1. (c)

2. (b)

3. (c)

4. (c)

5. (c)

6. (c)

7. (a)

8. (a)

9. (a)

10. (b)

11. (c)

12. (c)

13. (c)

14. (c)

15. (b)

16. (b)

17. (b)

18. (b)

19. (b)

20. (d)

21. (b)

22. (d)

23. (b)

24. (c)

25. (c)

26. (a)

27. (a)

28. (b)

29. (a)

30. (b)

31. (d)

32. (d)

33. (b)

34. (c)

35. (b)

36. (a)

37. (a)

38. (c)

39. (c)

40. (d)

41. (b)

42. (b)

43. (d)

44. (b)

45. (b)

46. (c)

47. (c)

48. (a)

49. (c)

50. (a)

51. (a)

52. (c)

53. (c)

54. (c)

55. (b)

56. (d)

57. (c)

58. (d)

59. (d)

60. (d)

61. (a)

62. (a)

63. (c)

64. (c)

65. (b)

66. (a)

67. (d)

68. (d)

69. (a)

70. (d)

71. (b)

72. (b)

73. (d)

74. (c)

75. (c)

76. (a)

77. (d)

78. (a)

79. (b)

80. (a)

81. (b)

82. (b)

83. (b)

84. (b)

85. (b)

86. (b)

87. (b)

88. (b)

89. (d)

90. (d)

91. (a)

92. (d)

93. (d)

94. (b)

95. (a)

96. (c)

97. (c)

98. (c)

99. (d)

100. (b)

101. (b)

102. (b)

103. (d)

104. (b)

105. (d)

106. (b)

107. (c)

108. (b)

109. (b)

110. (a)

111. (b)

112. (b)

113. (a)

114. (d)

115. (b)

116. (a)

117. (b)

118. (a)

119. (c)

120. (b)

Objective Question Practice ProgrameDate: 7th May, 2016

ANSWERS

IES M

ASTER

(2)

1. (c)Rotameter is a device that measures the flowrate of fluid in a closed tube. It measures thefluid flow rate by allowing the fluid to flowthrough a varying cross-sectional area tube.

2. (b)The core of the LVDT is made of highpermeability, nickel iron, which makes LVDTa high sensitive instrument.

3. (c)The radiations from a heated body at hightemperatures fall within the visible region ofthe electromagnetic specturm. Within thevisible region a given wavelength has a fixedcolour and the energy of radiation is interpretedas intensity or brightness. Therefore, if wemeasure brighness of the light of a given colouremitted by a hot source, we can have anindication of temperature.

4. (c)Active transducers are those which do notrequire an auxiliary power source to producetheir output. They develop their own voltageor, current output.eg. Thermocouple, piezoelectric pickup,photovoltaic.

5. (c)

Given, Strain = l l= 2×10–6

R = 120and, R = 60Since, Gauge factor,

Gf =l lR R

=

6

6120 10

60 2 10= 1

6. (c)

7. (a)

8. (a)• Diaphragm and Bourdon tube are used for

measurement of pressure.

• Orifice plate and venturimeter are use formeasurement of flow.

9. (a)

10. (b)Speed of the shaft = 900 rpmi.e. one minute 900 rotation 900 × 60 teeth 900 × 60 pulses

So, in one second 900 60

60

pulses

= 900 pulsesSo, frequency of output pulses = 900 persecond.

11. (c)When the core is displaced towards the positivedirection i.e. towards the top secondary coil,the voltage induced in the top secondary coilincreases and that of bottom secondary coildecreases and vice-versa.Given, the magnitude of voltage is 5.2 V fordisplacement of 0.5 inch.So, for a –0.25 inch displacement, the output

voltage will be 5.22 = 2.6V but of negative

direction.So, output voltage = –2.6V

12. (c)The synchros and circular potentimeters areused to measure angular position.But LVDT is used to measure lineardisplacement.

