EXP (1)

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Answer Keys:

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

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

15 0.456 16 4.8 17 -0.5 18 C 19 C 20 A 21 0.87

22 0.36 23 55.7 24 B 25 10 26 A 27 B 28 A

29 C 30 13 31 B 32 B 33 D 34 B 35 A

36 C 37 B 38 B 39 D 40 0.694 41 D 42 6.208

43 0.01 44 B 45 83.33 46 2.4 47 -6.25 48 B 49 B

50 33.43 51 182 52 B 53 D 54 C 55 B 56 D

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

64 A 65 D

Explanations:-

1. i is leading V (or) V is lagging behind i V=cos 4t-90

4. 9000

n 375 rpm24

Speed of motor n 10

375 10 385 rpm

( the disc appear to move forward at a rate of 10 rpm)

6.

7. Find out the thevenin voltage across the galvanometer

3 1 34 2 4th th

1 3 2 4 1 3 2 1

thg

th g

E.R R RER R RE ;R

R R R R R R R R

EI

R R

Time base

Generator

sweep

output

y

x(diagonal line)

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8. f(x,y)=3x2y, h=0.1, y0=1, x0=0

The value of y after the first step is p1 0 0 0 1 1

hy y f(x ,y ) f(x ,y )

2

p1 0 0 01.0015 where y y hf(x ,y ) 1

9. Power P=VI

logP logV logI

1 1 dV 1 dI.

P V dP I dP

dP dV dI

P V I

0.5% 2% 2.5%

10. 2Velocity v t r ' t 3t i j 2tk

11. 5 5 4dz dz

az bz 0 z z b ady dy

4

dv( 4b)v 4a is a linear equation

dy

wherev z

13. A

CLG CLG: closedloopgain1 AB

;

If AB 1;A 1

CLGAB B

15. t0

I I e

1.12t0.6 e

t 0.456mm

16. n k 120

21.9 rad / sM 0.25

2

2 2

Acceleration, a .x

(21.9) (0.01) 4.8 m/s

17. 1z 0.5z

x (n) ROC z 1z 0.5 (z 1)

1z 1n

z (z 1)(0.5z)x (n) lim(z 1) 0 0.5 0.5

(z 0.5) (z 1)

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18. Roc z 4 z 1 z 2 z 4 z 2 2 z 4

19. 2 S2S 3S 4 y S R S 1 2eS

2

Y S 1 2e

R S 2S 3S 4

20. Area of capillary tube, 2 2 6

tA 5 19.635 mm 19.635 10 m4

Volume of bulb and capillary tube, 6 3V 80 10 m

26 32

t

6

19.635 10 25 10A hP 153.4 m

V 80 10

21. VQ C P

V

V

2 0.71 30 30C 0.71

2 20C 20 = 0.87

22. oe gtp

3 2100 0.055 5 10 P 0.36 MN/m

23. TD dD

V10 0.7 2 0.0259I 0.93mA, r 55.7

10k I 0.93

24. A B C Y

0 0 0 0

0 0 1 0

0 1 0 1

0 1 1 0

1 0 0 0

1 0 1 1

1 1 0 1

1 1 1 1

25. N-bit ring counter is N:1 divider, so, output of ring counter will be

200KHz20KHz

10

MOD-N counter is divide-by-N counter. So, output of MOD-2 counter is

20KHz10KHz

2

0 0 0 1

0 1 1 1

0

1

BC A 00 01 11 10

Y BC AC

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26. i 0 ER r 1 R

m m 0

i 0 E i E

0 E

v

S i

2mAg 0.08ohms ; g r r 1.56k

25mv

R 500k; 1 R R r 500 1.56 k R 3.96k

1 R 126 3.96kA 0.988

R R 505k

27. The - equivalent circuit is

B 1 2Let R R ||R 50k

The contribution of RE to the internal impedance as seen across the terminals BE

is

m E

m E

vg v R

rg r R

vr

m E m E

r ' r g r R 1 g R

The resistance seen by CB is

S B

S

B

m E E

S B

L B

B

R R || r '

