GATE MOCK QUESTIONS

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Answer Keys

1 D 2 C 3 A 4 D 5 40 6 D 7 616

8 C 9 D 10 17.5 11 B 12 -4.61 13 B 14 D

15 128 16 D 17 4 18 A 19 C 20 D 21 C

22 117.3 23 A 24 A 25 C 26 A 27 C 28 B

29 C 30 11.12 31 61 32 C 33 44 34 14.4 35 A

36 138.02 37 A 38 B 39 B 40 599 41 B 42 A

43 18.55 44 D 45 A 46 A 47 B 48 C 49 D

50 57.5 51 D 52 B 53 0 54 85.3 55 250 56 C

57 C 58 C 59 D 60 C 61 B 62 B 63 D

64 D 65 C

Explanations:-

1. Skew- symmetric of even order, is a perfect square (non-zero)

Hence, option (d) i.e., 9 is a perfect square.

(or)

Using expansion, we get 9.

2. We have

x

y

x x r y y

1.5x 10 0.6 y 20

2

x 0.45 y 1

3. We know that f (z) u iv is analytic then u v u v

andx y y x

verify the options, only

(A) sasitfies

4. 1

Let x27

1

2 3

1 2f x 27 0, f x

x x

The Newton-Raphson formula is

3n n n

n 1 n n 11

n

f x 3x 27xx x x

f x 2




5. 2 2 2

2

2 2 2x y z

2 4 3 4 3 2 2

2

1, 1,1

12xy z 4x z 24x y z

12 4 24 40

6. 1 resistor is removed from fig, and the circuit configuration is shown in fig below

1

1

2

OC 1 2

I 2 1 1A

V 1A 2 2V

4 5 20Also V 1A V

4 5 9

2V V V V

9

Deactivating the constant sources, THR from fig below is obtained as

TH

0.c

TH

4 5 38R 2

4 5 9

2V 29I A 42.55mA

38R 1 471

9

7. Rotor losses = slip x air gap power

Slip = synchronous speed – rotor speed/synchronous speed = 0.04

Rotor losses = 0.04 x 15.4 x 1000 = 616 W

9. O

O180 180Slot angel= 30

S 12

P 2

10. r 6

0.35 0.35t 17.5nsec

B.W 20 10

OCV

542 121V

1I

2V

54

THR

I

THR

OCV 1




11. Now reducing the given circuit, getting thevenin’s equivalent at the both electrodes of the

diode, we will get the following result.

So the diode is in reverse bias and from its characteristics, it is clear that it should be replaced

with 5k resistor.

Hence ~I 0.3mA

12. Here, both BJTs can’t be ON at a time because it violates KVL between +5 and -5 sources.

Now let’s assume that upper BJT is ON and lower one is OFF. In this case, current flows into

the ground via 1k so emitter is at positive voltage and base is at (emitter voltage)-0.7V. But

for the current to flow into its base, it must be at voltage lower than -5 V. So this is not

possible. Now if upper BJT is OFF and lower one is ON then current follows the path as

shown below.

Now applying KVL over this path and putting E BI 101 I . We get EI 3.91mA

Also E BV V 0.7. Hence we get BV 4.61V

15. i0

ff ; where N MOD number of the counter

N

256kHz2kHz N 128

N

5V BV

10K

5V

5V

1K

BION

EI

OFF

2 1.5

6V4.5V

I

2 1.5

6V4.5V

I

5K




17. Since the ROC is left sided before doing the long division it has to be arranged in ascending

order of z. In the quotient expression, the coefficient of z represents the sample value

at n = -1.

18. jX e d 2 x 0 2 1 2

19. Let the overall impulse response of the system be h[n], then by observing the block diagram

given in figure

1 2 3 4h(n) h n h n h n h n nu n u n 4 u n (n 4) a u(n)

n n nu n 4 (n 4) a u(n) u(n 4 4) a u(n) [1 a ]u(n)

20. Steady state errors for various inputs

ss ps 0

p

ss as 0

v

1e unit step 0; as K limG ' s

1 K

1e ramp 0.5; as K limsG ' s

K

2

ss as 0

a

1e parabolic 0.5; as K lims G' s 2

K

Steady state error when r(t) is applied is 0 (4 0) (10 0.5) 5

21. Because derivative controller acts only on rate of change of an error.

So it is effective in transient only.

