ANSWERS - IES MasterQ S 1 R a r Surface of sphere For Gaussian surface S 1 Q 0 enc E.ds 0 Q enclosed...

1 (a)

2 (b)

3 (b)

4 (b)

5 (b)

6 (a)

7 (b)

8 (d)

9 (c)

10 (d)

11 (d)

12 (a)

13 (c)

14 (a)

15 (d)

16 (d)

17 (d)

18 (d)

19 (c)

20 (b)

21 (a)

22 (a)

23 (d)

24 (c)

25 (d)

26 (b)

27 (a)

28 (c)

29 (a)

30 (d)

31 (c)

32 (c)

33 (d)

34 (d)

35 (c)

36 (a)

37 (b)

38 (c)

39 (b)

40 (c)

41 (c)

42 (c)

43 (b)

44 (c)

45 (d)

46 (c)

47 (b)

48 (c)

49 (d)

50 (a)

51 (b)

52 (d)

53 (c)

54 (d)

55 (d)

56 (d)

57 (b)

58 (c)

59 (d)

60 (b)

61 (c)

62 (b)

63 (d)

64 (c)

65 (c)

66 (b)

67 (d)

68 (c)

69 (b)

70 (a)

71 (a)

72 (c)

73 (a)

74 (b)

75 (b)

76 (d)

77 (d)

78 (c)

79 (b)

80 (a)

81 (c)

82 (d)

83 (c)

84 (d)

85 (d)

86 (a)

87 (d)

88 (a)

89 (b)

90 (c)

91 (a)

92 (c)

93 (c)

94 (c)

95 (b)

96 (c)

97 (c)

98 (d)

99 (a)

100 (a)

ESE-2019 PRELIMS TEST SERIESDate: 30th December, 2018

ANSWERS

101 (d)

102 (a)

103 (b)

104 (c)

105 (c)

106 (a)

107 (c)

108 (b)

109 (d)

110 (b)

111 (b)

112 (c)

113 (c)

114 (c)

115 (b)

116 (a)

117 (c)

118 (a)

119 (a)

120 (b)

121 (c)

122 (a)

123 (c)

124 (d)

125 (c)

126 (c)

127 (b)

128 (c)

129 (a)

130 (b)

131 (c)

132 (a)

133 (d)

134 (d)

135 (c)

136 (c)

137 (d)

138 (d)

139 (b)

140 (a)

141 (b)

142 (a)

143 (a)

144 (a)

145 (a)

146 (a)

147 (a)

148 (c)

149 (c)

150 (a)

IES M

ASTER

(2) Full Syllabus

1. (a)2. (b)3. (b)

Under rated operating conditionVt = Ea + Iara

or Vt = m m a aK I r

Given, Vt = 200V,

m2 1000

60Ia = 10 Amp and ra = 2

200 =

m2 1000K 10 1

60or Km= 1.81 V–s/radFor rated motor torque, armature current = 10A V0 = Vt = m m a aK I r

2 2 230 cos =

2 5001.81 10 160

or

650.53 cos = 104.77

or cos =104.77

650.53cos = 0.505 = 59.61º

4. (b)5. (b)6. (a)

Minimum braking speed is

min =a a

m

I rK =

300 0.21.2

min = 50 rad/s or 477.46 rpmMaximum braking speed is

max =s a a

m

V I rK =

400 300 0.21.2

max = 383.33 rad/s or 3660.06 rpm

Note : 0 sV Vwhen 0 then V0 = 0 then minimum breakingspeed occur.when 1 then V0 = Vs then maximum breakingspeed occur.

7. (b)

Circulation of a vector is given by A.d l

602 2

q0 0 0

ˆ ˆ ˆA.d a A.d a A.d a

2 2

0 0

cos d 0 cos d

2 22 2

0 0cos0 cos60

2 2

12 – 2 12

8. (d)

A vector field A is said to be solenoidal.

Divergenceless if A 0

and irrotational orpotential if A 0

. For a scalar field V. Laplace

equation is 2V 0 .9. (c)

Total charge Q will be uniformly distributed overthe surface of the conducting sphere.

S2Q S1

R

ar

Surfaceof sphere

For Gaussian surface S1 encQ 0

enclosedQE.ds 0

E 0

for 0 < r < R.

For Gaussian surface S2,

enclosedQE.ds Q

2QE

4 r

2

1E for R rr

10. (d)Electric field exists from high potential to lowerpotential.

VP > VQ and hence VPQ is negative (VPQ = VQ– VP)Therefore VPQ is negative, there is a loss inpotential energy in moving unit charge from P toQ and hence work is being done by the electricfield.

11. (d)Equipotential surface is a surface which is cut by

IES M

ASTER

[EE], ESE Prelims Test Series |FLT - 05| 30th December 2018 (3)

the electric field perpendicularly.

Sphere

+q

The equipotential surface due to a point chargewill be concentric spheres as shown above.

12. (a)

Tangential component of E

is continues acrossboundary.

E1t = E2t

1E sin = 2E sin ...(1)

Normal component of D

is continues acrossboundary.

