Self Quiz EXP1

7/17/2019 Self Quiz EXP1

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Answer Keys

General Aptitude

1 D 2 C 3 60 4 D 5 D 6 D 7 B

8 B 9 299.7 10 B

Instrumentation Engineering

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

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

15 C 16 D 17 0.5157 18 A 19 30 20 0.8 21 C

22 B 23 C 24 8.7 25 B 26 0.52 27 D 28 B

29 C 30 -2.02 31 4.8 32 4 33 D 34 600 35 A

36 6.66 37 30 38 A 39 C 40 B 41 A 42 C

43 -29.5 44 A 45 A 46 B 47 B 48 12.2 49 0.66

50 C 51 67.12 52 D 53 A 54 A 55 C

Explanations:-

General Aptitude

1. Barrack, duplex and gazebo are spaces used for specific purposes while imbrue (which meanssoaked) does not fit in the group.

2. The man was treated as madman, hence the word in the first blank will be ‘deranged’; theothers (many) considered themselves to be sober and wise hence ‘prudence’ fits in the second

blank.

3. Let breadth = x metres.

Then, length = (x + 20) metres.

5300Perimeter m

26.5

200m 2 x 20 x 200

2x 20 100

2x 80

x 40.

Hence, length = x + 20 = 60 m.






4. Cataclysmic and catastrophic means disastrous.

5. The answer is option D. Option A uses past continuous tense which is not required since atruth is mentioned which should be in present tense. Option B is wrong because there is

unnecessary use of present continuous tense. Option C is wrong because ‘so’ changes the

meaning of the sentence and ‘downward on the surface’ should be replaced with ‘downwardfrom the surface’.

6. Time taken by A to fill the tank = 72 min

Time taken by B to fill the tank = 90 min

Time taken by C to empty the tank = 60 min

Pipes A and B are opened for 14 min.

Part of the tank filled is

=1 1 7

14 x ( )72 90 20

Remaining part of the tank =

7 13

(1 )20 20

Now all the pipes are opened

1 1 1 13t x ( )

72 90 60 20

t = time taken to fill remaining part of the tank

t = 78 min

Total time = 78 + 14 min

= 92 min = 1 hr 32min

7.

4 x 6 8 y 0 must be divisible by 3

18+x+y must be divisible by 3

and, x 8 0 4 6 y

x y 2 must be either 0 or 11

x-y-2=0 y x 2

18 x x 2 16 2x

x can be only 7

y 5

8.

6 LCM of 2&3 ; 12 LCM of 3,4

20 LCM of 4,5 ; 30 LCM of 5,6

So the missing numbers P = LCM of (12, 60) = 60

and Q = LCM of (60, 60) = 60






9. Distance between A and B is 600km

Speed of first train = 54km/hr

Speed of second train = 66km/hr

But 2nd train started 1 hr after the first. So first train might have covered 54 km

Relative speed = 54 + 66 = 120km/hr

546Time taken 4.55

120

First train travels 54 × 4.55 km i.e. 299.7km by the time they meet

10. The answer is option B. Option A is wrong because ‘intellect’ is not given in the paragraph.

Option C can be eliminated because it represents ‘long process of growth of aesthetic ideas’as a hurdle by using the word ‘though’. Option D is wrong since it was not instinc t that was

discovered.

Instrumentation Engineering

1. Since f(x) and g(x) satisfied the conditions of cauchy’;s mean value theorem in [1,2]

f (x) f (2) f(1)

g (x) g(2) g(1)

8 22 f (x) 2g (x)

g(2) 2

g(2) 5

2.2 1

A B4 0

4 3A B

2 6

6 2 3 12A A

6 6 3 3

2 1 2 1 3 1 1 2

A B B4 0 4 0 3 3 1 3

3 1 1 2 2 9AB

3 3 1 3 0 15

54

54

66

600 54

546






4.

even

x n x nx n

2

;

odd

x n x nx n

2

5. n n

n

x z a z

n

1

n

az

Since the summation is of infinite summation on both sides, the summation does not existhence x(z) does not exist.

7. Let us apply nodal Analysis to find VX.

O O

x x xV 20 0 V V 20 90

0 j10 25 j10

O

xV 70.7 45 3 O

xV t 70.7cos 10 t 45

8.1 2 2

V AV BI

1 2 2

1 Lin

1 L

I CV DI

V AR BZ

I CR D

2 2 LV I R

9. The Thevenin’s impedance at (a, b) is 10 j15 || j10

THZ 8 j14

The value ofL

Z for which maximum power is observed is *

L THZ Z

13. If inI increases, G D1V increases, I increases. Hence 0V decreases so that D2I decreases,

allowing GV to increase. Since retuned signal enhances the effect produced by inI , It is

positive feedback.

14. H Lf 50kHz T 20 s T T

H L

6

L B B 12

H A B A

Duty cycle 75% T 15 sec and T 5 s

5 10For astbale M.V, T CR ln 2 R 10.6k

680 10 ln 2

T C R R ln 2 R 21.2k

A B

C D

inZ

LR

1V

2V

1I

2I

0.1mF xV

10mH

25 2S

V ~

~ 1S

V

j10 xV

25 O20 90 ~

~

O20 0

j10






15.