22. Rating: 220V, 1000 rpm, 15A

t a a a m m

1000V E I R 220 K 2 15 1 K 1.9576

60

a

15N 500 rpm, I 7.5A

2 ; t

500V 1.9576 2 7.5 94.99V

60

mo t

2 V 2 230 2V V cos cos

O117.3

21Z 1

4

1

1

z 1

z 4z

( )

1 4z

4 z

represents

x(n) at n 1




24. As I=0, the bridge is balanced.

x 1 2 3

x

x

Z Z Z Z

(10k )(100 )Z 31.3 j11.4

30k 20

L 11.4H 12 H

26.

3

2

2

3dx 33log x 2x 2

dyx 2 3y 12 x 2 where x 2

dx

dy 3.y 12 x 2

dx x 2

3P Q 12 x 2

x 2

I.F e e x 2

Solution is

3 2 3

5

6

3

6

3 3

y x 2 12 x 2 x 2 dx c

12 x 2 dx c

x 2y x 2 12 c

6

1y 3 2 2 1 12 c c 0

6

x 2y x 2 12 y 2 x 2

6

27. s a

; Magnitude is same for any value of a and frequency, hence it is a All pass filters a

-1Phase =180-2tan ; At =a ; phase =90ºa




28.

1 1

2 22 2

s 1 1 1L L

4 s 2s 2 s 2s 2s 2s 2 s 2s 2

1 1

2 2

t 1 t 1

2 2

1 1 1L L

4 s 1 1 s 1 1

1 1 1 1e L e L

4 4 s 1 s 1

t t

t t

1e sin t e sin t

4

1sin t e e

4

1sin t sin ht

2

29.

X 1 2 3 4 ------ n

P(X) 1

n

1

n

1

n

1

n

------ 1

n

1 1 1 1Mean 1. 2. 3. .... n.

n n n n

n 1

2

30. 1 1 16 12

W 10 10 L 11.12mH1212 L 90 10LC L 90 10

31. primary e relayI nI I N

Since number of primary CTS is 3, n = 3, rrelay

V3000 112N 600;I 0.056

5 R 2000

primaryI 3 0.015 0.056 600= 61A




32. For t < 0, the switch is in close position and network is in steady state when the inductor is

short circuit.

L

12i (0 ) 1.5 A

8

But for t>0, switch is opened and here the circuit becomes a source free circuit. So Li ( ) 0 ,

as no source is present in the network.

t

L Li (t) i (0 )e

1

1

eq

t

2t

L L

V 8I 3V 0

V 32I V 8I

VR 32

I

L 16 1Z

R 32 2

i (t) i (0 ) e 1.5 e

12 V

8 8

Li (0 )13V

1V

(t 0)

8

13V

1V

I

V

(t 0)




33. As the network is linear, by superposition RI is because of linear combinations of I and V.

From 1st statement

1 26 k (1) k (5)

1 2k 5k 6 … (1)

From 2nd

statement

1 210 k (5) k (30)

1 25k 30k 10 … (2)

Solving Equation (1) & (2) 1 2

R

R

k 26& k 4

So, I 26I 4V

So when I 2A & V 2V

I 26(2) 4(2) 44A

34. C mP 10W and P 10W as no load components of power are equally divided

2 2

OC OC C h

C

2 2

C m

1 C s

160 V VV HV 160V V LV 80V as P and P

2 R W

80 80 50P at 40Hz 10 6.4Wand P at 40Hz 10 8W

100 100 40

W P P 6.4 8 14.4W

35. If the speed controlled by armature circuit resistance method of an external resistance is

inserted in series with the armature circuit. In shunt motor, field flux remains unchanged.

Therefore, with reduction of armature current from I1 to I2 electromagnetic torque decreases

from a 1 a 2k I to k I . Since cT T , the speed decreases, back e.m.f also decreases. As a

result of it, armature current Ia increases and gets equal to initial value I1, so that initial

electromagnetic torque is developed again.