D1n = D2n

r1 inE = r2 2nE

r1 1 r2 2E cos E cos ...(2)

Dividing (1) by (2), we have

r1

tan =

r2

tan

tantan

=

r1

r2

13. (c)

a

b

L

Gaussiansurface

Applying Gauss’ law to a arbitrary Gaussiancylindrical surface of radius a b , wehave

Q = E.ds E 2 L

E =Q a

2 L

V =a a

b b

Q ˆ ˆE.d a .d a2 L

l

V =a

b

Q d Q bn2 L 2 L a l

Capacitance, C = Q 2 L

bV lna

14. (a)

4

P(–4,6,4)

y=24

yz

r

x

y=6

Electric field due to infinite uniform line,

E = Ro

charge densitya 10 nC / m (given)2 r

Here, x yˆ ˆr ( 4 0)a (6 2)a

= x yˆ ˆ4a 4a

2 2r ( 4) (4) 4 2

Rra

| r |

Hence, 9

x y9

ˆ ˆ( 4a 4a )10 10E4 2102 4 2

36

= x yˆ ˆ22.5( a a ) V/m

15. (d)

Work done, W = B

A

Q E dl

= B

x y zA

5 (y dx x dy) dl dxa aya dza

= B

A

y x 15 [(x 1) dx x dx] dy dx

IES M

ASTER

(4) Full Syllabus

= x 2

x 1

5 (2x 1) dx

=

22

1

2x5 x2

= –[(4 – 1) + (2 + 1] × 5

= –30 Joule

16. (d)

y

z

x

(0,2,–1)x=0, z=0

x=0, z=–22A

2A

As the point (0, 2, –1) lies exactly between thegiven two lines. Hence, the magnetic field intensityby each line are opposite to each other.

Hence, net magnetic field intensity at point (0, 2,–1) will be zero.

17. (d)

x

y

z

1A1

21A

Right Hand Thumb Rule :Thumb is in the direction of current, fingerswill represent the direction of magnetic field.Due to wire 1 , H field will be in the xadirection and due to wire 2 , H field will be inthe

za direction. Thus, x and z components

are non zero.18. (d)

The net capacitance can be considered as twocapacitances in series.

1

1AC

d 2

and 2

2AC

d 2

C =

21 2 1 2

21 21 2

C C 4 A2AC C dd

C =1 2

1 2

2Ad

19. (c)

Given, 3 2 3 2 2 1A ,B ,C matrices then order of (7A –9B)3×2 is 3 × 2.

Order of (7A – 9B)3×2C2×1 is 3 × 1.

20. (b)

Given A =

3 41 1

By observation, for n = 1 only

1 2n 4n

n 1 2n =

3 41 1 is true

Let, An = 1 2n 4n

n 1 2n

Now, An+1 = AnA

=

1 2n 4n 3 4

n 1 2n 1 1

=

3 6n 4n 4 8n 4n3n 1 2n 4n 1 2n

An+1 =

1 2(n 1) 4(n 1)

n 1 1 2(n 1)

by induction, n 1 2n 4nA n 1 2n is true

Note : Student are advised to check options puttingn = 1, 2 or 3 (some values) in examination andsubsequently many options will be eliminated.

21. (a)

f(t) = t1(1 e )t

tL(1 e ) =

1 1s s 1

t

s

1 e 1 1L dst s s 1

= s( ns n(s 1))

IES M

ASTER


=

s

sln1 s

= 0 –

sns 1

=

s 1ns

22. (a)

I = 1 1 1 (x y z)0 0 0

e dx dy dz

I = 1 1 1x y z0 0 0e dx e dy e dz

= (e – 1)(e – 1)(e – 1) = (e – 1)3

23. (d)

For f(x) to have a minimum

dfdx = 0 and

2

2d f 0dx

24. (c)

x dy ydx = 2 2x y dx

dydx =

2y y1x x

Homogeneous differential equation

25. (d)

Sample space = {HHH, HHT, HTH, HTT, TTT, TTH,THT, THH}

Probability of getting no head = 31

2

Probability of getting at least one head

= 1 – 31

2

= 78

26. (b)

Probability density function

f(t) = 1 t, 1 t 01 t, 0 t 1

Mean, E[t] = tf(t) dt

=0 1

1 0t(1 t) dt t(1 t) dt

=0 12 21 0(t t ) dt (t t ) dt

=1 1 1 1(0 1) (0 1) (1) (1)2 3 2 3

E[t] = 0 ...(i)

Mean square value,

E[t2] =0 12 21 0t (1 t) dt t (1 t)dt

=1 1 1 1(0 1) (0 1) (1) (1)3 4 3 4

=2 1 13 2 6

Variance = E [t2] – (E [t])2 = 16 – (0)2

=16

Standard deviation = Variance

Standard deviation = 16

27. (a)

Mid value of the class interval = 40Class size = 6

Lower limit = 40 – 6 372

Upper limit = 40 + 6 432

28. (c)29. (a)30. (d)

– Induction generator runs at speed more thansynchronous speed, hence it is cal ledasynchronous machine like induction motor.

– Since, speed of induction generator N > Ns.

Hence, slip. s =s

s

N N0

N

– Induction generator supplies active power togrid but takes reactive power from grid.

– Since its speed may change time to time, it isused in windmill generator.