13. (c)Gauge factor,

Gf = 1 2v

If piezoresitive effect is neglected

= 0

then, Gf = 1 + 2v= 1 + 2 × 1.5= 4

IES M

ASTER

(3)14. (c)

fr =1

2 LC

I =L L

V VX 2 f L

L rHere, f f

I =1/2CV

L

15. (b)

PowerPhase = 690W 230W

3

PhI = 230W230V

= 1AIL = PhI 1A

16. (b)Let R be the resistance between each phase

RT = R 2R 2 RR 2R 3

Given RT = 12

R = 18

17. (b)

• 01LC

when C increases, 0 will decrease.

Bandwidth= R thus constantL

18. (b)A resonance XL = XC

Current = VR

= 1045

= 0.22

19. (b)

90

–90

X >XC L

X >XL C

0

As from the figure, the current is leading, so

C LX X and frequency is less than 0 .

20. (d)

BW = 2 1 2 456 434

= 2 22

0 = 02 f Q BW

= 20 2 2

f0 = 20 2 222

= 440 Hz

21. (b)

Q = 1 LR C

Q' = 1 2L2R 2C

= 1 L2R C

= Q2

Q' = 1002 = 50

22. (d)at 0 the circuit is

Rs

Vs RLV0

V0 = L

L S

RR R

at the circuit is

Rs

Vs RLV0

V0 = L

S L

RR R

at resonant frequency the circuit is

IES M

ASTER

(4)

Rs

Vs RLV0

V0 = 0This is band reject filter.

23. (b)Voltage across capacitor

= Q × supply voltage

Q = rf 2000KHz 10BW 20KHz

VC = 10 × 10= 100V

24. (c)

Since, Load factor = Average load

Maximum loadSo, Average load = 0.5555 × 15 KWand average daily energy consumption of theconsumer

= 0.5555 15 24 200 KWh

25. (c)For economic allocation of load, incrementalfuel cost of each unit should be equal.

So, 1IC = 2CI

140 0.2P = 225 0.3P

1 20.2P 03P = –15 ....(1)

and total load, P1 + P2 = 200 ...(2)Solving equation (1) and (2), we get

P1 = 90 MWP2 = 200 – 90 = 110MW

26. (a)

Penalty factor of the plant = L

i

1P1P

So, if L

i

PP

increases, then L

i

P1P

decreases

and hence its reciprocal i.e. penalty factor willincreases.

27. (a)The transmission loss for a two-power plantsystem is given as,

PL = Gm mn Gnm mP B P

where m = n, and is equal to number ofgenerating units or, plants.

PL = 2 21 11 1 2 12 1 2 21 2 22P B P P B P P B P B

= 2 21 11 1 2 12 2 22P B 2P P B P B

(since, B12 = B21)

28. (b)Synchronous condenser is basically an over-excited Synchronous motor which is used forpower factor improvement equipment.

29. (a)The capital cost of the power station dependsupon the capacity of the power station. Lowerthe maximum demand of the power station,the lower is the capacity required and thereforelower is the capital cost of the plant. With agiven number of consumers the higher thediversity factor loads, the smaller will be thecapacity of the plant required and consequentlythe fixed charges due to capital investmentwill be much reduced.

30. (b)The principle of incremental cost is employedto decide the economical operation of generators.It decides which unit should be operated atwhat load. It is not useful in deciding thesequence of addition of units and total plantcapacity to be operated.

31. (d)

32. (d)– By use of load curve, we know the

maximum demand and hence, totalinstalled capacity can be decided.

– According to load curve, we can alsodecide the size of the units to be installedand also in preparing the schedule ofoperation of the generating units.