1 a g1 a 1 2m1 m2

a a a a

V I r RV I r E Ew ; w

K K K K

;

1 a gm2 2 2

m1 1 1 1 a

V I r Rw n E

w n E V I r

36. 2 2

st sc fl fl 2

1000I 1000 x I 1000 x 5 I I 312.5A

0.8 5

Auto transformer starter X = 0.8

Maximum permissible induction motor rating

3 400 312.5 0.75 0.85 138.02 kW




37. 2

ot

S S

VEVQ cos 0.5 36.87

x X

S

o

S

S

EVP sin 1p.u if power isincreased by 2%, then

x

EV1.02 sin 37.7

x

P EVcos 1.318 p.u

s x

38. The given amplifier is a series-shunt feedback amplifier. Now for approximation

we can take voltage gain with feedback

f

A 1A

1 A

f

o

V 1

V 10

1A 10

39. Given circuit both Op-amps are in inverting mode

So, the output of 1st

Op-amp is 0 0

20V 1 2Sin t 10 1 2Sin t ,V 10 20Sin t

2

By applying nodal at second Op-amp ‘-‘terminal

The second one acts as a summer so,

0

10 1 2Sin t 10V 20 mV

1 1

200 1 2Sin t 200 200 400Sin t 200

400 400Sin t 0.4 1 Sin t mV

40. C T T 0.5 Cd real measured 599C T C 0.5

measured d

299.5

299.5

2Q

9k

3Q

4.7K

2K

0V

2mA1k

12V

inV

fV

1Q




41. If any input is HIGH, (-0.8V) then corresponding transistors will be ON and common emitter

voltage will become - 1.6V at emitter terminal of all the transistors of differential amplifier

(Q1 to Q5). Second terminal of RE is at -5.2V. So

E

1.6 5.2 1.6 5.2I 4.62 mA

779 779

42.

MVI A,5EH ; A 5EH

ADI A2H ; A A A2H

5EH A2H

00H ;CY 1

MOV C,A ; C A 00H

43. Given, S L cV 220V ; T 1ms ; R 1.4 ; x 9 ,x 8

SO n

n 1,3 n

4Vi sin nwt

n Z

Fundamental component

S01

1

4V 1i

z 2

where

2 2

1z 1.4 9 8 1.72 4 220 1

115.125A1.72 2

22

01 o1P i R 115.125 1.4 18.55kW

L Cx x , Forced commutation is required

44. Current commutated chopper, CP O S q OI 2I ; V 220V ; t 30 s ; I 180A

O c

1

S

xI tC

1V 2sin

x

where x = CP

O

I2

I

6

1

2 180 60 1046.87 F

1220 2sin

2

6

S C

1 1

O

V t 220 60 10L 17.5 H

1 1xI 2sin 2 180 2sin

x 2




45. x n 1 2 3 ; y n 1 4 8 8 3

1N 3 which starts at ‘-1’ and terminates at ‘+1’

N=5 which starts at ‘-2’ and terminates at ‘+2’.

We know,

1 2 2N N N 1 N 3 i.e the number of samples in h[n].

By observation, it can be stated that 2'N ' ranges from ' 1' to ' 1'.

1 o 1h[n] a a a

1 1 o 1 0 1 o 1 1a , 2a a , 3a 2a a , 3a 2a , 3a 1 4 8 8 3

1 o o

1 1

a 1; 2 a 4 a 2

3a 3 a 1

h n 1, 2,1

46. Given, R = 20 S; V 400Vph

O120 Mode: Ph SV 0.40825V 0.40825 400 163.3V

ph

ph

ph ph

V 163.3 163.3I 8.165A

R R 20

O180 Mode : ph SV 0.4714V 0.4714 400 188.56V

ph

ph

ph

V 188.56I 9.428A

R 20

1 o 1

1 o 1

1 o 1

1 o 1

a a a

1 a a a

2 2a 2a 2a

3 3a 3a 3a




47. By dominant pole technique the reduced open loop transfer function is

100

G(s)s(s 1)(s 2) 10

To determine the value of ω when the polar plot intersects with the real axis is nothing but

the phase crossover frequency.

pc

1 1

G(j ) 180

90 tan ( ) tan 1802

2 rad / sec

pc

pc

2 2

G(j ) M gain margin

101.67

1 4

48. After writing the characteristic equation Option (C) becomes 3 2s 2s 2 0 , which missed

out s-term. So it is unstable.