IES M

ASTER

(6) Full Syllabus

31. (c)Synchronous speed

Ns = 120f

P

=120 50 1500 rpm

4

Slip at full speed

s =s

s

N N 1500 1440N 1500

= 60 0.04

1500

Slip at speed of 900 rpm,

s' = 1500 900

1500

=600 0.4

1500

Since rotor ohmic losses,= (I'2)

2r2'

So,22 2

22 2

I r at 900 rpmI r at full load =

slip at 900 rpmfull load slip

=0.4 100.04

22 2I r at 900 rpm =10 × 2

2 2I r at full-load

32. (c)

Due to magnetic saturation, the crossmagnetizing armature mmf resulted intodemagrotizing effect which causes the reductionin total flux per pole. The reduction is flux perpole depends upon the degree of magneticsaturation. So, that reduction in flux per poledecreases the voltage generated.

Due to armature mmf, the zero crossing of theflux density wave is shifted from the interpolaraxis, hence there are some emf induced in thecoils undergoing commutation. This resulted isa delaying in the reversal of current and showpoor commutation.

The cross-magnetizing effect of armature mmfincreases the flux density under one half thepole and which is much more than thedecrease in the flux density under the otherhalf of the pole. This causes increase in ironlosses.

33. (d)From the Ward-Leonard method of speed controlof dc machines :

N<N0 N=N0 N>N0

(Base Speed)

Speed

TorP T

P

T

P

0

Vcontrol

a Field control

As the speed is increasing from base speedor, rated speed N to 3N, so it is possible onlyin field control.Since, power = torque × speed

aI N

aI N ;

which is constant

34. (d)If the pole tips are chamferred, there is leastgap between pole shoe and armature at thepolar axis, and the gap increases while movingto the pole tips. Correspondingly, reluctance isminimum at the polar axis and increases whileapproaching to pole tips. So, the flux densityis maximum at polar axis and minimum atinterpolar region and the air-gap flux densitydistribution achieve nearly a sinusoidalwaveform.

35. (c)When the speed of synchronous motor is morethan synchronous speed, then the motordamper winding will act like induction generator.

36. (a)

qX = min

max

V 96 8I 12

dX = max

min

V 108 10.8I 10

37. (b)Ns is same for both

1

1

120 fP

= 2

2

120 fP

IES M

ASTER


120 2510 =

2

120 60P

P2 = 24

38. (c)

KVA

KW

KVAR

KVAR = KVA sin

= 12400 sin(cos 0.65)= 1825 KVA

39. (b)

Open circuit emf f

Neglecting saturation

If

1

2

VV = 1

2

f 1

f 2

I fI f

V2 = 2

1

f 21

f 1

I fV

I f

= 2 25 3203.2 50

= 100V

40. (c)

Synchronous impedance = 1160

3200

= 3.35Resistance = 0.5

X = 2 23.35 0.5 = 3.31

41. (c) Control rods are meant for controlling the rate

of fission. these are made of Boron, Cadmiumetc.

Moderator are used to slow down the fastneutrons to a value that increases the probabilityof fission occurrence

42. (c)

When the line is open-circuited or lightly loadedthe receiving end voltage becomes higher than

the sending end voltage. This phenomenon iscalled Ferranti effect. This occurs due to theflow of capacitive or charging current.

43. (b)Flux linkages due to internal flux

= 0I

8

So, inductance per unit length

L = 0 i e a constant=I 8

. .

44. (c)Due to shunt capacitances to ground, currentleaks through it. Hence current through eachof the disc is not identical i.e. voltage acrosseach of the disc is different.

45. (d)As a 1 120 Positive sequence components,

120º

120º

120º1aI

1bI

1cI

l 1 1b a aI I aI1 120

l 1 1

2c a aI I a I1 240

Negative Sequence Components :

120º

120º

120º2aI

2bI

2cI

2 2 2

2b a aI I a I1 240

2 2 2c a aI I aI1 120

46. (c)The stored energy in the rotor = GH

IES M

ASTER

(8) Full Syllabus

=100 × 8

=800 MJ

47. (b)

Units generated per annum = Max. demand ×Load factor × Hours in a year

= (20 × 103) × 0.40 × 8760

= 7008 × 104 kWh

So, coal consumption per annum

= 4 60.9 7008 10 63 10 kg

48. (c)

Demand factor = Maximum demand 25

Connected load 45

= 0.555 i.e. 55.5%49. (d)

Sodium chloride NaCl Na+ + Cl– The compound NaCl is formed using

ionic bond. Silicon has 4 elements in outermost

orbit. Hence it forms covalent bond withother atoms.

Si Si

Si

Si

Copper (Cu) is a metal Argon is an inert element. Hence only

Van der Waals force is present.50. (a)

Dipole moment per unit volume

P = 2r 0( E – E ) ( )E – E)

P = 0 rE( –E)

51. (b)Usually, dielectric materials havepermanent dipoles. As temperatureincreases, the molecules in the dielectrichave more thermal energy and therefore,the amplitude of random motion is greater.This means that the molecules are lessclosely aligned with each other (even in thepresence of an electric field). Hence, thedielectric constant reduces.

52. (d)

A B

G F, G F, Given that G = 10dB F = 10dBThe overall noise figure (Feq) of two stageamplifier is given as

Feq = F 1F

G =

10 11010 = 10 + 0.9

Feq = 10.953. (c)

Resolution - It is the ability of TV system toclearly represent even a tiny structure ofthe object in the image. Higher theresolution, higher will be the picture quality.This depends on the number of framesscanned per second. Higher the no. of framesscanned per second, higher will be theresolution.