All the 1’s are grouped in two loops only so need to make a group of 1’s shown by the dottedline. So, the term represented by dotted line is called redundant term.

i.e., Redundant term C D

16. The range for the 2’s complement representation is n 12 to n 12 1 where

n number of bits. For n = 4 the range is -8 to +7. so , we can’t represent +8 in 4-bit digital

system.

17. Let f(x)=7x-3cos x-1

Consider x :01 2... f (x) 7 3sinx

f (x) : 4 4379

Choose an interval is

(0,1)

Let 0

0 1x 0.5

2

By Newton-raphson method;

01 0

0

f(x ) f (0.5) 0.132747685x x 0.5 0.5 0.5157

f (x ) f (0.5) 8.438276616

18. Eigen values:

1I A det

2 3

1 2 0

1,2

Eigen Vectors:

1

CD AB 00 01 11 10

1

1

1

1

1

1

1

1 10

11

01

00






x1 1 01

2 2 x 02

Therefore eigen vector is

x x 11 2

2x 2x 11 2

1Similarly; for 2 The eigen vector is

2

19. Maximum phase shift11sinm1

G s is a lead compensator c

T 0.3, T 0.9

111 11 1 o33, sin sin 30m 1 21

3

20. The input is 1 4t u t r t

2 2

2

s 0

1 4 s 4R s

s s s

R s 1 s 4 Now, E s

101 G s H s s1 1s s 2

e lim sE s 0.8ss

21. Bandwidth requirement of SSB is half of the double sideband. Thus bandwidth requirement is

less.

Power requirement is also less as only half of the spectrum is used.

22. f maxf k m t

max

m t 5 3 8volts

f 10 8 80Hz.

100

Bandwidth of m t Hz

m

100 200Bandwidth of FM 2 f f 2 80 160 Hz






23. To suppress errors that persists for longer times, the ITAE criteria will tune the controller better because the presence of large time‘t’ amplifies the effect of even small error in the

value of the integral.

24. The sensitivity of meter movement for d.c. =dc 3

fs

1 1S 1000 / V

I 1 10

For half wave rectifier, the ac sensitivity,acS 0.45 1000 450 / V

Total resistance of circuit of ac operation =acS V 450 20 9000 ;

s ac m dR S V R R 9000 300 0 8700

26. Let probability of winning C be ‘x’ Let probability of winning A is ‘2x’ Let probability of winning B is ‘2x’

x 2x 2x 1

1

x 5

We have

P B C P B P C P B C

P B P C P B .P C

2 1 2 1 13. 0.52

5 5 5 5 25

27. divp .p

2 2 3(4x 8y z) (3x 3x) (8y z 2x )

0

3

p is not solenoidal.

2Q xyz p

2 2 2 2 3 2 2 2 3xyz (2x 8xy z)i xyz (3x y 3xy) j xyz (4y z 2x z)k

divQ .Q

2 2 3 3 4 2 2 2 3 3 4 26x yx 16xy z 6x yz 6x yz 16xy z 6x yz

0

Q is solenoidal

28. 1 1

3 2

1 1L L

s s s(s 1)

Using partial fractions






2 2

1 1

2 2

1 A Bs C

s(s 1) s s 1

solving, A 1, C 0, B 1

1 sL L

s(s 1) s s 1

1 1

2

1 sL L

s s 1

1 cost

29. Let z = x + iy then

2 22 2

2 z 1 z i 2 x 1 iy x i y 1

4 x 1 y x y 1

2 2 2 2 2 2

2 2 2 2

4 x 2x 1 y x y 2y 1 u iv u v8 2

3x 3y 8x 2y 3 0 x y x y 1 03 3

The region is interior of a circle

30. The above figure can be redrawn as

1 11 1 1 1 1

v v2.5 0.2v v 0.2v 2.5 6.67 v 50.5V i 2.02A

6.67 25

31. Power is the rate at which energy expended.

Total energy expended E, is the integration of power supplied during the 24minutes interval.

Total energy24

0 pdt

1 1

5 6 10 10 7 10 60 115 602 2

Total energy 115 60

Average power 4.8WPeriod sec 24 60

32.1

ΔPsensitivity

2 1

i

25

1

0.2v 10

2.5A

100

1v

1v 1

6.67

10.2v

2.5 A

2






A A

B B

ΔP ΔS 8Ratio of pressure = 4

ΔP ΔS 2

33.1

p , n 18

2

1q

2

P X 8 P(X 0) P(X 1) .... P(X 8)

o 18 0 1 18 1 8 18 8

18 18 18

o 1 8

1 1 1 1 1 1C C ... C

2 2 2 2 2 2

0.04

34.in m V m D D

m

1 10R 60 g 16.66 mS A g R 10; R 600

g 16.66mS

35. 2H sat sat

1 2

R V V V

R R

2

2 1 2 2 1 1 2

1 2

R 60.2; R 0.2R 0.2R 0.8R 0.2R R 4R

R R 15 15

36. 1V

1 1 1

1C 1

A 15 6.6 Mrad / secR C R






37.

