ELECTRICAL ENGINEERING · 2020-01-06 · (c) Shunt with inverting input (d) Shunt with non-inverting input Ans. (a) Sol. Opertational amplifier gives finite output even if input is

ELECTRICAL ENGINEERING

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Detailed SolutionSET - B

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1. In a tunnel diode, the width of the junctionbarrier is(a) Directly proportional as the square root of

impurity concentration(b) Inversely proportional as the square root

of impurity concentration(c) Directly proportional as square of impurity

concentration(d) Inversely proportional as square of

impurity concentrationAns. (b)Sol. A tunnel diode is a heavily doped P-N junction

its depletion width is very narrow to allowtunneling effect.In diode depletion width w is given by formula

w =

s

a d

2 1 1 Vq N N

where Na = Acceptor concentrationNd = donar concentration

If Na is very large.

Then w

d

2 1q N

So, w d

1N

Hence option (b) is correct.

2. In a grounded-emitter transistor, when emittercurrent becomes zero in cut-off region theemitter potential is called(a) Floating Emitter Potential(b) Breaking Emitter Potential(c) Cascading Emitter Potential(d) Cut-off Emitter Potential

Ans. (*)

3. When maximum reverse-biasing voltage isapplied between the collector and base terminalsof the transistor and emitter is open circuited,breakdown occurs due to(a) Avalanche breakdown(b) Avalanche multiplication(c) Punch-through(d) Reach-through

Ans. (b)Sol. In transistor emitter is highly doped. Base

region is lightly doped and collector region ismoderatly doped with high area.So, junction between emitter and base regionis prone zener breakdown and junction betweenbase and collector is prone to avalanchebreakdown.As in question emitter is open.So, base and collector part will act as diode.Increasing reverse bias will cause avalanchemultiplication. Which is region of breakdown.

4. In a Field Effect Transistor (FET), themaximum voltage that can be applied betweenany two terminals is given by(a) Low |VDS| causing avalanche breakdown(b) Low |VGS| causing avalanche breakdown(c) |VDS| = 0 when gate is reverse-biased(d) |VGS| = 0 when gate is reverse-biased

Ans. (a)Sol. The maximum voltage that can be applied

between any two terminals is upto it reach abreakdown region

V = 0GS

V = –1GS

V = –2GS

V BDSVDS–4 –1 0

IDSS

ID(mA)

As low |VDS| causing avalanche breakdown.So, option is (a).

5. A depletion-type MOSFET can be operated inan enhancement mode where negative chargesare induced into n-type channel by applying(a) Positive Gate Voltage(b) Negative Gate Voltage(c) Positive Drain Voltage(d) Negative Drain Voltage

Ans. (a)Sol. As depletion-type mosfet are operating in

enhancement mode.We need to induce n-type channel for this we





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need to provide positive gate voltage to attractnegative charge.

6. The double base diode which is operated withthe emitter forward biased and a smalleremitter junction is called(a) Field Effect Transistor (FET)(b) Uni-Junction Transistor (UJT)(c) Bipolar Junction Transistor (BJT)(d) Metal Oxide Semiconductor Field Effect

Transistor (MOSFET)Ans. (b)Sol. UJT has three terminals: an emitter(s) and

two base (B1 and B2) so, it is some times knownas double-base diode.As area of emitter is very small so, its junctionis very small.

B1

B2

NP

Fig: UJT

7. Which one of the following is not a distortiontype that exists either separately orsimultaneously in amplifiers?(a) Linear distortion(b) Non-linear distortion(c) Frequency distortion(d) Delay distortion

Ans. (a)Sol. As in diagram the characteristics if mosfet

amplifier is shown

IDSaturation Linear

ArainCurrent

Brackdownregion

DVDS

There is no distortion exist either separatelyor simultaneously in amplifier in linear region.

8. Transistor noise caused by the recombinationand generation of carriers on the surface of thecrystal is called(a) Thermal noise (b) Excess noise(c) White noise (d) Shot noise

Ans. (a)Sol. Shot noise: It is caused by the random arrival

of current carriers (holes and electrons) at theoutput element of electronic device.It is due to discrete nature of current.Excess noise: It is flicker noise occur inresister it is 1/f where f is frequency.Thermal noise: At the surface crystal due tophoton or heat there is generation of carrierwhich produces noise.So, thermal noise is answer.

9. During a low frequency response of an amplifierwhich is invariably of RC-coupled type, thereis a range of frequency characteristics overwhich the amplification is constant and delayis also constant, called(a) Low band frequency(b) Mid band frequency(c) High band frequency(d) Hyper band frequency

Ans. (b)Sol. Characteristics of RC coupled amplifier is shown

below

A

FL FH

A = gainf = frequency

So, in mid band frequency amplification anddelay is constant.

10. In a crystal oscillator, especially whenpiezoelectric crystal like quartz is applied, then

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the inductor L, capacitor C and resistor R arethe analogs of the mechanical system as(a) Mass, compliane, viscous-damping factor(b) Mass, spring constant, viscous-damping

factor(c) Mass, momentum, viscous-damping factor(d) Mass, displacement, viscous-damping factor

Ans. (b)Sol. A quartz crystal can be modelled as :

R

L

C1

C2Crystal

where, analogy can be given asMass InductanceStiffness (spring constant) capacitanceViscous-damping factor Resistance

11. In a phase shift oscillator using an FET, at acertain frequency if the phase shift introducedby the RC network is 180°, then the total phaseshift from gate around the circuit and back tothe gate will be(a) 0° (b) 90°(c) 180° (d) 270°

Ans. (a)Sol. As we know in phase shift oscillator 180° phase

shift is provided by amplifier and 180° isprovdied by 3 RC circuits.

A

+ 360°

+ 180°

RC feedback circuit

Vo

+ 180°Vi

So, total phase shift from gate around thecircuit and back to gate will be 0°

12. In a feedback amplifier, which configurationincreases bandwidth, decreases non-lineardistortion and improves transconductance withnegative feedback?

(a) Voltage-series (b) Current-series(c) Voltage-shunt (d) Current-shunt

Ans. (b)Sol. We know in case of transconductance with

negative feedback.Zi = input impedance = maximumZo = output impedance = maximumSo, feedback configuration is series – series.Series – series = curent series.

13. In order to balance the offset voltage of anoperational amplifier, a small DC voltage isapplied to input terminals where the connectionis(a) Series with both inverting as well as non-

inverting input(b) Series with non-inverting input(c) Shunt with inverting input(d) Shunt with non-inverting input

Ans. (a)Sol. Opertational amplifier gives finite output even

if input is zero this is because of offset voltage.This is eliminated by applying a voltage inseries with both inverting as well as noninverting input.

V V = 0

Fig: Offset voltage compensation of OPAMP

14. Multivibrator circuit that remains in stablestate until a triggering signal causes a trnsitionto quasistable state and returns to stable stateafter certain time is called(a) Astable multivibrator(b) Monostable multivibrator(c) Bistable multivibrator(d) Unstable multivibrator

Ans. (b)Sol. Monosable multivibrator: It is one shot

multivibrator to generate a pulse for finite timewhen trigger pulse is applied.





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t0

quasi stable state

Fig: without trigger

Stablestate

quasi stable

15. In a paraphase amplifier, where two amplifiersare connected in cascade, the output from secondstage(a) Equals signal input without change of sign(b) Equals signal input with change of sign(c) Does not equal to signal input and has no

sign change(d) Does not equal to signal input but has sign

changeAns. (c)Sol. Paraphase amplifier is based on the concept of

the distributed amplifier it gives gain and 180°phase shift.In question two paraphase amplifier arecascaded.So, Net phase shift = 180° + 180° = 360°Gain of amplifier = A × A = A2

So, best option is output signal is not equal tosignal input and has no sign change.Output signal = A2 × input signal

16. Which one of the following statements is notcorrect for an active filter used in the field ofcommunication and signal processing?(a) It is more economical(b) It does not cause loading of the source or

load(c) It is easier to tune or adjust(d) It exhibits insertion loss

Ans. (d)Sol. As active filter is more cheaper than passive

filter because of use of OPAMP and absence ofinductor.

It doesn’t cause loading effect because of use ofhigh input impedance OPAMP.It is easier to the tune in wide range.It exhibit negligible insertion loss because ofamplifier present unit.

17. A two-step procedure in a typical diffusionapparatus to obtain the complementary-error-function Gaussian distribution involves the firststep and second step respectively as(a) Predeposition and Drive-in(b) Predeposition and Drive-out(c) Drive-in and Postdeposition(d) Drive-out and Postdeposition

Ans. (a)Sol. In VLSI technology

Predeposition also known as infinite sourcediffusion is done before drive-in (finite source)diffusion.