54. (d)For making interlacing easier, the numberof scanning per frame is chosen to be oddnumber.

55. (d)Given signal is

g(t) = 210cos cos50 t 100 t

or g(t) = 1+cos300 t10cos 50 t 2

= 5 cos50 t + cos50 t.cos300 t

= 55cos + cos + cos50 t 350 t 250 t2

1 = 150 f = 25

2 = 2350 f = 175

3 = 2250 f = 125 Now sampling frequency =2 × (Maximum frequency component)

= 2 × 175 = 350 samples/sec56. (d)

When transmission path is long, then signalgets attenuated and noisy.Repeater receives the signal, removes thenoise, amplify the signal and retransmit itto the channel.

57. (b)Given that minimum distance of hammingcode is dmin = m. Let it can detect errorsand correct ‘t’ errors.

IES M

ASTER


Then mind s +1 or m s +1 orm s – 1Again mind 2t +1 or m 2t +1

or t –m 12

58. (c)In a multistage amplifier,

(i) Common-emitter configuration is used forrealizing bulk of the gain of the multistageamplifier.

(ii) Common-collector configuration is required forproper impedance matching. This configurationis not used in the intermediate stages ratherthey are used in the last stage of the multistageamplifier.

59. (d)The circuit diagram of Wien bridge oscillator is shownbelow:

–+ V0

R1

R2

R

R

CC

The above circuit uses negative feedback throughR1 and positive feedback through a series RC anda parallel RC network.

60. (b)The given circuit compares the input voltage withthe reference voltage of 4V and input is applied atthe non-inverting terminal. Thus it is a non-invertingOp-Amp comparator.

8

T/2T12

+Vsat

–Vsat

v0

5T12

vi

t

t

4

0

Diodes D1 and D2 are provided in the circuit to

protect the op-amp against damage due to excessiveinput voltage. Because of these diodes, the differentialinput voltage, Vd, is clamped to either +0.7V or –0.7V, hence these diodes are called clamped diodes.

R1 is used to limit the current through the protectiondiodes D1 and D2 while resistance R is used toreduce the offset problem.

Duty cycle of output voltage

= ON

ON OFF

T100%

T T

=

5T T12 12 100%

T

= 33%

61. (c)Differential gain,

Ad,dB = 60 dB

20 log AAd = 60 dB

Ad = 103

Similarly, CMRRdB=80dB i.e. CMRR = 104

Vd = V1–V2 = 12 mV – 8mV = 4mV

VCM = 1 2V V2

= 10 mV

Output voltage,Vo =CM

d dd

V1A V 1CMRR V

=3

41 10 mV10 4m 1

4 mV10

= 4 + 0.001 m

= 4.00100 V

62. (b)

u1/s x2 x1 5 y

–3 –4x3

1/s

–22

1/s

2x

3x

1x

1x = 1 24x x

2x = 23x u

3x = 32x u

1

2

3

x Ax Bu

x 4 1 0 0x 0 3 0 X 1 Ux 0 0 2 1

A =

4 1 00 3 00 0 2

IES M

ASTER

(10) Full Syllabus

63. (d)Non linearity may be introduced by any of thefollowing given.

64. (c)

State transition matrix –1 –1t L (sI – A)

| sI – A | =

s 0 0 2 s –2sI A –

0 s –2 0 2 s

|(sI – A)| =2s –2

s 42 s

(sI – A)–1 = 2

s 2–2 sadj(sI – A)

sI – A s 4

(sI – A)–1 =2 2

2 2

s 2s 4 s 4

–2 ss 4 s 4

therefore

L–1 [(sI – A)–1]=cos2t sin2t– sin2t cos2t

65.(c)

66.(b)

Corner frequency w = 1 rad/sec, 20 rad/sec, 40rad/sec

• w1 = 1 rad.sec, change in slope is 0 – (–20) =20 db/dec so one zero exist at w1 = 1 rad/dec.

• Initially slope is –20 db/dec so one pole exist atorigin.

• At w2 = 20 rad/sec, change in slope is (20 – 0)= 20 db/dec, so one zero exist at w2 = 20.

• At w3 = 40 rad/sec, change in slope is (0 – 20)= –20 db/dec, so one pole exist at w3 = 40.

G(s) = 1 2

3

s sk 1 1w w

ss 1w

= sk 1 s 1

20ss 1

40

G(s) =

2k s 1 s 20

s s 40

for calculation of k

line with slope = –20 db/dec

y = M log w + 20 log k

at w = 1, y = –9 and M = – 20 db/dec

–9 = –20 log (1) + 20 log k

k = 0.35

0.7(s 1) (s 20)G(s)s(s 40)

67.(d)

G(s) H(s) =K (1 s)(1 – s)

given K < 0 – |K|

G(s) H(s) =–|K| (1+ s)

(1 – s)

G(s) H(s) =(1+ s) |K|

(s – 1)

Put s j and H(s) = 1

G( j ) H( j ) =(1 j ) | K |

( j – 1)

2

2

1 KGj( )1

By putting different values of , we get

w = 0 1 180w = 0.01 1 –178.85w = 0.1 1 –168.57w = 1 1 –90w = 10 1 –11.521w = 100 1 –1.145w = 1000 1 –0.1145w = 1 0