R 0.3k

R 50k

B 150r ' 1 g R r 1 R r 1 0.4 2 62k

r 2

R R R || r ' 0.3k 50k || 62k 28k

1 1f C 0.28 F

2 R C 2 28k 20

28. Angle of Asymptotes are 90 and270.

P Z 2

But Z 0

P 2

C CPole zero Pole zero poles 0

2P Z P Z 2 0

poles 4

BC B C

mg vv

ER

0.4k

1 2R ||R

50k

5R

0.3k

SV

1.5k

BC

SV

SR

r 'BR

E

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29.

o n o so

o n 2 1 2

o s 1 2 1 2

2 1 2 1 2 1 2o

1

o

S /N v / v / v / v

v / v A / 1 A A

v / v A A / 1 A A

Hence

S /N A / 1 A A / A A / 1 A A

1/ A

Hence S /N would be independent of

30. As R1 and R2 are in series, the same current is flowing through then and as they

are equal, the voltage drop across each is V so that the circuit becomes

V 8V 5V 13V

SV

1

1

1

1 2 2V SV

1

1

1

23

2V

SV

1

1 5V SV5

85V

SV

8V

5

85V

3V

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31. As there are no independent sources the Thevenins and Norton equivalent will

have 0V and 0A sources. To find RTH, a 1A source is connected as x

TH

VR

1

Writing a nodal equation at n,

x x x x

xx

x x

x x x xx

x x x

x

xTH

V V V 1000 i1

100 3000 100V

V 1000V V 30001

100 3000 100V V V V2

1 i100 3000 3 100 3000

0.01V 0.0003V 0.0067 V 1

V 58.82 V

VR 58.82

I

32. n 2dsin

o60 24'1 2(302.9)sin

2

2(302.9) 0.503

304.7 pm

33. 2.52.5 3d

8 90 8Q C 2g H tan 0.6 2 10 0.8 1 0.8192 m /sec

15 2 15

Now, 1.5

d

3

dQ 8 dQ dHC 2g 2.5H or 2.5

dH 15 Q H

0.02dQ 2.5x0.8192x 0.0512m /sec

0.8

34. P x'y'z' x'yz xy'z xyz' x y z x' y z

x y z x' y z ' x y z '

35. Meter constantNo.of revolutions

Energysupplied

Energy supplied = VIcos t 220 10 0.85 6 11.22 kwh

5329.5Meter cons tant 475 rev /kwh

11.22

For 2375 rev, energy consumed is 2375

5kwh475

35 10 220 12 cos 4 cos 0.4734

100

1000 xi

n

xV 1A

A

B

xi

3k

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36.

37. Given that d=0.4

Required probability = P(x>2) 1 P(x 2) 1 P(x 0) P(x 1) P(x 2)

0.4 0 0.4 1 0.4 2

0.4

e (0.4) e (0.4) e (0.4)1

0! 1! 2!

1 e (1.48) 1 0.9921 0.0079

38. 2x

x 42 h

2

1

4 2

2

x

x x

x 2, 4

xA x 4 dx 18

2

A 'y ' of upper curve ' y ' of lower curve

39. The singularities are z = 3i, -2i lies outside the square

Value of integral = 0, using Cauchys integral theorem

40. Possible mode wavelength

2

9 2

6

L cavity length between reflecting surfaces2Ln

n cavity refractive index.(1000 10 )

2 200 10 3.6 wavelength

0.694 nm

2, 2 4, 8

y x 4

2xy2

2 1

2 1

2 1

2 1

2 1

2 1

2 1

oI

1I

2I

3I

4I

5I

6I

1I

c

B

A

fB

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41. e o1 1

x n x n x n , x n x n x n2 2

42. 1 frame will have 1bit of every sample , totalno of bits 96 8 8000

96 8 8000number of synchronous bits required 64000

96data rate total number of bits number synchronous bits required

6.208Mbps

43.