18. In AM modulation, the equation of themodulating signal is given by f(t) =

m mA cos t . If the amplitude of the carrierwave is A and there is no over-modulation, themodulation efficiency will be(a) 33.3% (b) 38.6%(c) 43.3% (d) 48.6%

Ans. (a)Sol. The modulation efficiency of AM signal

=2

2 1002

with no over modulation 1

So maximum efficiency 1

% = 1 1002 1

= 1 1003

= 33.3%

19. For a binary phase-shift keying modulator witha carrier frequency of 70 MHz and nput bitrate of 10 Mbps, the maximum Upper SideFrequency (USF) and minimum Lower SideFrequency (LSF) are respectively(a) 85 MHz and 65 MHz(b) 75 MHz and 65 MHz





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(c) 55 MHz and 45 MHz(d) 75 MHz and 45 MHz

Ans. (b)Sol. The Spectrum of BPSK. is

bc

Rf2

–fc fcbc

Rf2

bc

Rf2

bc

Rf2

Given fc = 70 MHzRb = 10 Mbps

So, Maximum upper side freq.

(USP) = bc

Rf2

= 1070 75MHz2

Minimum lower side frequency (LSF)

= bc

R 10f 702 2

= 65 MHz20. In modulation system, the energy per bit-to-

noise power density ratio b

o

EN is

(a) bfCN B (b)

b

N BC f

(c) b

C BN f (d) bfN

C BWhere:N = Noise power of thermal (W)B = Bandwidth(Hz)C = Carrier power (W)fb = Bit rate (bps)

Ans. (c)Sol. The signal to noise ratio (SNR) of received signal

CN = b b

0

E fN B

b

0

EN =

b

C BN f

so b

0

EN =

b

C BN f

21. Which one of the following is not a transmissionparameter of a private line data circuit thatutilizes public telephone network?(a) Geographical parameter(b) Bandwidth parameter(c) Interface parameter(d) Facility parameter

Ans. (a)Sol. Transmission parameter are divided into three

broad categories. Bandwidth parameters:- It include

attenuation distortion and envelope delaydistortion.

Interface parameter:- It include terminalimpedance, in band and out of band signalpower, test signal power and groundisolation

Facility parameter:- Which include noisemeasurements frequency distortion, phasedistortion amplitude distortion, andnonlinear distortion.

22. The Shannon limit for information capacity Iis

(a) 2

SBlog 1 –N (b)

2SBlog 1N

(c) 10

SBlog 1 –N (d)

10SBlog 1N

Where:N = Noise power (W)B = Bandwidth (Hz)S = Signal power (W)

Ans. (b)Sol. Shannon limit for Information capacity

I = 2SB log 1N

where N = noise powerB = Band widthS = Signal power.

23. In a time division multiplexing, there are 8000samples for a digital signal-0 channel that uses8 kHz sample rate and 8 bit PCM code. Theline speed will be





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(a) 56 kbps (b) 64 kbps(c) 76 kbps (d) 84 kbps

Ans. (b)Sol. Given sampling rate.

fs = 8 KHzNo of bits n = 8 bit

Line speed Rb = n fs

Rb = 8 × 8 Kbps= 64 Kbps

Hence line speed is Rb = 64 Kbps.

24. For an 8-PSK system, operating with aninformation bit rate of 24 kbps, the baud ratewill be(a) 16,000 (b) 12,000(c) 8,000 (d) 6,000

Ans. (c)Sol. Given Rb = 24 Kbps.

Band Rate BT = b

2

Rlog M

Give M = 8 for (8-PSK)

BT =2

24 kbpslog 8

BT = 24 kbps3

BT = 8 kbpsBT = 8,000

25. During transformation of independent variable,if two signals identical in shape are displacedrelative to each other, then the difference inpropagation time from point of origin oftransmitted signal results in(a) Time shift (b) Time reversal(c) Time scaling (d) Time reduction

Ans. (a)Sol. During transformation of independent variable,

if two signals identical in shape are displacedrelative to each other, then due to difference inpropagation time from of origin of transmittedsignals, results into time shifting.

26. Which of the following statements is/arecorrect?

1. A continuous-time system is a system inwhich, continuous-time input signals areapplied, resulting in continuous-time outputsignals.

2. A system is said to be linear if it followsthe superposition theorem.

3. A system is said to be non-linear if it followsthe superposition theorem.

(a) 1 only (b) 1 and 2 only(c) 2 only (d) 1 and 3 only

Ans. (b)Sol. A system is said to be linear if it satisfies

superposition principle, which comprises ofadditivity and homogencity.

Addivity : If 1 1x t X s

& 2 2x t X s

homogeneity then

1 2 1 2ax t bx t aX s bX s

So (2) is correct & (3) is false,A non-linear system doesn’t follow superpositiontheorem.A continous time signal is one which exists forentire continuous time duration.Whereas a system which takes continous timesignal as input & yields a continous time signalas output is termed as a continous timesystem.

27. A discrete time signal is said to be a unitsample sequence if

(a)

n = 1for n =0 = 0 for n 0

(b)

n = 2for n=0 = 0 for n 0

(c)

n = – 1for n=0 = 0 for n 0

(d)

n = – 2for n =0 = 0 for n 0

Ans. (a)Sol. A discrete time signal is said to be unit sample

sequence. if (a)

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n 1 for n = 0

0 for n 0All other signals also have only one samplesequence, but their magnitude is not equal to1.

28. A signal is said to be1. Deterministic if there is no uncertainty over

the signal at any instant of time.2. Deterministic if it is expressible through a

mathematical equations.3. Random or non-deterministic if there is

uncertainty over the signal at the instantof time.

4. Random or non-deterministic if it is nonexpressible through a mathematicalequation.

Which of the above statements are correct?(a) 1, 2 and 3 only (b) 1, 2 and 4 only(c) 3 and 4 only (d) 1, 2, 3 and 4

Ans. (d)Sol. A deterministic signal is a signal about which

there is no uncertainity with respect to itsvalue at any time. These signals can bemodelled as completely specified functions oftime or mathematical expression.Whereas, a random signal is a signal aboutwhich there is uncertainity before which itoccurs. This can be viewed as a signal. Whichis an ensemble or a group of various signalswith different waveform, hence it is not possibleto model these signals in a mathematicalexpression.

29. Linear time-invariant systems that aredesigned to pass some frequencies essentiallyundistorted and significantly attenuate oreliminate others are(a) Frequency-shaping filters(b) Frequency-selective filters(c) Time-shaping filters(d) Time-selective filters

Ans. (b)Sol. Those linear time-invarient systems which are

designed to pass desired frequency components

in a signal without distortion and attenuate oreliminate other frequency components are calledas frequency relective filters.

30. If the input signal x(t) and impulse responseh(t) of a continuous-time-system are describedasx(t) = e–et.u(t) and h(t) = u(t–1), the output y(t)will be

(a) –3 t–11 1 – e3 (b) –3t1 1 – e

3

(c) –3 t–11 1 e3 (d) –3t1 1 e

3Ans. (a)Sol. Given,

x(t) = e–3t . u(t)h(t) = u(t – 1)To calculate y(t)y(t) = x(t) * h(t)Taking Laplace transform Y(s) = X(s) . H(s)X(s) = 3te u tL

X(s) = 1

s 3

H(s) =

s

1 eh ts

L

Y(s) =

se 1 A Bs s 3 s s 3

By partial fraction

0

s 0

e 1A lim Y s s3 3

3t

s 3

eB lim Y s s 33

So,

31 eY s3s 3 s 3

31 1 eY s3 s s 3

3 3t1y t u t e e u t3

3 t 11y t 1 e u t3





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31. Consider an LTI system whose response to theinput

x(t) = [e–t + e–3t] u(t) is

y(t) = [2e–t – 2e–4t] u(t).

The system’s impulse response will be

(a) –2t –4t3 e e u t2

(b) –2t –4t3 e – e u t2

(c) –2t –4t1 e e u t2

(d) –2t –4t1 e e u t2

Ans. (a)Sol. Given,

LTI systemx(t) = (e–t + e–3t) u(t)y(t) = (2e–t – 2e–4t) u(t)To calculate impulse responsey(t) = x(t)*h(t) y(s) = X(s).H(s)

y(s) = t 4t2e 2e u tL

y(s) =

2 2 6

s 1 s 4 s 1 s 4

X(s) = t 3te e u tL

X(s) = 1 1

s 1 s 3

=

2s 4

s 1 s 3 =

2 s 2

s 1 s 3

Now, H(s) =

Y sX s

=

6s 1 s 4

2 s 2s 1 s 3

=

6 s 3

2 s 2 s 4

H(s) =

3 s 3

s 2 s 4By partial fraction

H(s) = A B

s 2 s 4

A =

s 2

3lim s 2 H s2

B =

s 4

3 3lim2 2

H(s) =

3 1 12 s 2 s 4

2t 4t3h t e e u t2

32. Consider an LTI system with a system function

H(z) = 1

11 – z – 14

Its difference equation will be

(a) 1y n – y n –1 x n2

(b) 1y n – y n –1 x n4

(c) 1y n y n – 1 x n2

(d) 1y n – y n 1 x n4

Ans. (b)Sol. Given,

LTI system

H(z) = 1

111 z4

H(z) =

1

Y z Y z 11X z X z 1 z4

The difference equation is

11y n 1 z4 = x(n)

1y n y n 1 x n4

Note: Using property of shifting

If x n X z

then 0n0x n n X z z

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33. It is assumed that quantization error, e(n) isa sequence of random variables where1. The statistics do not change with time.2. It is a sequence of uncorrelated random

variables.3. It is uncorrelated with the quantizer input

x(n).4. The probability density function is

uniformly distributed over the range ofvalues of quantization error.