G (j ) G(j )

w = 0 w = U

jVGH plane

68.(c)Given Open-loop transfer function

So characteristic equation is

IES M

ASTER


1 + G(s) H(s) = 0

G(s) = 2K (s – 2)(s 1)

given H(s) = 1

2

k (s – 2)1(s 1)

= 0

(s + 1)2 + k(s – 2) = 0

s2 + 2s + 1 + ks – 2k = 0

s2 + s(k + 2) + 1 – 2k = 0

The routh’s tabulation is

1 1–2kk+2 0

1–2k

s2

s1

s0

For stability K + 2 > 0

k > –2 ...(1)

1 – 2k > 0 1k2

...(2)

so from (1) and (2)

For –2 < k < 12

system is stable

69. (b)Derivative control will increase the damping ofthe the system and thereby transientperformance is improved.Intergral control changes the system order asit adds a pole to the system at origin andthereby reduction in steady state error. Henceit improves the system performance of steadystate.It changes the order of the system from secondorder, to third order as a pole at origin is addedHence statement 1, 3 is true and 2 is false.

70. (a)The effect of stray magnetic field in AC bridgescan be reduced by shielding the bridges with theuse of high permeable material around the bridge.High permeable material provide low reluctancefor stray magnetic field and hence screened themagnetic field to affect the bridge.Wagner earthing device is used in the ac bridgesto eliminate the effect of stray capacitance.

71. (a)

XCXL

R

R/Xe/XL

Z

30 4020Hz r

0

60Hz

72. (c)V = IR R

= 200 0.02 30

= 4 30

L LV 4 30I I

j L 200 0.5 j

=4 30100 j

= 0.04 60

C C 6

V 4 30I I1 10j C 200 50j

= 0.04 120 I = IR + IC + IL

= 0.02 30 0.04 60 0.04 120

I = 0.0230°

73. (a)

60 V 2

i (t) 2

i (t)L

At t = , inductor behaves as short circuit

Req = 1

i(t) =601

= 60 AA

IES M

ASTER

(12) Full Syllabus

Hence IL(t) =60 2 30 A4

For finding time constant, we short circuit the voltagesource. Hence circuit becomes like

2 H2

2

=L 2 2secR 1

74. (b)

BW =fr 150 3KHzQ 50

fUpper = BW 3fr 1502 2

= 151.5 KHz

fLower = BW 3fr 1502 2

= 148.5 KHz

75. (b)

For the positive half-cycle of the input, diode D1is ON and D2 is OFF. Capacitor C1 is charged toVm.

v = Vi m sin( t)

–

+

V0

~ C1

C2

D1

D2

Vm

Vm

+ –

+ –

For negative half-cycle of the input, diode D1 isOFF and D2 is ON. Capacitor C2 is charged toVm with polarity as shown in figure above.

Output voltage, V0 = c1 c2V V

= m mV V = 2 mV

76. (d)Ic = collector current

Ie = emitter current

Ib = base current

Apply KVL in the emitter-base loop, we get

1 e e EB b bV I R V I R 0

1 EB c b e b bV V I I R I R 0

c e b e b 1 EBI R I R R V V

Differentiating above equation w.r.t. Ib, we get

ce e b

b

IR R R 0

I

b e

c e b

I RI R R

Stability factor, S c

bco

c

I 1II 1I

S =e

e b

1R1

R R

77. (d)

78. (c)The point where asymptotes of a root locus meetis called centroid hence statement-I is false

Root locus always starts from open loop pole (k=0)and end at open loop zero (k = ). Hencestatement-II is true.

79. (b)

Hall effect occurs when a transverse magneticfield is applied to a conductor carrying current.Due to the current, a longitudinal electric fieldnormally present in the direction of flow of current.

80. (a)

• Superconductors are used for generating verystrong magnetic field.

If a superconductor caries a current I, thendeveloped magnetic field

H =I

2 rwhere r is the radius of superconductor wire.• Magnetic bubble memories are built using

ferrites.

IES M

ASTER


81. (c)

ABCABC

ABC

ABC

ABC

ABC

ABC

A B

C

Let shaded area = Y

Y = ABC ABC ABC ABC= m 3,5,6,7

1

11

ABC

A

BC BC BC BC

A 1

0 1 3 2

4 5 7 6

Y = AB BC CA

82. (d)

A

B

C

Y

AB

BC

x

X = AB BC

= AB.BC AB.BC ABC

Then, Y = ABC A C

= A C1 BC

= A C

83. (c)

• When switch is open, corresponding input willbe high.

• When switch is closed, corresponding inputwill be low.