33. (b)Power loss,

PL = 2 21 11 1 2 12 2 22P B 2P P B P B

= (50)2 × (0.002) + 2× 50 × 60

IES M

ASTER

(5)× (–0.0005) + (60)2 × (0.001)

= 5 – 3 + 3.6= 5.6 mW

34. (c)

Irating =

ON0

TIT

=

0 ONV E TR T

=

sV ER

rdId =

s2 V ER = 0

=s

E2V

r maxI =

s

s

s

EV E2V E

R 2V

=2

s

E2V R

= 50 50

4 100 10

= 58 = 0.625 A

35. (b)

Type Voltage CurrentType ATypeBTypeC ,TypeD ,TypeE , ,

36. (a)

Modulation index = r

c

VV

= 100150 = 2

3 = 0.66

cfN 12f if zero of triangle wave coincides

with zero of reference.

N = 1500 1500 = 2

37. (a)Expn.

T = 1f

Duty cycle = ONTT

= onT f

= 2ms × 200= 0.4

38. (c)

half

full

PP = 1

4

Pfull = 4 × Phalf

= 4 × 20= 80 W

39. (c)Vs = 50V – 2V = 48V

Maximum rms output voltage at fundamentalfrequency,

V0 =

s4V2

=

4 482

= 43.2 V

40. (d)

Per unit ripple

0.5 Duty ratio

The graph of a typical ripple versus duty ratio.

41. (b)In a PWM method of chopper switching, thefrequency is kept constant and TON is varied.Accordingly TOFF is also varied.

IES M

ASTER

(6)42. (b)

The maximum ripple occurs at D = 0.5, and isgiven by

mxI = sV4fL

= 3100

4 1 10 0.1 = 0.25 A

43. (d)In a voltage source inverter, voltage is fixedand the output voltage wave form depends onlyon the switching sequence. Current waveformdepends on the load impedance.

44. (b)DC chopper can be controlled by two methods,one is fixed frequency and varying ON andOFF time, other is variable frequency.

45. (b)Multiple PWM technique is used to reduceharmonic content as compared to single pulse.

46. (c)Given output power = 24W and efficiency = 0.8

Input power = 24 30W0.8

Output voltage = 240V and 0.95

Input voltage = 240 (1–0.95)= 12 V

Input current = 30 2.5A12

47. (c)

I = ripple current

= dc battery ONV V TL

= 3

30.3 1 1050 10

10 10

=40 0.3 1.2A

10

48. (a)Using the norton’s theorem

25V+– 3 5

+–24 10 50VVa

i1

a a aV V 50 V 25 03 15 6

Va = 13.88

i1 = a25 V 1.85A6

49. (c)

SCi 5A [Given], we can find Rth

8 4 4

8 4Rth

Rth = 8 8 4

Norton equivalent = SC thI R & 4

Max power = 2 thSC RI 25 4 25W4 4

50. (a)By using thevenin theorem at left and rightside, and by replacing with therenin equivalent.

+–+–

1 2 2 2

4 12VV1

+

–5V

1 11 V 5 V 12V 04 3 4

V1 = 5.6 V

51. (a)

1 1 1 11 2V 0.5V I 2 aII I

1 1 2V I 4I6 2a

2 11 2V 2 aII I

2 a 2V I 2I2 a

IES M

ASTER

(7)Reciprocal if

12Z = 21Z4 = 2 + aa = 2

52. (c)If only one network is taken

P =

2th

th

V RR R& if two networks are taken

taken P' =

2th

th

VRR R

2

=2

th

th

V4 R2R R

comparing P and P', P' would be in between Pand 4P.

53. (c)

By applying KVL in various loops, we can getvoltage V1

A B

C D50V

+10V

–+

40V–

+ 70 –

+ 30 –

Applying KVL in ABCD Loop,V1 + 100 – 10 V – 70 V – 10 + 50 = 0

V1 = –60 V

54. (c)

By applying cut principle

4A

1

5A

1A

2

5 cut

cut

principle states that the current passing through this should be zero.

V = 5 1A

= –5V

55. (b)

612 3

5A

Va Vb

a a a bV V 30 V V

30 12 6 = 0

a b17V 10V = 150 ....(1)

and,

b a bV V V 120 56 3 = 0

a bV 3V = 210 ....(2)

From (1) and (2)Va = 62.2 V, Vb = 90.7 VV1 = 30 – 62.2 V

= –32.2 V

56. (d)Millman’s theorm gives either equivalentvoltage or current source on simplification.