Which of the above statements are correct?(a) 1, 2 and 3 only (b) 1, 2 and 4 only(c) 3 and 4 only (d) 1, 2, 3 and 4

Ans. (b)

34. For a given differential equation,

2

2d y t dy t4 5y t 5 t

dtdt

with

––

0

dy ty o 1 and 2dt

and input x(t) = u(t).

The output y(t) will be(a) 2u(t) – 2e–2t sin t(b) u(t) + 2e–2t sin t(c) u(t) – 2e–t sin t(d) 2u(t) + e–t sin t

Ans. (b)Sol. Given,

Differential equation

2

2d y t dy t4 5y t 5x t

dtdt

with

0

dy ty 0 1 & 2dt

Input x(t) = u(t)Taking laplace transform of differentialequations2y(s) – sy(0) – y'(0) + 4(sy(s) – y(0)) + 5 y(s)= 5x(s) y(s) (s2+4s+5) = sy(0) + y'(0) + 4y(0) + 5x(s)x(t) = u(t)

X(s) = 1s

(s2+4s+5)y(s) = 5s 2 4s

= 5s 6s

Y(s) =

2

2

s 6s 5s s 4s 5 =

2A BS CS s 4s s

By partial fraction & comparsionA + B = 14A + C = 65A = 1

A 1 B 0 C 2

y(s) = 2

1 2s s 4s 5

y(s) =

2

1 2s s 2 1

Taking inverse laplacey(t) = (1 + 2e–2t sint) u(t)

2ty t u t 2e sin t u t

35. The Nyquist rate for the signal

x(t) =

1 cos 4000 t2 cos 1000 t will be

(a) 5 kHz (b) 10 kHz(c) 15 kHz (d) 20 kHz

Ans. (a)Sol. To evaluate the Nyquist rate of the signal.

1x t cos 4000 t cos 1000 t2

1x t 2cos 4000 t cos 1000 t4

1x t cos 3000 t cos 5000 t4

Using cos(A–B) + cos(A+B) = 2cosA.cosB m 5000

m

5000f 2500Hz2

So, by low pass sampling theorem Nyquist rate 2fm





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for fs = 2(2500)fs = 5000 Hz

sf 5KHz

36. Which of the following statements is/arecorrect?1. A system is said to be Finite Impulse

Response (FIR), if the output samples ofthe system depend only on the presentinput and a finite number of past orprevious input samples.

2. If the output of a system y(n) depends onlyon the present input and past inputs, butnot on past outputs, then it is called anon-recursive system.

3. If the output of a system y(n) depends onlyon the present input and past inputs, butnot on past outputs, then it is called arecursive system.

(a) 1 only (b) 1 and 2 only(c) 1 and 3 only (d) 3 only

Ans. (b)Sol. 1. A finite impulse response (FIR), is an

impulse response which has only finitenumber of samples. Which depend only onpresent input and a finite number of pastor previous input samples. Whereas an IRRfilter consists of infinite number of samples.

2. For a non-recursive system, current outputdoesn’t depend upon the previous or pastoutputs; but only upon present & pastinputs. Whereas in a recursive system, itdepends upon past outputs also.

37. The value of the steady state error for first

order system,

1sT 1 with Unit Ramp Function

will be

(a) 1T (b) T

(c) 1–TT 1 – e (d)

1–T1 e

TAns. (c)

Sol. C(s)= 1 R s1 ST

Since, input is unit Ramp R(s) = 1/s2

C(s) = 21 1

ST 1 S

C(s) =2

21 ST T

1 STS

C(s) = 21 T 1T. 1SS S

T

Taking Laplace inverseC(t) = t – T + T.e–t/T

Error signal, e(t) = t – [t – T +T.e–t/T]( r(t) = t u(t))

e(t) = T(1 – e–t/T)

38. If the number of zeros are less than the numberof poles, i.e., Z < P, then the value of thetransfer function becomes zero for s .Hence we say that there are zeros at infinityand the order of such zeros is(a) P+ Z (b) P – Z(c) Z – P (d) Z

Ans. (b)Sol. If a system has no. of zeros less than no. of

poles i.e. z < P it is called as proper system.In such caseNumber of zeros at infinity = P – zWhere, P = No of finite poles

z = No. of finite zeroes.

39. The method of determination of the stability ofthe feedback systems as a function of anadjustable gain parameter which does notprovide detailed information concerning locationof closed-loop poles as a function of gain K iscalled(a) Root locus method(b) Nyquist criterion method(c) Bode plot method(d) Routh-Hurwitz criterion method

Ans. (d)





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Sol. Routh-Hurwitz criterion method is used fordetermination of stability of the feedback systemas a function of an adjustable gain parameter.This method only tells about number of rootslying in right half plane and doesn’t provideconcerning location of closed poles.Root locus method provides exact locations ofroots of characteristics equation .

40. Consider the sinusoidal transfer function intime-constant form

22 1 jG j

j110

Asymptotic log magnitude characteristic offactor 1 j is straight line of

1. 0dB for 1

2. + 20 dB/decade for 1

3. – 20 dB/decade for 1

Which of the above relations is/are correct?(a) 1 only (b) 1 and 2 only(c) 1 and 3 only (d) 2 only

Ans. (b)Sol. For Asymptotic log magnitude characteristic

of a factor a1 j T

Corner frequency =a

1T

and before corner frequency i.e.a

1T

contribution of particular factor is 0dB.

here, for 1 j , Ta = a

11 1T

0 dB for 1

and every single order zero contributes for+20dB/decade of slope after it’s cornerfrequency.So, 1 and 2 are correct.

41. A graphical technique for plotting the closed-loop poles of a rational system functions as afunction of the value of gain for both continuous-time and a discrete-time system is

(a) Root locus method(b) Nyquist criterion method(c) Bode plot method(d) Routh-Hurwitz criterion method

Ans. (a)Sol. Root locus method is a graphical technique

which gives location of closed loop poles of arotation system as a function of value of gain.Root locus technique is used for both continuoustime and a discrete time system.Nyquist criterion and Bode plot method arenot directly function of value of gain, althoughboth can be used for discrete-time as well.Routh-Hurwitz criterion is not a graphicaltechnique.

42. Which of the following statements regarding‘Root locus’ is not correct?(a) By addition of poles to left half, the root

locus shifts towards right hand side andstability of system decreases.

(b) By addition of zero towards left, the rootlocus shifts towards left half, since rootlocus shifts towards left half, the relativestability of control system increases.

(c) By addition of zero towards left side, theroot locus shifts towards left half, therelative stability remains same.

(d) By addition of poles to the left half, thesystem stability decreases, while byaddition of zeros towards left half, thestability of system increased.

Ans. (c)Sol. Effects of addition of poles to left half

1. Root locus branches shifts toward righthand side and relative stability decreases

2. System becomes more oscillatory.Effects of addition of zeros to left half1. Root locus shifts towards left and relative

stability of system increases2. System becomes less oscillatory

43. In time domain, the relative stability ismeasured by maximum overshoot and dampingratio. In frequency domain, the relative stabilityis measured by

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(a) Steady state error (b) Damping ratio(c) Resonant peak (d) Bandwidth

Ans. (c)Sol. In the time domain, relative stability is

measured by parameters such as maximumovershoot and damping ratio. In the frequencydomain, the resonant peak is used to indicaterelative stability.

44. Consider a feedback system with thecharacteristic equation

11 K

s s 1 s 2 = 0 for root locus.

The angles of asymptotes A and the centroid

of the asymptotes A are respectively

(a) 60°, 120°, 180° and –(b) 45°, 90°, 300° and 0(c) 60°, 180°, 300° and –1(d) 45°, 90°, 180° and 0

Ans. (c)Sol. Given

11 K

s s 1 s 2

= 0

then,

G(s) H(s) = K

s s 1 s 2

Angle of asymptotes

(QA) = 1802K 1P z

Where, P = No. of finite polesz = No. of finite zeros

andK = 0, 1, 2, ......,((z – P–1)

Hence

QA = 1802K 13 0

= (2K + 1) × 60

QA = 60°, 180°, 300°

Centroid =

real parts of poles of G s H s– real parts of zeros of G s H s

P z

= 0 1 2 0

3 0

= –1

45. Which one of the following statementsregarding an effect of phase lead network isnot correct ?(a) The velocity constant is usually increased(b) The slope of the magnitude curve is reduced

at the gain crossover frequency, as a resultreltive stability improves

(c) Phase margin increased(d) the bandwidth decreased

Ans. (d)Sol. Effects of phase lead network

1. Velocity constant Kv increases2. The slope of the magnitude plot reduces at

gain crossover frequency, as a resultrelative stability improves.