For option (c)

1 2S S A

1 1 0

Then,S1S2 = 1.1 = 1

S2.A = 1.0 = 0

Y = 1 2 2 1 0 1 0S S S A

84. (d)

85. (d)

For multiplexer,

Y = 0 1B.I BI

= B B.A1

Hence, X = 0B AB

= .0B AB

= .1B.AB

= B. A B

= BA86. (a)

Basic causes for non-sinusoidal nature of no loadcurrent of transformer are:

1.Hysteresis of core material

2.Saturation / non-linearity of core material

87. (d)

In a moving iron instrument 2I

1

2

=2

1

2

II

= 22mA

1mA

= 4

2 = 14

=200V

4 = 50V

88. (a)h – parameter equations :V1 = h11I1 + h12V2I2 = h21I1 + h22V2h-parameter model :

1I 2I

2V1 21I h12 2h V

11h

1V

+

–

+

–

22

1h

IES M

ASTER

(14) Full Syllabus

Compare the given circuit to h-paramer model,we get

11h = b e 12r r , h 1

h21 = cb 22e d

1, hr r

89. (b)Output voltage

V0 = m Sg V V600Now, apply voltage division rule in input loop,

V = in1000 V

1000 500

V = s2 V3

From (1)

V0 = m2g 6003

V0 = m s400g V

or, 0

S

VV = –400 gm

or, 0

S

VV = –400 × 0.1

= – 40

90. (c)For stability of negative feedback closed loop controlsystem output linearly varies with input, whenany non-linearity should not be introduced by thecomponent of control system.

91. (a)

× × G1G2

H

R(s)++–

+N(s)

C(s)

In a feedback control system, if subjected to noiseN(s), then R(s) = 0

× G1G2

–H

++

N(s)

C(s)

Transfer function

C sN s =

1

1 2

G1 G G H

92. (c)

1010 C1R1 × 10R2 –+

C2

1

1

C 10 10 100R

2

2

C 10 10R 1 10 11

If the forward path gain is reduced by 10% ineach of the system then

90 C1R1 × 9R2–

+C2

1

1

C ´ 90R

2

2

C 9 9´R 1 9 10

Variation in C1 Variation in C2

1 1

1 1

1

1

C CR RVariation 100C

R

90 100 100 10%100

2 2

2 2

2

2

C CR RVariation 100C

R

9 1010 11 100 1%10

11

93. (c)To measure the higher voltage, resistance to beconnected in series with the moving ironinstrument.

750V

IfsI =50mAfs R

R =10m

750V = fs m fsI R I R

750 = 3 350 10 10 50 10 R

R = 3

750 0.5 15k50 10

94. (c)

I(s) = 2s

(s 2) = 2s 2 2(s 2)

= 2 2(s 2) 2(s 2) (s 2)

= 21 2

(s 2) (s 2)

Taking Inverse Laplace transform, we get,

IES M

ASTER


i(t) = e–2t – 2t.e–2t

= 2te (1 2t)

95. (b)

For the above circuit

i = Rt/Lm mI sin t I sin e ;

where, 1 LtanR

When, 0 or, ie. when voltage applied ispasses through its maximum, there will be notransient.

96. (c)Off the total 70V, 10V is across R2, so 60V isacross R1. 30V is across 60 resistor so, current

= 3060 = 0.5A

R = VI = 60

0.5 = 120

97. (c)A capacitor does not resist any abrupt change inthe current flow. It resists abrupt change involtage.

98.(d)

Quality factor,

Q =resonant frequency

Bandwidth

=0ff

=

6

320 10

100 10

= 200

99. (a)When in series, we have to add the Z-parametersto combine the two, two network.

100. (a)L-C network has only energy storage element.Power dissipation elements ‘R’ is not there, so thepoles and zeros lie on the imaginary axis. Networkfunction should be determined by zero initialcondition.

101. (d)

For n-bit Analog to Digital conversion,

1. Successive Approximation Register (SAR) ADCrequires ‘n’ clock cycles.

2. Dual slope integrating ADC requires

approximately n 12 clock cycles.

3. Counter method ADC requires (2n – 1) clockcycles.

4. Simultaneous ADC requires only 1 clock cycle.

Note: Counter type ADC is also known as Ramp

type ADC

Dual slope integrating type ADC is the slowestamong all ADCs. It is the most accurate ADCand thus, is used in Digital voltmeters.

The number of clock cycles (given above) forvarious ADCs is the maximum conversion timerequired.

102. (a)V(s) = Z s .I s

=

2 s 1.s s 2 s 3

si t 2u t ; I 2 / s Initial Value :

t 0LimV t =

sLims.V s

=

s

2 s 1Lims. 0s s 2 s 3

Steady State Value :

tLimV t =

s 0Lims.V s

=

s 0

2 s 1 2 1Lims.s s 2 s 3 6 3

103. (b)

A potentiometer is used as a error detection device.Its input is the position error and potentiometerconvert it to corrersponding voltage.

104. (c)BW of colour TV transmission = 6 MHz

105. (c)

t s 0Limy LimsGt s

and the input is given at instant t = 2

= 2ss 0LimsG es

IES M

ASTER

(16) Full Syllabus

= 2s

2s 0

1Lims es s 2s 2

=12 = 0.5

106. (a)XRA A instruction execution resets contents ofaccumulator and hence sets zero flag.