57. (c)Ferroelectric materials exhibits thephenomenon of hysteresis.

58. (d)

0.52.5 =

1 X5 Y where X is 1 mA and Y is 10 mA

The meter with higher current reading will haveless resistance, so ratio of internal resitance is 5:1.

59. (d)The emf of a Weston standard cell can bemeasured by a potentiometer only becauseat balance the current drawn from the cellis zero.

60. (d)

90

0.9 I

0.1 II

x

0.9I x = 0.1I 90

x = 909

= 10

IES M

ASTER

(8)61. (a)

By using thevenin theorem at left and rightside, and by replacing with therenin equivalent.

+–+–

1 2 2 2

4 12VV1

+

–5V

1 11 V 5 V 12V 04 3 4

V1 = 5.6 V

62. (a)

Now given that l l0n and the volume remainssame

l l0 0 0a an or 0a a n

0

RR =

ll

0 2

0

a n n na

R = n2R0

63. (c)Measurement of capacitance is done by(i) De-Sauty’s Bridge, and(ii) Schering bridge

64. (c)

65. (b)Hay’s bridge is used for Q > 10

As, Q = 0LR

and, time constant, LR

So, for high time constant coil, Hay’s bridge isused.

66. (a)Since, both the input are same. So there is nophase displacement between them and hence astraight line will appear on screen of CRO.

67. (d)

R=1M

Vn

eq

1R =

m

1 1 1R R R mas, R

Req = R = 1M

So, voltage reading, Vn = eqI R

= 650 1 10

= 650 1 10

= 50V

68. (d)

Resolution = nVoltage range

10where, n is number of digit of display

So, resolution = 31 1mV

10

69. (a)

70. (d)

In 132 digital voltmeter, there are three active

digits and one digit for over-ranging.So, in case of over-ranging it will display as

1 9 9 9

Over-rangingActive digits

71. (b)

T(s) =

G s1 G s

=

2

2K s 1s 3s1 K 2 K

All the coefficients of characteristic equationshould be non zero and positive.

1 K 0 K 1

2 K 0 K 2

Common to both is K > –1.

72. (b)Given characteristic equation

3 2s 4s s 6 0

by applying routh’s method,

IES M

ASTER

(9)

3

2

1

0

s 1 1s 4 6

4 6s 2.5 04

s 6

There is two times sign change, so there aretwo roots in right of s-plane. The total numberof roots being 3, the number of roots to left are(3 – 2) = 1.

73. (d)

Angle of asymptote is 2K 1Poles zeros

= 2K 1 2K 14 1 3

K = 0 Angle = 3

K = 1 Angle =

K = 2 Angle = 53

74. (c)

Percentage overshoot = 2/ 1pM e

If Mp is same then is also same. = cos constant

= constant75. (c)

The characteristic equation is2Ks s 6 0

2 1 6s s 0K K

Comparing with 2 2n nns 2 s 0

n12K

and 2n

6K

62 0.5K

= 1K

K = 16

76. (a)Nyquist stability test can give inputs on thestability and how to stabilize.

77. (d)Equation for constant-N circle is

2 2 2N 11 1x y 4N2 2N

Here radius =2N 1

2N

Putting N = 1

=21 1

2 1

= 22

=12

78. (a)When Nyquist plot of G(s)H(s) passes through 1, j0 then it implies that

gain cross over frequencyG Hs s 180 Phase margin = 180° – 180° = 0°

79. (b)

Open loop transfer function = 10

s s 2Putting s j ,

G j = 10

j j 2

G j =

1

1 1

tan 0

tan tan2

= 1 1tan tan2

= 1tan2 2

0 G j 2

0 G j

PM = to2

= 0 to2

80. (a)Speed of rotor flux = ns (at stand still)and speed of rotor flux at nr is sns. Then w.r.t.stator speed of rotor field is

s s shr sh n sh n1 s

IES M

ASTER

(10)81. (b)

Nsa = 3000 r.p.m.Na = 2500 r.p.m.

sFL = 16

With V/f control

Nsb = 303000 180050

New Speed =11800 1 0.66

= 1620 r.p.m.