3. Phase margin increases4. Response becomes faster5. Increases the system bandwidth.Hence option (d) is not correct.

46. A lead compensator1. Speeds up the transient response.2. Increases the margin of stability of a

system.3. Helps to increase the system error constant

though to a limited extent.Which of the above statements are correct?(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3

Ans. (d)Sol. A lead compensator causes the following effect

1. System error constant increase2. Seeps up the transient response3. Increases stability margin (pm) of system4. Increases band width of system.Hence all of given statements are correct.





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47. Which of the following statements are correct?1. The pair (AB) is controllable implies that

the pair (ATBT) is observable.2. The pair (AB) is controllable implies that

the pair (ATBT) is unobservable.3. The pair (AC) is observable implies that

the pair (ATCT) is controllable.4. The pair (AC) is observable implies that

the pair (ATCT) is uncontrollable.(a) 1 and 3 only (b) 1 and 4 only(c) 2 and 3 only (d) 2 and 4 onlyWhere: A, B and C are having their standardmeanings.

Ans. (a)Sol. If pair [A, B] is controllable, then pair

[AT, BT] will be observable.If pair [A, C] is observable, then pair[AT, CT] will be controllable.

48. Bounded-input bounded-output stability impliesasymptotic stability for1. Completely controllable system2. Completely observable system3. Uncontrollable system4. Unobservable systemWhich of the above statemetns are correct?(a) 1 and 4 only (b) 1 and 2 only(c) 2 and 3 only (d) 3 and 4 only

Ans. (b)Sol. Bounded-input bounded-output stability implies

asymptotic stability for system with no-polezero cancellation i.e. completely controllable andcompletely observable system.

49. The degree of humming level of the noise causedin the transformers may be reduced by(a) Magnetostriction(b) High flux density in core(c) Tightening of core by clamps(d) Quality of transformer oil

Ans. (c)Sol. Transformer noise is basically due to a

phenomenan called magnetostriction. When atransformer core is magnetised, it’s dimensions

change slightly with the magnetisation. Atlower flux densities in transformers, theamount of magnetostrition is also low. Hencefrom the given options the noise can be reducedby tightening of core by clamps as transformeroil has nothing to do with humming noise.

50. A transformer with a 10: 1 ratio and rated at50 kVA, 2400/240 V, 50 Hz is used to stepdown the voltage of a distribution system. Thelow tension voltage is to be kept constant at240 V. If the transformer is fully loaded at 0.8power factor (lag), the load impedance connectedto low-tension side will be nearly(a) 3.15 (b) 2.60(c) 1.15 (d) 0.60

Ans. (c)Sol. Rated so =50 kVA

Rated vol = 2400V/240VRated current on LV side = I2

So = V2I2

I2 =

3

o

2

S 50 10V 240

Load impedance

ZL =

2

32

V 240I 50 10 240

=

2

3240 1.15

50 10

51. A DC shunt generator supplies a load of 7.5kW at 200 V. If the armature resistance is

0.6 and field resistance is 80 , the inducedemf will be(a) 224 V (b) 218 V(c) 212 V (d) 204 V

Ans. (a)

Sol.

R =0.6aR = 80f

If Ia I

E V = 200V

I

P = 7.5L

Load current

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IL =

3LP 7.5 10 37.5 A

V 200

Fld current If = f

V 200 2.5AR 80

Ia = I + If = 40AE = V + IaRa

= 200 + 40 × 0.6 = 224V

52. A 4-pole DC motor is lap-wound with 400conductors. The pole shoe is 20 cm long andthe average flux density over one-pole-pitch is0.4 T, the armature diameter is 30 cm. Whenthe motor is drawing 25 A and running t 1500rpm, the torque developed will be nearly(a) 30 kN (b) 40 Nm(c) 50 Nm (d) 60 Nm

Ans. (a)Sol. P = 4 Z = 400

Length l = 20 cm = 0.2Bav = 0.4 T

D = 0.3 mFlux per pole

= Bav × Area/Pole

= Bav × DP

l

= 0.4 0.3 0.24

= 0.006 wbLup winding A = P

Ka =ZP

2 A

=

400 2002

For Ia = 25 AT = a aK I

=

200 0.006 25 = 30 Nm

53. In an unloaded shunt generator, when switchis closed, a small field current is producedwhich leads to generation of still larger voltagesdue to addition of

(a) Armature voltage(b) Residual flux voltage(c) Generated voltage(d) Voltage drop

Ans. (c)

Sol.+

–E

S I = 0

UnloadsIf

If = 0 Q = Qres E = Eres Residualvolume when S-closed fI 0 due to small fieldcurrent Q > Qres E > Eres

54. The generator efficiency of a shunt generatorwill be maximum when its variable loss isequal to(a) Constant loss(b) Stray loss(c) Iron loss(d) Friction and windage loss

Ans. (a)Sol. Power output Po = VI

Constant losses = Pc

Variable loss i.e. Cu–loss = I2R

Efficiency =

o o2

in o c

P PP P P I R

= 2

c

VIVI P I R

= c

VPV IRI

For max Dr = cPV IR minI

cPd V IRdI I = 0

c2

P RI

= 0

Pc = I2R





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55. Induction motor can be regarded as ageneralized transformer due to certainsimilarities except rated(a) Frequency (b) Flux(c) Speed (d) Induced emf

Ans. (a)Sol.

3-supply

V, f

Stator (Primary)

f = sf2

Rotor (Secondary)In transformer frequency some on both sidesi.e. rated.In I.M. frequency m rotor side f2 = sfwhere slip ‘s’ depends upon frequency.

56. A 3-Phase, 400/200 V, Y – Y connected wound-rotor induction motor has v rotor resistanceand 0.3 standstill reactaince per phase. tomake the starting torque equal to themaximum torque, the additional resistancerequired in the rotor circuit will be(a) 0.24 / phase (b) 0.34 / phase

(c) 0.42 / phase (d) 0.52 / phase

Ans. (a)Sol. To make starting Ts = Tm Mhz torque

Slip for Tm: sm = 1

sm 2

2

R 1X

ext0.06 R

0.3 = 1

Rext = 0.24

57. Potier triangle method is helpful in obtainingthe voltage regulation of synchronous machinesby determining the armature(a) Leakage reactance and its reaction mmf(b) Leakage reactance and air-gap flux(c) Resistance and its reaction mmf(d) Resistance and air-gap flux

Ans. (a)

Sol. Sol.Potier- CBA

CB = a alI x drop

BA = FaArmature reaction mmf

B A

CI xa al

Emf

V I = constanta

I = 0a

ZPFC

OCC

F (AR. MMF)a

mmfF0

58. In a synchronous motor, the magnitude ofstator back emf, Eb depends on(a) Speed of the motor(b) Load on the motor(c) Both the speed and rotor flux(d) Rotor excitation only

Ans. (d)Sol. Stator induced emf i.e. excitation emf.

Ef = ph w f2 N K f

f Supply frequency

f Rotor field flux

Speed N = 120fP decided by frequency.

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59. A stepper motor has a step angle of 2.5°. Ifthe shaft is to make 25 revolutions, the numberof steps required will be(a) 1800 (b) 2200(c) 2800 (d) 3600

Ans. (d)Sol. Numer or revolution = 25

Angle rotated = 25 × 360°Step Angel = 2.5°

Number of steps =

25 360

2.5 = 3600

60. In which of the following respects, theservomotors differ in application capabilitiesfrom large industrial motors?

1. They produce high torque at all speedsincluding zero speed.

2. They are capable of holding a staticposition.

3. they do not overheat at standstill or lowerspeeds.

4. Due to low-inertia thye are not able toreverse direction quickly

(a) 1, 2, 3 and 4 (b) 1, 2 and 3 only(c) 1, 2 and 4 only (d) 3 and 4 only

Ans. (b)Sol. Statement-4 wrong. because due to low inertia,

the reversal will be quick.However statement-I also not completely correctas speed torque characteristic.i.e. N-increase then torque decreases.