107. (c)

x1

e f g

a x2 b x3 c x4 d x5

L1 L2 L3

Applying Mason’s gain formula

5

1

xx =

n

K KK 1

P

Forwards path :

P1 = abcd, 1 1

Loops : L1 = be, L2 = cf, L3 = dg

Non-touching Loop :

L1L3 = bedg

5

1

xx =

1 1

1 31 2 3

P1 L LL L L

5

1

xx =

abcd1 bedgbc cf dg

108.(b)R1 = 500 ± 1% R2 = 1000 ± 1%

R = 1 2R R (Parallel combination equivalentresistance)

Nominal value of R = 500 1000 = 1000

3

Tolerance (or limiting error) for R1 and R2respectively

1

1

R1% 0.01

R

1 1R R 0.01

2

2

R1% 0.01

R

2 2R R 0.01

Hence R 1 = ± 5

R 2 = ±10

Now1R

= 1 2

1 1R R

2

1

R 1R R =

2

2 211 2

R1 1RR R

2

RR

=

1 22 2

2

R RR R

Hence, the limiting error in the resultant

R =

2 2

1 21 2

R RR RR R

= 4 15 109 9

= 2.22 + 1.11 R = 3.33

Tolerance =RR

= 3333 9.99 101000

3

= 0.99% ±1%109. (d)

i. The sacle is uniformly divided

ii. The power consumption is very low (as 25 W

to 200 W )

iii. The torque to weight ratio is high which giveshigh accuracy

iv. Since the operating forces are large on accountof large flux densities which may be as high0.5 Wb/m2 the errors due to stray magneticfields are small.

110. (b)Common base amplifier has excellent high-frequency. Thus, common base amplifiers haveinvariably high cut-off frequency.

111. (b)

Conduction current density, CJ E

Displacement current density, dDJt

D E

dJ

=E j Et

IES M

ASTER


If CJ

= dJ

Then, =

=

f =02 3

=

2

8

1012 3 10

360

f = 60 MHz

112. (c)

The material required for resistance thermometershould possesses :

(i) high temperature resistance coefficient sothat a small change in temperature producesa large change in resistance.

(ii) high resistivity so that minimum volume ofmaterial is used for the construction ofthermometer.

113. (c)

114. (c)

Td = n(IPd)FF

Total propagation delay, Td = 4 × 20 ns

= 80 ns

So, counting time T 80ns

So frequency of counting, 1fT

i.e. counting speed,

f 91 1T 80 10

61000 1080

12.5 MHzSo, maximum counting speed in the given option= 10 MHz.

115. (b)

67 H 0110 0111=81H 1000 0001=

1110 1000

Flag = S Z AC P CY

(because result is positive)

So sign, S = 0

Zero, Z = 0

AC, AC = 0

CY, CY = 0

116. (a)

To reset all the flip-flops

Q1 = 1, Q2 = 0, Q3 = 1

(101)2 = (5)10

The given circuit is a Mod-5 counter becauseafter the counter state is 101, the counter resetsand starts counting again from 000.

117. (c)Power transformers use silicon steel as amagnetic material and alnico is a permanentmagnet.

118. (a)

In paramagnetic materials interaction betweenneighbouring dipoles is negligible.

119. (a)

120. (b)

121. (c)Depletion region is created due to diffusion ofmajority charge carriers across junction so itdoes not contain any free charge (electron orhole). It only contains ions, negative ions on p-side and positive ions on n-side.

122. (a)

1mA

5V

CRCV 2V50k

5V

Given,IE = 1 mA

IC = E EI I1

=75 1mA

75 1

IES M

ASTER

(18) Full Syllabus

IC = 0.987mA

123. (c)LED will glow only when output from OR willbe ‘0’ and it is possible only when both inputof OR gate will be ‘0’ i.e. output of NOT gateand EXNOR gate should be zero. It is possibleonly when S1 is closed and S2 is open.

0

S = open i.e. 12

S = closed i.e. 01

00

1

124. (d)Number of chips required

=

Required sizeAvailable size

=32K 81024 8

=32 1024 8

1024 8

= 32

125. (c)

CLK Q– 0 Q2 Q1 Q0

0 – 0 1 0

1 1 1 0 1

2 0 0 1 0

So for 2 clock pulses, the values of Q2Q1Q0

resets, so frequency = 18KHz2 = 9KHz

126. (c)For 4-flip flops the count is 24 = 16, So the no.of unused or missed states are

16 – 10 = 6 counts.127. (b)

We know

X(z) =n

nx[n] z

=n 3 3

n

15

u[n-3] z–n–3 + 3

3 (n 3)

3 (n 3)

n

1 1z u[n 3] z5 5

n–3 = n'

n'–3

–n'

n'=–

z 1× u[n'] z125 5

Since u[n'] = 0 n 0

X(z) =

3

1

z 1 1, | z |1125 51 z5

So X(z) =n '3

n '

0

z 1 z125 5

= 0 1 23z 1 1 1 ...

125 5z 5z 5z

= 23z 1 11 ...

125 5z 5z

X(z) = 3z 1 1provided 1

1125 5z15z

128. (c)

0s 6 j8 s 6 j8

2 28 0s 6

2 2s 2 6 s 10 0

Comparing with 2 2n ns 2 s

n = 10

= 2 62 10

= 0.6

129. (a)

130. (b)In the dual trace C.R.O., two inputs can bedisplayed simultaneously as there are two seperatevertical input channels and they use seperateattenuator and preamplifier.Aquadag is an aquous solution of graphite used tocollect the secondary electrons produced duringbombarding of electrons on screen.

131. (c)Bifilar winding is most commonly used method ofwinding arrangement to make noninductive coil.

IES M

ASTER


This method is used to reduce the inductive effectof coil. In this method, two wires wound side byside carrying current in opposite direction producetwo equal and opposite magnetic fields and sincetwo wires are very close to each other, the netmagnetic field is almost zero. Therefore, the netinductance of the coil is almost negligible.