82. (b)

Rotor power factor = 2´2 ´

2

RcosZ

If slip increases, then rotor frequency increases.This will increase ´

2Z and since R2 is constant,p.f. will decrease.

83. (b)Mechanical power developed = (1 – s) × power

input to rotor= 10001 0.02

= 980 W

84. (b)Airgap input power = Pair

Rotor loss= s × PairPower to rotor = air air airP sP P1 s

h = efficiency = air

air

P1 s 1 sP

This efficiency is without considering otherlosses, so the efficiency would be less than(1–s) if frictional and other losses are considered.

85. (b)Plug setting multiplier,

PSM = / CTratioPr imary fault current

Secondary relay current

= 200400 5 15

= 2000 5400

86. (b)Expression for the restriking voltage

V = maxt1 cosVLC

So, rate of rise of restriking voltage

RRRV = maxVdV tsindt LC LC

RRRV will be maximum when tsin 1LC

or

t LC2

.

Given, 0 = 41 10LC

t = 41LC

2 2 10

= 41 sec.

2 10Average RRRV

= Peak Restriking Voltage

Time to reach peak restriking voltage

=6

4

100kV1 10 sec

2 10

= 2200 10 kV sec.

= 2KV sec

87. (b)Due to doubling effect, the rms value ofbreaking current=1.8 × (symmetrical breaking capacity)But as making current is taken in maximumvalue or, peak value. So, the making capacity

= 2 1.8 symmetrical breaking capacity

= 2.55 symmetrical breaking capacity

= 2.55 2000= 5100 MVA

88. (b)In biased differential relay, the bias or fixedpercentage is the ratio of the differentialoperating coil current to the averagerestraining coil current.

89. (d)

Ripple factor = ripple voltage rms

dc voltage

IES M

ASTER

(11)

= 2 1

502= 0.028

90. (d)In half wave rectifier

Idc =

m

L

E 1 cos2 R

Idc = 200 312 50 2

= 1.1879

Power = 2dc LI R

= 70.6 W

91. (a)

I = SVR

VL = dILdt

0V = LdIV Ldt

= Sd VLdt R

V0 = sdVLR dt

= sdVdt

92. (d)

Rif = iR 1 A

= 2 100K1 49

93. (d)

V0 = i i10K 10K 10K1V V10K 10K 10K 10K

= i i1V 2V2

= i iV V

= 0 V

94. (b)As op-amps are ideal -current through 2 resistor is

2 12 = 0.5A

95. (a)The input voltage is added in series to the feedback voltage. So it is a voltage series circuit.

96. (c)The advantage of immediate addressing likeMVI B, 2EH is immediate execution ofinstruction. 8085 microprocessor has to doother operations like decoding and fetching inimmediate addressing also.

97. (c)Capacity of RAM = 8 K × 8 bit

= 23 × 210 × 8 bit= 213 × 8 bit

So, number of address lines required = 13Starting address =

F FC 8

1010 1001 0000 0000 A900H1 1111 1111 1111

1100 1000 1111 1111

i.e. C8FF H

98. (c)

Interrupts Priority Triggering Mask Location

(1) TRAP(2) RST 7.5(3) RST 6.5(4) RST 5.5

1st2nd3rd4th

(0024)(003C)(0034)(002C)

H

H

H

H

99. (d)STAX B - has register indirect addressing modeas address of data is stored in content of register(HL) pair [A-3].DAD D - has register addressing mode as datais available in content of register.JMP 2010H - Immediate addressing mode asaddress of data is given in instruction itself.CMA - has implicit addressing mode asoperation is performed on content ofaccumulator and address of operate is notrequired.