N

T

61. If is eigenvalues of A, and A is idempotentmatrix, then(a) 0 (b) 1

(c) Either 0 or 1

(d) 0 and 1

Ans. (c)Sol. A matrix said to be idempotent if

A2 = A 2 =

2 – = 0

1 = 0

= 0, 1

62. the eigenvalues of the matrix 5 41 2

are

(a) 5 and 2 (b) 1 and 4(c) 1 and 6 (d) 7 and 5

Ans. (c)Sol. Characteristic equation of given matrix

5 2 4 = 0

2 7 6 = 0 = 1, 6

63. Using Runge’s formula of order 2, when x =

1.1, given 2dy 3x ydx

and y = 1.2 when x =1, taking h = 0.1. The value of y will be nearly(a) 1.3 (b) 1.5(c) 1.7 (d) 1.9

Ans. (b)Sol. Comparing

dydx = 3x + y2 with

dydx = f(x, y)

f(x, y) = 3x + y2

x = 1

o

x = 1.1 1

y = 1.2

o

y = ? 1

NAK = hf(x , y )

= hf(1, 1.2)= 0.1(3×1)+(1.2)= 0.444

K = hf(x +h, y +k )

= 0.1f(1.1, 1.644)= 0.60027

1 0 0

2 0 0 1

2

x y Calculation





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y1 = o 1 21y k k2

= 11.2 0.444 0.600272

= 1.7221 1.7

64. The expression 2 x

x2 x

Eee .E e

(the interval ofdifferencing being h) is(a) ex–h (b) ex+h

(c) ex (d) 2ex

Ans. (c)

Sol.2 x

x2 x

EeeE e

2xe

E

= 2 1 xE e

= 2 1 xE e

= 2 x he e

= h 2 xe e

2 xx

2 xEee .

E e

= e–hEex

= e–hex+h

= ex

65. The solution of differential equation

(x2y – 2xy2) dx – (x3 – 3x2y) dy = 0, is

(a)x 2log x 3log y cy

(b) y 2log y 3log x cx

(c)x 2log x 3log y cy

(d) y 2log y 3log x cx

Ans. (a)Sol. M =x2y – 2xy2

My = x2 – 4xy

N = –(x3 – 3x2y)

Nx = –3x2 + 6xy

My

Nx

So, I.F. = 1

Mx Ny

I.F. = 2 2 3 21

x x y 2xy y x 3x y

I.F, = 2 21 16 x y

This can be taken as 2 21

x y only

By multiplying 2 21

x y in given D.E

2 2 3 2

2 21 x y 2xy dx x 3x y dy

x y = 0

2

1 2 x 3dx dyy x yy

= 0 ... (i)

Equation (i) exact D.E so solution is

1 2 3dx dyy x y = C

x 2 n x 3 n yy = C

66. If u = x log xy, where x3 + y3 + 3xy = 1, thendu isdx

(a)

2

2x x y1 log xyy y x

(b)

2

2y x y1 log xyx y x

(c)

2

2x x y1 log xyy y x

(d)

2

2x x y1 log xyy y x





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Ans. (c)Sol. u = x log xy ...(i)

(given)x3 + y3 + 3xy = 1 ...(ii) (given)Differentiating (i) w.r.t x

ux =

1 ylog xy y xy x

ux =

1 ylog xy y xy x

ux =

x ylog xy 1y x

...(iii)

Differentiating (ii) w.r.t x

2 2 y y3x 3y 3 y xx x = 0

yx =

2

2x yy x

...(iv)

Substituting value of yx in equation (iii)

ux =

2

2x x ylogxy 1y y x

ux =

2

2x x ylog xy 1y y x

ux =

2

2x x y1 logxyy y x

67. The solution of differential equation3 3 3

x 2y3 2 3z z z3 4 e

x x y y

is

(a) 1 2 3z f y x f y 2x xf y 2xx 2ye27

(b) 1 2 3z f y x f y 2x xf y 2x x 2ye23

(c) 1 2 3z f y x f y 2x xf y 2x

x 2ye27

(d) 1 2 3z f y x f y 2x xf y 2x

x 2ye23

Ans. (a)

Sol. 3 2D 3D D 4D y = ex+2y

3 2 3D 3D D 4D y = ex+2y

Auxiliary exhaustionm3 – 3m2 + 4 = 0

m = 1, 2, 3C.F = f1(y–x) + f2(y+2x) + x

f3(y+2x)

P.I = 3 3 131

D 3D D 4D ex+2y

= x 2y1 e1 6 32

=x 2ye27

z = f1(y–x) + f2(y+2x) + x

f3(y+2x) + x 2y1 e27

68. If the imaginary part v = ex (x sin y + y cosy) is part of analytic function f(z) = u + iv,then f(z) is(a) (1+z)ez + c (b) zez + c(c) ze2z = c (d) (1 – z)ez + c

Ans. (b)Sol. If f(z) is analytic then

f(z) = ux + ivx

f(z) = vy + ivx [Using CR]

v = x xx.e sin y e .y cos y

vx = siny[ex + xex] + exy cosyvy = xex cosy + ex[cos y-ysiny]vy = xex cosy + ex cos y-exysiny

f(z) = xex cosy + excosy–exysiny + i(xexcosy + ex

cosy – exysiny)Replacing x by z & y by 0

f(z) = zez + ez + i (zez + ez)integrating both side

f(z) = zez – ez + ez + i(zez – ez +ez) + c

f(z) = zez + izez + c

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f(z) = zez (1+i) + c

f(z) = zez + c

1 i

zf (z) ze k

69. The first four terms of the Taylor series

expansion of f(z) = z 1

z 3 z 4

, when z = 2

is

(a) 2 311 27 59z 2 z 2 z 2 ...4 8 16

(b) 2 311 27 59z 2 z 2 z 2 ...4 8 16

(c) 2 33 11 27 59z 2 z 2 z 2 ....2 4 8 16

(d) 2 33 11 27 59z 2 z 2 z 2 ..2 4 8 16

Ans. (c)

Sol. f(z) =

z 1z 3 z 4

f(z) =

4 5z 3 z 4

f(z) =

2 24 5

z 3 z 4

f(2) = 114

f(z) =

2 38 10

z 3 z 4

f(2) = 274

f(z) =

4 424 30

z 3 z 4

f(2) =177

8Taylor series expansion

f(z) = f(a) +(z-a) fa

fa 2(z a)2!

3(z a) f "'(a) .....3!

f(z) =

23 11 27z 2z 22 4 2! 4

3 17727 z 2 ...3! 84

f(z) = 3 11 27

z 22 4 8

(z–2)2 + 359 ....z 316

70. The mean deviation about mean of a normaldistribution is nearly

(a) 35 (b) 5

3

(c) 45 (d) 5

4

Ans. (c)Sol. The mean absolute deviation of the normal

distribution

= 2.

= 0.7978= 0.8

Option (a) 35 = 0.6

Option (b) 53 = 1.67

Option (c) 45 = 0.8

So, Option (c) is correct choice.)

71. Consider the following regression equationsobtained from a correlation table:

y = 0.516x + 33.73

x = 0.512y + 32.52

the value of the correlation coefficient will be(a) 0.514 (b) 0.586(c) 0.616 (d) 0.684

Ans. (a)Sol. y = 0.516x + 33.73 ...(i)

x = 0.512y + 32.52 ...(ii)byx = 0.516bxy = 0.512





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r = byx bxy

= 0.516 0.512= 0.514

72. If the probability of a bad reaction from acertain injection is 0.001, the chance that outof 2000 individuals, more than two will get abad reaction will be(a) 0.72 (b) 0.54(c) 0.32 (d) 0.14

Ans. (c)Sol. P = 0.001 n = 2000

= np = 0.001 × 2000= 2

P(x > 2) = 1 – P (x < 2)= 1 – (P(x = 0)+P(x=1)

+ P(x = 2))=

0 – 1 2e e e101 11 21

= 0.323

73. As per de Broglie’s relationship, the wavelength related to its mass m and velocity v is

(a) hmv (b) hv

m

(c) hmv (d) mv

hWhere:h = Planck’s constant

Ans. (a)Sol. As per Einstein’s equation,

E = mc2

Where, E = energym = mass

and, c = speed lightAlso, as per Plank’s theory, every quantumof a wave has a discrete amount of energy.

E = hWhere, E = energy

h = Plank’s constant = frequency

As per de Broglie, particles and wave havethe same properties. So, he hypothesized thatthe two energies would be equal.

mc2 = hHowever, real particle cannot travel at thespeed of light, hence he replaced speed oflight (c) by velocity (v),

mv2 =hv

h = c=

mv2 =hv

mv =h

h=

mvi.e., option (a).

74. Which of the following statements regardingan atom are correct?

1. If two atoms with similar ionizationpotential form a bond, then this bond will mostprobably be either covalent or metallic.