132. (a)“Guard circuit” is used in the high resistancemeasurement to eliminate the error due to leakagecurrent over high resistance. “Guard Circuit” isconnected in parallel with the measuring ammeter.So, the leakage current passes through the GuardCircuit and hence does not affect the reading ofammeter.

133. (d)Forbidden energy gap of Si and Ge are given as1.12 eV and 0.72 eV and as we know

2in = gE /KT3

0A T e

ni gE /KTehigher the band gap lower the intrinsicconcentration, which results into low conductivity

= i nn q niTherefore, Si has higher forbidden energy gapbut low conductivity.

134. (d)In case the moving coil instruemnt is used as avoltmeter a large series resistance of negligibletemperature coefficient (made of material likemanganin) is used. This eliminates the errordue to temperature. This is because the coppercoil forms a very small fraction of the totalresistance of the instrument circuit and thusany change in its resistance has negligible effecton the total resistance.

135. (c)The phasor diagram of a wattmeter for a laggingpower factor is given below

V

IP

I

Here,V Voltage applied to pressure coil circuitIP Current in pressure coil circuitI Current in current coil circuit Angle by which IP lags V

Angle by which I lags V(load angle)

As the know,True power

Actual wattmeter reading

=cos

cos cos( )

× Actual wattmeter reading

[For lagging loads]

136. (c)

Clipper circuit is used to remove the certainportion of a waveform. A clipper can be made ofresistor and diode e.g. half wave diode can beused to remove one half (positive or negative) ofsinusoidal waveform.

137. (d)

Sequential circuits :• It gives outputs depending on the present input

and past input also.• It contains atleast one feedback path.• It contains some memory.• It exhibit cyclic nature i.e. its output sequence

repeats after some clock cycle.

138. (d)

With increase in temperature mobility of chargecarriers decreases as vibration of atom’s inmolecule increases but at the same time carrierconcentration increases.

2in = gE /KT3

0A T e

in increases with temperature increase.Now conductivity of intrinsic semiconductor .

= i nn qAs with temperature intrinsic concentration niincreases and mobility decreases, but increasein carrier concentration is much more thandecrease in carrier mobility therefore conductivityeffectively increases. Hence resistivity decreases.

IES M

ASTER

(20) Full Syllabus

139.(b)Shunt enhancement allows the ammeter to measure

currents higher than the meter current.

I Im

Rsh

I sh

Rm

A

I is measured, which is larger than the full scalemeter currents Im.The voltage drop across the meter for full scaledeflection always remains ImRm.

140. (a)

For 3 induction motor,,

rotor frequency fr = sfs

Also, core losses depend on frequency asPc f + f2

As the slip of motor is very low, the rotorfrequency will be very low. Hence, core lossesin rotor of 3 induction motor is taken asnegligible. Core losses occur only in stator.

141.(b)Power factor in synchronous machine isimproved by varying the excitation of rotor.A large gap decrease the power factor, butit is kept to increase stability of the machine.

142. (a)

For a series motor, aI

At no-load, Ia is very low

As, speed, b b

a

E ENI

As Ia is very low at no-load, a series motorattains dangerously high speed. To avoid this,we never run dc series motor at no-load.

As swineburne’s test is a no-load test, it cannotbe done on dc series motor.

143. (a)For medium and large rotating machines, if weuse rotating armature, large currents have tobe flown through slip rings. This would makethe operation unreliable.

144. (a)

For a required power, if the voltage is increased,the current would decrease and I2R losseswould decrease and the efficiency increases.

145. (a)

Zero-sequence currents can never flow in thelines connected to a delta-connected windingas no return path is available for zero-sequencecurents. Zero-sequence currents can, however,flow through the delta connected windingthemselves if any zero sequence voltages areinduced in delta.

a

b

c

I = 0a0

Iab0Ica0

I =0b0

I =0c0

Ibco

146. (a)The transient-stability can be improved in bestway by using high speed breaker because thecircuit breaker clears the fault quicker.

147. (a)Because of fixed air pressure, the air blastcircuit breaker exerts the same extinguishingforce irrespective of the magnitude of currentto be interruped which leads to “current-chopping” phenomenon.

148. (c)A shunt resistor is generally connected acrossthe contacts of a circuit breaker in order todamp out the restriking transient. If resistoris not connected across the circuit breakercontacts, a high voltage may appear acrossthe contacts of circuit breaker after arc-quenching.

149. (c)

If a line is terminated by its characteristicimpedance Zc, then the reflected wave of eithervoltage or, current is zero.

So, this line is called “infinite line” or “flat line”.

150. (a)

IES M

ASTER


For a long transmission line,

V(x) = x xR R C R R C

Incident Voltage Reflected Voltage

V I Z V I Z.e .e

2 2

While moving from receiving end to sending end,

the term xR R CV I Z.e

2

increases. Conversely,,

we can say the given term diminishes if we movefrom sending end to receiving term. Hence, thisterm is called “incident voltage”.

ANSWERS - IES MasterQ S 1 R a r Surface of sphere For Gaussian surface S 1 Q 0 enc E.ds 0 Q enclosed...

Documents

Transcript of ANSWERS - IES MasterQ S 1 R a r Surface of sphere For Gaussian surface S 1 Q 0 enc E.ds 0 Q enclosed...