49. AB bc1 RMS

sinV 0.9Edc. , E 600V

2

dcAB, RMS

AB,1 AB, 3RMS

For square wave output 180º

Third harmonic RMS Voltage

2EV 3 sin3 t sin3 t

3

V 270V V 90V

50. v mE k W2

1500 2.0 314.2V60

average aw aV E I R 314.2 40 0.12 319V 1

av

3EV cos

3 440 2cos 594cos

O3 440 2 319cos 319 cos 0.537 57.5

594

51. m2EAverage reverse recovery voltage =

LC 62 10kV

201.316 10 V1m 1




52.

S r

S r

S r

V V 0I

z

V VI

z z

V VI*

z z

Complex power transmitted over the line is Vr. I*

2

S S rrr

R R

2 2

S r S rR

V V VV VrV

z z z z

P jQ

V V V VVr Vr RP cos cos cos z

z z z z z

53. From stokes theorem A.dl A.ds

x A =

x y z

2 2 2

a a a

x y z

x y z

= 0

A.dl 0

54. Real power flow, where V1 – sending and V; V2 – receiving end V 2

1 2

L L

V V V sinP sin

X X

2

6 O10k sin 1

10 10 sin 305 2

Reactive power drawn by the line at load end = reactive power supplied by capacitor

2 2

O 6

L

V vQ 1 cos 1 cos30 2.68 10 VAR

X 5

Since 2Q V C

65

22

Q 2.68 10C 8.53 10 F

V 10k 2 50

= 85.3 F

SV rV 0

Z R jX Z




55. n=10 x=10, b=5

eqn

eqn

b x bn xZ b x x

n 1 n 1

1 1 n 1

Z x bn x

Short circuit kVA eq

Base kVA 1 n 1Q 100

Z x bn x

1 10 1

10 100 250 kVA10 5 10 10

56. 8 145 140 135 130 125 120 115

28.77%10 150 150 150 150 150 150 150

57. Exaggerate – blowing out of proportion, overstate

58.

i. , False ii. , True

iii. , True iv. , False,

Hence (C) only (II) and (III)

59. Exulted- to express great pleasure or happiness, especially at someone else's defeat or failure

60. Reluctant – unwilling

M

H C

C

M

P

C

M

CE P

M

E




61. x 1,onlyif 1 x

1 1 1 1 1 800, 0............................................, 0

4 108 4 108 4 108

1 81 1 1881,.............................................................., 1

4 108 4 108

1 189

4 108

1 190 1 2162, 2,......................................., 2

4 108 4 108

For1,2,..................................., 80, x 0

For81,82,83,..........................,188, x 1 108x1 108

For189,190,...........................,216, x 2 28x 2 56

Hence the required sum 108 56 164

62. Let N 120120120120120120120120...............................................278times

We know that a number isdivisible by72 8x9 ,if it isdivisibleby8and9

Clearly thelast threedigitsof thegiven number is120,so it isdivisible by 8

Sumof digits 1 2 0 278 834,soif wedivide thenumber by9, then theremainder is6

N 8K 9

M 6 8K 9M 6

The values that fits in for 8K = 9M +6 = 24 or96

96The required remainder rem 24

72

Alternate Solution:

12048

72

12012024

72

1201201200

72

12012012012048

72

Forms a cycle,

1 - 48

2 - 24

3 – 0

278th - 24




63. Let the total work be 120 units

A B C D E F

Number of days 10 12 15 20 24 30

Units/day 12 10 8 -6 -5 -4

Any three alternate days, whatever way they be paired up, net work done would be the same

i.e.,30 – 15 = 15 units every 3 days.

Hence, (D)

64. 1 – P(x) ≥ 50%

0 7 1 6 2 5

7 7 7

0 1 2

0 8 1 7 2 6

8 8 8

0 1 2

1 2 1 2 1 21 C C C 0.4293 50%

3 3 3 3 3 3

1 2 1 2 1 21 C C C 0.5317 53.17%

3 3 3 3 3 3

65. (B) and (C) qualifies, but (C) best sums up the passage as it is clearly stated in the passage

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