2. When atoms with different ionizationpotentials form a bond, the bond will be mainlyionic.

3. If the atom or molecule already has itsouter shells completely full, then the bondingbetween the atoms or molecules will be asecondary bond when it solidfies.(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3

Ans. (d)Sol. It atoms have similar ionization potetial (or,

electro negativities i.e. same affinity forelectrons), covalent bonds are most likely tooccur. Since, both atoms have the same affinityfor electrons and neither have a tendency todonate them, they share electrons in order tofull-fill the octet configuration and becomestable.Probability of formation of ionic bond is highwhen there are atoms with different ionization





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potential, as ionic bond is formed by losing andgaining of electrons.It the atom or, molecule already has its outershells completely full, then the bonding betweenthe atoms or molecules will be a secondarybond when it solidifies. In such type of bonding,no major modification in the electronic structureoccurs.

75. A barium titanate crystal is inserted in a parllelplate condenser of area 10 mm × 10 mm. Theplates having a separation of 2 mm, give acapacitance of 10–9F. If the value of

12 10 8.854 10 Fm , the relative dielectric

constant of the crystal will be nearly(a) 2640 (b) 2450(c) 2260 (d) 2080

Ans. (c)Sol. For a parallel plate capacitor,

Capacitance, C = o r A

d

Where, o = permittivity of free space

r = relative dielectric constant

A = area of plated = separation between plates

Given, C = 10–9FA = 10 mm × 10 mm

= 100 × (10–3)2 m2

= 100 × 10–6 m2

= 10–4 m2

d = 2mm = 2 × 10–3 m

Since, C = o r A

d

r = o

C .dA

=–9

–3–12 –4

10 ×2×108.854×10 ×10

= 0.225886 × 104

= 2258.86 2260i.e., option (c).

76. A transformer core is wound with a coilcarrying an alternating current at a frequencyof 50 Hz. The hysteresis loop has an area of60,000 usnits when the axes are drawn in unitswhen the axes are drawn in units of 10–4 Wbm–2 and 102 Am–1. If the magnetization isuniform throughout the core volume of 0.01m3, the power loss due to hysteresis will be(a) 300 W (b) 350 W(c) 400 W (d)450 W

Ans. (a)Sol. Energy lost in hysteresis, per unit volume,

= Area of hysteresis loop= 60000 × 10–4 × 10+2 Jm–3

= 600 Jm–3

So, power loss due to hysteresis

Ph = Energy lost per unitvolume of core

time

=600 0.01 1T1 f

50

= 600 × 0.01 × 50= 300 W

i.e. Option (a)

77. When ferromagnetic of ferrimagnetic materialsare magnetized, the direction of magnetizationin any domain will be rotated from itspreferrential direction. This will show ananisotropic behaviou. On removal of themagnetizing force, the total magnetizetion willin general have a non-zero value. Thisbehaviour is due to(a) Crystal anisotropics(b) Stress anisotropics(c) Shape anisotropics(d) Crystal, stress and shape anisotropics

Ans. (d)Sol. When ferromagnetic or, Ferrimagnetic

materials are magnetized, the direction ofmagnetization in any domain will show ananisotropic behavior. On removal of themagnetizing force, the total magnetization will

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in general have a nonzero value. This behavioris due to crystal, stress and shape anisotropic.1. Crystal anisotropy:- Depending upon the

crystallographic orientation of the magneticmaterial, the magnetization reachessaturation at different fields.eg.[111] = Easy direction of magnetization.[100] = Hard direction of magnetization[110] = Intermediate direction ofmagnetization

2. Stress anisotropy:- Under a compressiveforce, the material can be magnetized mosteasily in the direction magnetization andunder tension they magnetize easily in adirection perpendicular to the direction ofmagnetization.

3. Shape anisotropy:- Ferromagnetic materialis magnetized most easily in the directionof its largest dimension.

78. The paramagnetic susceptibility varies inverselywith the absolute temperatrue for ordinaryfields and temperatures. It is given by therelation

x = C.T

The relation is known as(a) Phenomenon of magnetostriction(b) Curie law of parmagnetism(c) Hall Effect(d) Diamganetism

Ans. (b)Sol. Magnetic susceptibility of paramagnetic

material depends on temperature, as

m =

2m o

rNpM–1 = =

H kT

or, mC=T

Where, C = 2m oNpk

is called Curie

constant and this law is referred as CurieLaw.

79. If the interaction between the atomicpermanent dipole moments is zero or megligibleand the individual dipole moments are orientedat random, the material will be a(a) Ferromagnetic material(b) Ferrimagnetic material(c) Paramagnetic material(d) Antiferromegnetic material

Ans. (c)Sol. Magnetic materials which exhibit permanent

magnetism dipoles are: Paramagnetic,Ferromagnetic, antiferromagnetic andferrimagnetic. However, out of these, onlyparamagnetic materials are such type ofmagnetic material in which magnetic dipolesdo not interact with each other and arerandomly distributed in the absence of themagnetic field.

On the other hand, ferromagneticantiferromagnetic and ferrimagneticmaterials have dipole moments whichinteract between nighbouring dipoles.Dipole arrangements for different magneticmaterials are represented as:

Magnetic Materials Dipole arrangements

1. Paramagnetic2. Ferromagnetic

3. Antiferromagnetic4. Ferrimagnetic

80. The magnetic moments of diamagneticmaterials are mainly are mainly due to(a) Electron spin angular moment(b) Nuclear spin angular momentum(c) Orbital angular momentum of the

electrons(d) Centrifugal angular momentum

Ans. (c)Sol. — Diamagnetic materials have zero

resultant magnetic moment.— Electrons in an atom orbiting around

nucleus have orbital angularmomentum just like a current carryingloop.





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— When an external field is applied todiamagnetic material, those electronshaving orbital magnetic moment in thesame direction slows down and those inthe opposite direction speeds up, as perLenz’s Law. Thus, the diamagneticmaterial develops a net magneticmoment in the direction opposite to thatof the applied external magnetic field.

81. The inductance of an air-cored coil isproportional to

1. The square of the number of turns

2. The diameter of the coil.

3. A form factor, F, dependent on the ratio ofcoil radius to coil length plus windingdepth.

Which of the above statements are correct?(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3

Ans. (b)

Sol. Since, LI = N

L = B.A

N NI I

=BA

= H .ANI B = H

= A.HNI

However, as per Ampere’s Law,

H = NI

HI =

N

So, Inductance,

L = AHNI

= A.N

N

= A

2N

2NL =

S

Where, S = A

; reluctance

i.e., Inductance N2

A D2(or, square of diameter)

Hence, statement (2) is wrong.

82. Light is capable of transferring electrons tothe free- state inside a material thus increasingthe electrical conductivity of the material.When the energy imparted to the electrons isquite large, the latter may be emmited fromthe material into the surrounding medium. thisphenomenon is known as(a) Photoemissive effect(b) Photovoltaic effect(c) Photoconductivity effect(d) Photo absorptive effect

Ans. (a)Sol. Photoemissive effect: Electrons or, charge

carriers absorb light, get excited and ejectedout of the material.Photovoltaic effect: Electrons or, chargecarriers absorb light, get excited but stillcontained within the material.Photoconductivity effect: After absorbinglight, the number of free electrons or chargecarriers increases and hence leads to increaseelectrical conductivity.

83. Which of the following statements is/arecorrect?

1. Conductor contains a large number ofelectrons in the conduction band at roomtemperature. No energy gaps exist and thevalence and conduction bands overlap.

2. A semiconductor is a material is whichthe energy gap is so large that practicallyno electron can be given enough energy tojump this gap.

3. an insulator is a solid with an energy gapsmall enough for electrons to cross rathereasily from the valance band to theconduction band.





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(a) 1 only (b) 2 only(c) 3 only (d) 1, 2 and 3

Ans. (a)Sol. Conductors: Valence band and conduction

band overlap and there is no energy gapbetween them.Semiconductors: Energy gap between valenceband and conduction band is small, such that,the electrons will only require a small amountof energy to become excited and moved into theconduction band. Energy gap (Eg) forsemiconductor is normally 1 eV.Insulators: Energy gap between valence bandand conductor is higher than semiconductorand electrons do not leave the valence band. Ininsulator, passage of electrons from valenceband to conduction band may only be accidentaland occur because of some defect in thestructure of insulating material. Energy gapfor insulator is approx 6 eV.

84. Which of the following statements regardingsuperconducting materials are correct, when alarge number of metals becomesuperconducting below a temperature?

1. The resistivity of the superconductor iszero.

2. The magnetic flux density B vanishesthrough the substance.

3. Ferromagnetic and Antiferromagneticmetals are good examples ofsuperconducting materials

(a) 1, 2 and 3 (b) 1 and 3 only(c) 1 and 2 only (d) 2 and 3 only

Ans. (c)Sol. Properties of superconductors:

1. It possesses zero resistivity.2. Magnetic flux density (B) inside the

superconductor is zero i.e. superconductorsexpel the magnetic flux through it.

3. For superconductors,Magnetic flux density (Bin);

Bin = o H M

0 = o H M inB 0So, magnetic susceptibility

m =

M M 1H M

Negative susceptibility indicates that thesuperconductors are diamagnetic in nature.