At the first stage delay by Ex OR 10ns

Inside the IC package of carry lookahead Generator, there will be two-level AND and OR

gates to represent SOP of carry generator and carry propagation. So, the delay inside the IC package = 5ns (for AND) + 5ns (for OR) = 10ns.

At the output stage one more Ex-Or gate will make the delay of 10ns.

To appear the sum bits total delay required is = 10 5 5 10 ns 30ns .

38. 101k RAM 2

10 address lines required for 1K RAM designated by o 9A to A

15 14 13 12 11 10 9 OA A A A A A A A Address Range

0 0 1 0 1 0 0 0 2800H

0 0 1 0 1 0 1 1 2BFFH

39.

n 1

n

S R Q0 0 Q

0 1 0

1 0 1

1 1 *

3B

3A 3P

3G

2B 2A

1B

1A

0B

0A

0C

2P

2G

1P

1G

0P

0G

0C

carry

lookahead

generator

4C

4C

3P

3S

2P

2S

1P

1S

0P

0S

2C

1C

0C

3C

0 1

10

01

00 00






40. N 1

kn

N

n 0

x k x n . W

N 1

*kn

N

k 0

1x n x k w

N

1

N N N N N NX w x x w X

*

N N N

1x w X

N

1 * * *

N N N N N N N N N N

1 1w X w X I W W W W NI

N N

41. h t x T t

h t

*

The range of convolution is 0 0 to T T

y 0 0

y 2T 0

42. The characteristic equation is given by 1 G s 0 2 2 1 B

Ts s B 0 s s 0T T

T = Constant

A = Variable gain of the system comparing with standard form

2 2s 2 s 0n n

B 1n

T 2 BT

1 10.3 ; 0.9

2 B T 2 B T1 2

B0.9 11 B B2 10.3 B 9

2

43. Poles at s 0, 20, 30 j95.5 ; Zeros at s 40

Angle of Departure

2T

T

2

Ty t

T

2T

T






95.5 95.5 95.5o o 1 1 o 1180 180 tan 180 tan 90 tan30 10 10

o o o o o o180 107.4 96 90 84 29.5

44.

1

Gain Margin 200G j H j pc

oarg G j H j 180 at 10 pc

o oPhase m arg in 180 arg G j H j 89.89

G j H j 1 when 0.01gc

47. Since the modulation scheme is double-side band suppressed carrier also known as AM with

suppressed carrier, Output of envelope detection is m t

48.1 1β

373 298100 25R = R e

1 1β

373 2981K = 10K e β =3412.55

1 1β

423 298150 25 150R = R e R = 339.1

Time to reach 5% of initial 3 Time constant = 3 RC

= 3 RC -6= 3 339.1 12 10

12.2msec

49. Given equation can be written as

2

2

2

dyx cos x y cos x xy sin x sin x 0

dx

dy dy cosx xsin x sin xx cos x y cosx xsin x sin x y

dx dx x cos x x cosx

I.F xcosx

dy 1 dy 1cot x.y cosecx.y cot x. cosecx

dx y dx y

1 1 dy dzLet z

y y dx dx






cosydx

sinx

dzcot x.z cosecx

dx

I.F e sin x

zsin x ( cosecx)sinxdx c zsin x x c

at x , y 24

1 1 1c c

4 4 22 2

sin xnow x c

y

sin xat x y

4 x c

1sin

4 2

y y 0.661 1

4 4 2 2 2

50. i C C fe b cV I R h I R

s f s f i fe c

s ie

i s f

V VV VV h R 200 3k

R h 1k 2k

V 200 V V

51. 1

2

n 1.421 1 0.014

n 1.4

For step index fiber, pulse dispersion= 1

8

n L 1.42 10000.014 67.12 ns/km

c 3 10

53. Beam speed =19

6a

31

2eE 2 1.6 103000 32.47 10 m / s

m 9.1 10

Deflection sensitivity, S2

d

3

a

L l 1 3 10m / V 0.833

2d E 2 6 10 3000

mm/V

Deflection factor =1 1

1.2S 0.833

V/mm

54.t

2

2e

1

1kbI

SV ieh

cI

f V

iV

CR

fe bh I






t

222

d 2 4jt.e j

d 1 1

55. Inductive reactance of pressure coil = 32 100 5 10 3.14V

Resistance of pressure coil = 1000

If the phase angle of pressure coil circuit is , then3.14

tan 0.00314 rad1000

=0.18º

If cos is the load power factor, then

cos

true power = reading of wattmeter cos cos -

….. (1)

But True Power = 2 2 2 240I R 5 Zcos 5 cos

5

Equation (1) becomes 5 240cos

cos23

cos cos -

89.08º

Percentage error = tan tan 100 19.56 18%

Self Quiz EXP1

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