6

|

I

I

VZ

I

VZ

I

We know

2 1 21

11 12

21 22

2

2 2

2 2

V V Z

Z ZSo

Z Z

45. Ans. B.

The number of complex multiplications required for DIF- FFT = (N/2 log2N)

(N/2 log2N) (20 sec) = 125 sec

46. Ans. C.

2

[ ] 5 [ ] 5 [ 2]

( ) 51(1 )

0,| ( ) | 0

j j

j

y n x n x n

H e e

At H e

At ;| ( ) | 0jH e

Thus, the filter is band pass filter

47. Ans. A.

1419 3

4

| |

10 01.6 10 0.001 800 26 10

0.5 10

c n n t

dn dnI qAD qA V

dx dx

| | 6.6cI mA

48. Ans. A.

1

1 2

21 21

1 2 1 1 2 2

1 2

.

4 202.5

4 2 20 1

E E

E Et tC

E E E t E t

t t

1 42.5 1.6II Eq

Eq Eq

EC t nm

t t

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49. Ans. A.

50. Ans. A.

3

11 6 6

2

26 6

11 1 2| | 10 10

1 1 1

1| | 1 1 2110 10

c

RA

j L

RA

R X

The overall voltage gain 1 2

outv

in

VA A A

V

12 1

2

out

in

V

V

51. Ans. C.

52. Ans. B.

50 (1 ) 550

0.015

1

1 11

5

157

2

in m

m

m

in

in

in in

C PF g R PF

g

R k

g R

R k

f kHzR C

53. Ans. B.

It is a 5-variable K-map = Q’SX’ QS’X

54. Ans. A. Case (i) When T = 0

Ttotal = delay of NOR + delay of 1st MUX + delay of 2nd MUX= 2+1.5+1.5 = 5ns Case (ii) When T = 1 Ttotal = delay of 1st NOT-gate + delay of 1st MUX + delay of 2nd NOR-gate + delay of 2nd MUX = 1+1.5+2+1.5 = 6 ns So, the maximum delay = 6 ns.

55. Ans. D. The time period of clock is 1nsec and that of nand gate is 2n sec. The time period of clock 1n sec and that of nand gate

is 2n.sec. If the nand gate would have 0 delay then the counter

must be MOD 6 because on 6th CP it will be rejected. 56. Ans. D. 3

2

1 2 3|2 4

kS

kS

From the table we can find characteristic equation 3 22 (2 3) 4 0s ks k s

For stability

2

(2 )(2 3) 4

4 6 2 0

1( 2) 0

2

k k

k k

k k

So, the conditions are 1

2k and k>-2

and combining k > – 2 57. Ans. A.

The set of equation of the system are

we can frame the state space of the system as

1 1

2 2

1

2

2 0 34

0 1 1

[1 0]

x x

x x

xy

x

A matrix is 2 0

0 1

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B matrix is 3

1

C matrix is 1 0

for controllability determinant of

| | 0B AB

3 63 6 3 0

1 1

so controllable

for observability determinant of 0a

C

C

1 00

2 0

so not observable

final controllable but not observable

58. Ans. B.

To find break point, from characteristic equation we need

to arrange k as function of s, then the root of 0dk

ds

gives break point

Characteristic equation is given by

2

2

2

2 2

2

2 2 2 0

( 2) ( 2 2)

2 2

2

( 2) ( 2 2) ( 2 ) ( 2)

( 2)

s s k k

k s s s

s sk

s

d ds s s s s s

dk ds ds

ds s

2

2

( 2)(2 2) ( 2 2)

2

dk s s s s

ds s

2 2

2

0

2 2 4 4 2 2 0

4 2 0

0.58 3.414

dk

ds

s s s s s

s s

s and

To find the valid break point we need to find that lies on root locus

3.414 lies on root locus

So break point – 3.414.

59. Ans. B.

60. Ans. A.

Information rate

For Error free transmission

C=2log 1

SB

N

B=4KHz

bb b

b

ES E R

T

2

.22

log 1b

N B B

SR B

N

2

2

log 1

log 1

31.5 /

b

b bb

b b

b

E RR B

B

E RE B

B

E mJ bit

61. Ans. A.

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62. Ans. A.

63. Ans. C.

2 2

2c

C m nf

a b

For 01TE 2

198.4

2 1.016 10 2c

C Cf

For 10TE 2

143.74

2 2.286 10 2c

C Cf

For 11TE 107.72

c

Cf

For 20TE 2

287.49

2 2.286 10 2c

C Cf

10 20 01 11TE TE TE TE

64. Ans. A.

65. Ans. D.

Electrical 0 d no d po

q qN x N x

Potential

2

0

0

( ) | |2

xx x

w

***