85. A voltage source-Series resistance combinationis equivalent to a current source-parallelresistance combination if and only if their

1. Respective open-circuit voltage are equal.

2. Respective short-circuit currents are equal

3. Resistance remains same in both cases.

Which of the above statements are correct?(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3

Ans. (d)Sol. Voltage series resistance combination can be

considered as thevenin equivalent voltage andits resistance in series

a

b

Vth

Now, current source-parallel combination canbe considered an Nortan’s equivalent currentand resistance in parallel current and resistancein parallel

a

b

IN RN

Both circuit will be equivalent when(1) Open circuit voltage for both circuit will

be equali.e. Vth = INRN ...(i)

(2) Short circuit current for both circuit mustbe equal

i.e. th

th

VR for first current must be

Equal to IN i.e for the second circuit





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th

th

VR = IN

Vth = INRth ...(2)(3) Resistance must be same in both case

i.e RN = Rth

So equation (2) becomesVth = INRN i.e equation

So; Both circuit to be equivalent, all thesethree condition must be satisfied

86. For a network graph having its fundamentalloop matrix Bf and its sub-matrices Bt and Blcorresponding to twings and links, which ofthe following statements are correct?

1. Bl is always an identity matrix.

2. Bt is an identity matrix.

3. Bf has a rank of b – (n – 1), where b is thenumber of branches and n is the numberof nodes of the graph.

(a) 1 and 2 only (b) 2 and 3 only(c) 1 and 3 only (d) 1, 2 and 3

Ans. (c)Sol. Bf =[I Bt]

So Bt is not an identity matrix

Bf has fundamental loop of = b – n + 1which is no of rowIt has rank of (b – n + 1).So option (c) is the correct answer

87. The resistance R of a conductor is

(a) EAJl (b) EJ

Al

(c) EJA

l (d) JAEl

WhereE = Electric field intesityA = Cross-sectional areJ = Current densityI = Length of conductor

Ans. (c)

Sol. R =

L

a A

J = E

= 1 E

J

So R = L E E.LA J AJ

Option (c) is the correct answer

88. Which of the following statements are correctfor an ideal constant voltage source?

1. Its output voltage remains absolutelyconstant whateverthe change inloadcurrent.

2. It possesses zero internal resistance so thatinternal voltage drop in the source is zero.

3. Output voltage provided by the sourcewould remain constant irrespective of theamount of current drawn from it.

4. Output voltage provided by the sourcevaries with the amount of curren drawnfrom it.

(a) 1, 2 and 4 only (b) 1, 3 and 4 only(c) 2, 3 and 4 only (d) 1, 2 and 3 only

Ans. (d)Sol. Statement I is the correct

i.e. output voltage of ideal voltage source isirrespective of the current of the load.Ideal voltage has zero internal resistance i.e.output voltage = V = E – i r r = 0 for ideal voltage source

r

EV

r = internal resistance

So V E

So statement 3 is the correct answer





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So option (d) is correct i.e. statement 1, 2 and3 are correct.

89. Which of the following statemetns are correct?

1. A lowpass filter passes low frequencies andstop high frequecies.

2. A highpass filter passes high frequeniesand rejects low frequencies.

3. A band pass filter passes frequencies withinthe band and blocks/attenuates frequenciesoutside a frequency band.

4. A band stop filter passes frequencies withinthe band and blocks/attenuates frequenciesoutside a frequency band.


Ans. (d)Sol. A low pass filter passes low frequency and stops

high frequency

H( )

A high pass filter passes high frequency andrejects low frequency.

H( )

A band pass filterStatement-2 is correct

H( )

BandA band stop filter stops frequency within bandand passes outside the band so statement 4 isincorrect so

H( )

Band

Option (d) is the correct answer

90. A point charge of 10–9 C is placed at a point Ain the free space. The potential differencebetween the two points 20 cm 10 cm awayfrom the charge at A will be(a) 40 V (b) 45 V(c) 50 V (d) 55 V

Ans. (b)

Sol.A–9 C B C

10 cm 10 cm

The potential at point due to point

charge = o

Q4 r

So VB – VC =

0

Q Q4 AB 4 AC

= 2

Q 1 14 0.10 0.2

= 10–9 × 9 × 109 [10 – 5]= 45 V

Note:1

4 = 9 × 109

91. Which of the following statements regardingsteam boilers are correct ?1. The boiler must be capable of quick starting

and loading.2. The boiler should have no joints exposed to

flames.3. The boiler must be capable of burning low

ash content coal efficiently.(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3

Ans. (d)

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Sol. Quality required for boilers:1. Quick starting and loading2. Efficient burning of coal3. Proper design so that no joints are exposed

to flames.

92. Which of the following are the main parts of apower system ?1. Generating stations2. Transmission systems3. Distribution networks(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3

Ans. (d)Sol. A power system consists of

1. Generating stations2. Transmission systems3. Distribution systems

93. Which of the following factors affect Corona?1. Atmospheric conditions, temperature,

humidity, moisture, ice and fog2. Current of conductor3. Waveform4. Condition of surface of conductors,

smoothness and dust(a) 1, 2 and 3 only (b) 1, 3 and 4 only(c) 1, 3 and 4 only (d) 2, 3 and 4 only

Ans. (b)Sol. Corona loss in over head line is given by

PL = -5 f 25 r242×10d

(V –Vc)2

kW/km/phase

Where, = 3.92P273 T

= air density factorP = atmospheric pressureT = ambient temperaturef = frequency

V = phase voltageVc = critical disruptive voltage

r = conductor radiusd = spacing between conductors

Critical disruptive voltage is given by

Vc = v v rdm g nr

l

Where, mr = surface irregularity factorg = breakdown strength of

surrounding airHence, it can be seen that the corona isindependent of the current in the conductor.

94. Bundled conductors that are used to increaseline voltage in EHV lines for raising criticalcorona voltage depend on(a) Number of conductors in the group(b) Voltage gradient(c) Optimum spacing(d) Communication interference

Ans. (c)

95. In a 275 kV transmission line with lineconstants A = 0.85 5 and B = 200 75 , ifthe voltage profile at each end is to bemaintained at 275 kV, the power at UnityPower Factor (UPF) will be nearly(a) 98 MW (b) 118 MW(c) 144 MW (d) 184 MW

Ans. (b)Sol. |V | = 275S SR

|V | = 275 kVR

A = 0.85 5°b = 200

75°

SL

Load pf = 1 L = 0

LS = L L LP j P

RS = L LS P

PR = PL & R 0

PL = S RV Vcos

B





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2

RA Vcos

B

PR =

2275cos 75

200

0.85cos 75 5

PR =

2275cos 75 0.29

200 ...(i)

QR =

S RV Vsin

2RA V

sin

0 =

2275sin 75

200

0.85sin 75 5

0 = sin 75 0.8

75 = 53° = 75° – 53° = 22°So from equation (1)

PR =

2275cos 75 22 0.29

200= 117.9 MW

96. When the sinusoidal steady state current iscalled the symmetrical short circuit current,then the unidirectional transient component iscalled(a) AC short circuit current(b) DC short circuit current(c) AC offset current(d) DC offset current

Ans. (d)

Sol.

R Lv = V sin ( t + )max

Short circuit current is given by

i = I sin( t + – ) + I sin ( – )emax max –t

Steady statecurrent

Transientcurrent

Where, Imax = max max

22

V Vz R L

= 1 LtanR

= LR

The first term in above expression, thesteady state current is called thesymmetrical short-circuit current.

The second term, unidirectional transientcomponent is called the DC offset current.

This transient current causes the totalshort circuit current to be unsymmetricaluntil transient decays

97. Consider the following balanced line-to-neutralvoltages with abc sequence:

Vp =

an

bn

cn

V 277 0V 277 120 volts.

277 120V

The values of V0, V1 and V2 are respectively

(a) 0, 177 0 and 177 0

(b) 277 0 , 0 and 0

(c) 0, 277 0 and 0

(d) 277 0 0 and 177 0Ans. (c)Sol. Given, Phase voltages

VP =an

bn

cn

VVV

=

277 0277 120 Volts277 120

Since the phase voltage are balanced, thenegative and zero sequence components ofvoltage will be 0 and only positive sequencevoltage will be present.

Van = 0 1 2a a aV V V





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0aV = zero sequence voltage

= 02aV = negative sequence voltage

= 01aV = positive sequence voltage

Van = 1aV 277 0

98. In a HVDC transmission, the DC outputvoltage can be controlled to get inverteroperation when the firing angle is(a) 0 (b) 0 90(c) 90 180 (d) 0 180

Ans. (c)

Sol.