2

i i

n n

i

i

i i

1Transfer function is

Ts 10Ts 1

1w 2 w 10

T

12 0.5 10

T

0.5T 0.1 T 0.01

5

44. FSD in

3 3

sh

sh

Given I 5mA; R 30

5x10 x30 95x10 xR

R 1.579

change in temperature, t 25 c

'

m m c

'

sh sh m

'' shm ' '

sh m

'

m

R R 1 t 30 1 0.004 25 33

R R 1 t 1.579 1 0.00015 25 1.585

R 1.

change in meter curren

585I I 100 4.583mA

1.585 33R R

It 4.583 5 0 .417mA

45. p

3r

V 100C 83.33 F

V fR 2 60 10 10

46. 1 1 2 2 1 220 10i 10 i i 1, i i 1 5 5i 0

29

i A13

22Power loss in the 5 resistor = i 5 2.4W

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47. For the input 1010; 0V i 5k

10 10

i 1mA 0.25mA10k 40k

1.25mA

0V 6.25V

48 & 49. c 2Ks k

Open loop transfer function G(s)G (s)S (S 40)

3 2

Ks kClosed loop transfer function of unity feedback system

S 40s ks k

Requirement:

2

p

1

%M 30%

e 0.3

0.357

s

n

n

d

t 2

42

5.60

5.23

n d

So dominant root will be

j

2 5.23

The standard second order equation 2 2n nS 2 s for comparing with the above

an insignificant pole must be added as for far away from the dominant pole. Let it

be n(S )

Now the equation

2

n n n n

2 2 3 2 2 2 2

n n n

3 2

S (S) 2 2

S 2 S 4s S 4s s 2 s 8 s 2

S s (4 2 ) s(31.36 8 ) 62.72

available

3 2S 40s ks k

by doing term wire comparison.

40 4 2

18

31.36 8 k

k 175.36

62.72 k

62.72 186.43

175.36

k 175.36

6.43

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50 & 51.

36A 1384

0.026

oo5

1 f

1f

o i

i

v 1384v 33.43V

0.026 1 (1384 5.55 10 )

D 7% , D 1%

DD

1 A

A 6

Av v

1 A

36 7v 0.182 V 182mV

1384

52. In Cromptons potentiometer,

slidewire3

voltage drop across the slidewire is ,

V R (I) [Given I 15mA]

10 15 10 0.15 v

V 0.15voltage of each division 1.5mv

no of divions 100

53.

1 1Resolution (0.0015) [ Given accuracy upto ]

3 3

0.0005 V

54. 1

X(j )200 j

c c c

c c

cc

f f f2

out 2 2 2 2 2

f f 0

ff

1

2 2 2

0 0

1 c

2

1 df 2 dfE X(j ).1 d

2 (200 ) (2 f) (2 ) (100) (f)

1 1 1 df 1 1 f. tan

100 100 100 1002 2f1

100

f1tan J

100200

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55. Energy of input= 2x (t)dt2( 200 ) 400 t

00

1 1e dt e

400 400

1 c

2

21 c

c

1 1 1As per given condition of energy of input .

4 4 400

f1 1tan

1600 100200

f.200tan

.1600 100

f 100 tan 41.42Hz8

60. As 1 day work 1

12

Bs 1 day work 1

16

(A +B) s 1 day work 1 1 4 3 7

12 16 48 48

(A +B) s 4 days work 7 7448 12

Remaining work 7 5

112 12

62. Let the tens & unit digit be 8x &x

Then 2 2

2 2

8 810x 18 10 x 10x 18x 8 80 x

x x

9x 18x 72 0 x 2x 8 0

x 4 x 2 0 x 2

Number is 24

63. Ratio of initial investment = 7 4 6

: : 105x, 40x, 36x2 3 5

150A : B : C 105x 4 105x 8 : 40x 1 2 : 36x 12

100

1680x : 480x : 432x 35 : 10 : 9

10Hence B 'share Rs. 21600 Rs.4000

54

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64. Since first & second are mixed with equal proportions

There average price = 75 85

Rs. Rs.802

So, the mixture is formed by mixing two varieties, one of Rs. 80 per kg and the other at say, Rs. x per Kg in the ratio of 2 : 2 ie., 1 : 1

Bu the rule of Allegation

x 115

1 x 115 35 x 150115 80

65. Required difference 16% 18% of 6000 18% 10% of 9000

34% of 6000 18% of 9000

2040 1620 420