100. (b)In memory mapped I/O scheme, some addressesare assigned to memories and some addressesto I/O devices.But in I/O mapped I/O scheme, the addressassigned to memory locations can also beassigned to I/O devices. Since the same addressmay be assigned to a memory location or, anI/O device, the microprocessor must issue asignal IO M to distinguish whether the

IES M

ASTER

(12)address on the address bus is for a memorylocation or, an I/O device. Hence, I/O mappedI/O scheme is used for large system.

101. (b)TRAP may be treated as RST 4.5So, vector location of TRAP = 4.5 × 8

= 36

16 3616 2 4 i.e. H0024

0 2

102. (b)XOR operation =• If both inputs are same, output will be low

i.e. ‘0’• If both inputs are opposite, output will be

HIGH or, ‘1’.Let Accumulator contain :

001011010000111100100010

OF.H =XRIOFH :

Lower Nibble

Complement of lower nibble 1101

103. (d)TRAP – Both edge and level triggeredRST 7.5 – Edge triggeredRST 6.5 – level triggeredRST 5.5 – level triggered

104. (b)

67 H 0110 0111=81H 1000 0001=

1110 1000

So sign, S = 0Zero, Z = 0AC, AC = 0CY, CY = 0

105. (d)After the execution of XTHL, the content ofthe register L are exchanged with the byteof the stack-top. The content of the Hregister exchanged with the byte below thestack-top.

106. (b)Execution of HLT instruction takes 5T statesinstead of 4T states. One extra T-state isused to provide high impedance state i.e.tristate.

Since, time period, T = 1f =

61

2 10

= 0.5 secSo, for 5T state, t = 5 × T

= 5 × 0.5 sec= 2.5 sec

107. (c)Band width is same for BPSK and APSK whichis equal to twice the signal bandwidth.

108. (b)For a sinusoidal input SNR(dB) in PCM isobtained by following formula

SNR(dB) = 1.8 + 6n= 1.8 + 6 × 8 [ n = 8]= 49.8 dB

109. (b)

step size = P2M 6.4 0.1VL 64

Quanization noise power = 2

12

= 0.1 0.112

= 4 28.33 10 V

110. (a)512 levels = 29, so 9 bits are required to encodea sample. As the sampling rate is 18KHz, ina second, there will be 18KHz × 9 bits = 162K bits.

The duration of each bit = 31

162 10 sec.

= 6.17 s

111. (b)

sf 10 2 3.4KHz 68KHz

0.1V 3

n 2 10

Amax = 3

s3

m

f 0.1 68 10 1.08V2 10

IES M

ASTER

(13)112. (b)

In PCM, the B.W. requirement = n×fm wheren = no. of bitsQuantization level increases from 4 to 16 means22 to 24

2 bits to 4 bits

1BW = 2 × fm

BW2 = 4 × fm = 2(BW1)= doubled

113. (a)Thermistors are generally composed of semiconductor materials which show negativetemperature coefficient of resistance ie. theirresistance decreases with increase intemperature. This allows the thermistor todetect very small changes in temperature.

114. (d)A series resonance is called voltageresonance. So the assertion statement is false.Reason is true where series circuit acceptscurrent at resonance.

115. (b)

A freewheeling diode is necessary to avoid anyhigh voltages due to sudden current switchingin load (RL). Inductor will not allow suddenchange of curent. So the free wheeling diodeprovides current path.

116. (a)In eddy current damping, there is an inducedemf in disc and hence an eddy current, whichinteracts with alternating flux and produce atorque opposite to the direction of motion. Toflow the eddy current, the disc should beconducting in nature.

117. (b)Both the statements are true, but reason isnot appropriate for assertion.

118. (a)A squirrel cage rotor slots are semi closed orfully closed to reduce teeth loss and magnetizingcurrent.

119. (c)

120. (b)