AC

+

–

V0 HVDC

I0

RectifierBridge

Voltage of the HVDC system (V0) is obtainedfrom the rectification of AC supply generatedfrom generating stations. For 3 phase full wave converter

V0 = mL3V cos

In inversion mode, power flow is inverted(i.e. P0 is negative). It is possible when V0is negative (negative current cannot flowthrough rectifier thyristors)

V0 = mL3V cos 0

cos 0

90 180

99. In an HVDC transmission mode, the link whichhas two circuits that are almost independentof each other is called(a) Monopolar link (b) Bipolar link(c) Homopolar link (d) Dualpolar link

Ans. (b)

Sol.

a

Load

L1

L2

dV2

dV2

In bipolar link with mid point grounded asshown, if there is fault etc in one of the link,it will remain continue to operate as monoplarlink (with earth return)

100. In a photovoltaic system, there is a thermallygenerated small reverse saturation currentwhich flows even in the absence of light, called(a) Photon current (b) Diode current(c) Leakage current (d) Dark current

Ans. (d)Sol. Dark current is the residual current flowing

in a photovoltaic device when there is noincident illumination (absence of light).

101. In a power system, due to interconnection orgrid formation and transmission lineredundancy, the ability to serve all powerdemands without failure over long periods oftime, is due to(a) Power system quality(b) Power system reliability(c) Computers and microprocessors(d) Reserve generating capacity

Ans. (b)Sol. In a power system, due to inter connection or

grid information, the ability to serve powerdemand over long periods without failureincreases, hence power system reliability isimproved.

102. In wind power, the speed which is consideredas the single most important parameter is(a) Wind speed (b) Peripheral speed(c) Tip speed (d) Blade speed

Ans. (d)





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Sol. The frequency of induced emf of generator isdecided by speed which is decided by blodespeed.

Tip speed ratio = Blade speedwind speed

103. Commutation circuitry is an extra circuit usedto turn off(a) Line-commutated thyristors(b) Phase-commutated thyristors(c) Forced-commutated thyristors(d) Reverse-commutated thyristors

Ans. (c)Sol. When load or line is not supporting the natural

turn off or when we want to switch-off thyristorforcefully than an intra circuit called commutedcircuitary is used to turn it off.

104. TRIAC as a bidirectional triode thyristor isused to control the output voltage by varyingconduction time or firing delay angle in(a) AC-DC converters (Controlled rectifiers)(b) AC-AC converters (AC voltage controllers)(c) DC-DC converters (DC choppers)(d) DC-AC converters (Inverters)

Ans. (b)Sol.

i0

R+

–V0

+

–Vs

TRIAC

2 Q

Vs

Vm

2 Q

V0

p

V (ac) can be controlled by controlling forming angle ofTrac

0

AC-DC converters (controlled rectifiers) Thyristor is usedAC-AC converters (AC Voltage controllers) Traic is usedDC-DC converters (DC Choppers) for highpower Thyristor is used for low power BJT is used.DC – AC converters (Inverters) Thyristors is used.As Triac is a directional switch so it canconduct both positive and negative currentsso it is used in AC-AC converters (AC VoltageControllers).

105. For large power output, multiphase rectifiersare used along with filters to reduce level ofharmonics by increasing the fundamentalfrequency in(a) Diode rectifier (b) Bridge rectifier(c) Star rectifier (d) Delta rectifier

Ans. (c)Sol. Multiphase rectifiers are connected in star

configuration to control the switching and asonly one device will conduct so non-symmetricalcurrent flows. To compansate this primary oftransformer is in delta connection.

106. In a Bipolar Junction Transistor (BJT) due tocurrent flow to small portion of the base, hotspots are produced causing localized excessiveheating and damaging the transistor. Thisswitching limit is called(a) Forward-Biased Safe Operating Area

(FBSOA)(b) Reverse-Biased Safe Operating Area

(RBSOA)(c) Power Derating(d) Second Breakdown (SB)

Ans. (d)Sol. Due to flow of large current, current clow to

small portion of the base, local hot spots areformed and large power dissipation takes place.This will cause damage to the BJT. This isknown as secondary breakdown of BJT.

Power Derating: Due to increase intemperature power dissipation capacity ofBJT reduces.





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107. In a three-phase inverter with 180° conduction,there are six modes of operation in a cyclewhere duration of each mode is(a) 90° (b) 75°(c) 60° (d) 45°

Ans. (c)

Sol. In 3 – inverters with 180° conduction, eachthyristor conducts for 180° of a cycle, but in 6modes in which three different thyristorsconducts.

6 modes 360°1 mode 60°So each mode conducts for 60°

108. In a closed-loop control of squirrel cageinduction motor, the field oriented controlstrategy implemented is(a) Scalar control(b) Vector control(c) Adaptive control(d) Frequency control

Ans. (b)Sol. Vector control and scalar control are two

different methods of variable frequency drive(VFD) controls.

Vector control is also called “Field OrientedControl (FOC)”.In vector control stator currents of a

3 – AC motor are identified as twoorthogonal components that can be visualisedas vectors.

Field Oriented Control

Direct Torque Control

Vector Control

109. In a DC motor drive, if the armature currentis reversed by keeping field current positiveproducing a braking torque, then the drive issaid to be operating in(a) Motoring mode(b) Regenerative braking mode(c) Dynamic braking mode(d) Plugging mode

Ans. (d)Sol. To brake the Speed, the torque developed ie.

Braking torque should be in the oppositedirection of rotation so, Torque is reversed byreversing Armature current (As arm inductanceis small).

N

TB

If armature current is reversed by keeping fieldcurrent positive than braking will occure. If aresistance is connected across the armatureterminal than dynamic braking mode willoccure and it a reversed voltage battery isconnected than plugging mode will occure.

110. If the induction motor drive is capable ofbidirectional power flow where limited range ofspeed control is required for large powerapplications, then this arrangement is called(a) Static conductance drive(b) Static Scherbius drive(c) Static compressive drive(d) Static reluctance drive

Ans. (b)Sol. Static scherbius drive is used to control the

speed both below and above synchronous speed.When speed is subsynchronous slip power isfeedback to supply. In supersynchronous speed,slip power is injected so we are havingbidirectional slip power.

111. In a DC-DC switched-mode converter, if theoutput voltage polarity is opposite to inputvoltage, then this inverting regulator is called(a) Buck regulator(b) Boost regulator(c) Buck-Boost regulator(d) Cuk regulator

Ans. (c or d)Sol. Buck convertor, Vo = DVin

Boost convertor, Vo = in1 V

1 – D





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Buck-Boost convertor, Vo = in–D V

1 – D

Cuk convertor, Vo = in–D V

1 – D

So, in Buck-Boost and Cuk convertor outputvoltage is of opposite polarity.

112. In a zero current switching resonant converter,the switching loss and noise are increased dueto presence of capacitive coupling called(a) Miller capacitor(b) Series resonant capacitor(c) parallel resonant capacitor(d) Switch capacitor

Ans. (a)Sol. ZCS can eliminate the switching losses at turn

off and reduce the switching loss at turn-ON.When power MOSFET are zero currentswitched-ON stored in the device’s capacitancewill be dissipated.

During turn-ON, considerable rate of changeof voltage can be coupled to the gate drivecircuit through the “Miller capacitor”, thusincreasing switching loss and noise.

ZVS eliminates the capacitive turn-ON loss.

113. Which one of the following devices is not aswitched-mode DC power supply?(a) Flyback forward converter(b) Full bridge converter(c) Push-pull converter(d) Resonant converter

Ans. (d)Sol. Flyback forward converter is equivalent to

Buck converter. It is a DC/DC converter.Full Bridge conterter: It converts full wave toDC.Push-Pull converter: It contrasts with Buck-Boost converters.Resonant converter: These are circuits whichemploy zero voltage and zero current switching.

114. The ideal core should exhibit very highpermeability in case of transformers andinductor core due to magnetic saturation causedby DC imbalance condition that can beminimized by(a) Low permeability core only(b) High permeability core only(c) Low and high permeability combination

core(d) No permeability core

Ans. (a)

Directions: Each of the next six (06) itemsconsists of two statements, one labelled as‘Statement (I)’ and the other labelled as‘Statement (II)’. You are to examine these twostatements carefully and select the answers tothese items using the codes given below:Codes:(a) Both statement (I) and Statement (II) are

individually true and Statement (II) is thecorrect explanation of Statement (I).

(b) Both Statement (I) and Statement (II) areindividually true, but Statement (II) is notthe correct explanation of Statement (I).

(c) Statement (I) is true, but Statement (II) isfalse.

(d) Statement (I) is false, but Statement (II) istrue.

115. Statement (I): In a substitutionalsemiconductor, atom is replaced by anoccasional foreign atom. The imperfections maybe deliberately controlled or created intransistor material.Statement (II): The lattice vacancies createdwhen certain atoms in a semiconductor aremissing are known as Schottky defects.

Ans. (b)Sol. Substitutional impurities replace the host

atoms e.g. B, AS, P are substitutionalimpurities in Si and replace Si atoms.This imperfection can be controlled deliberately,so, statement-I is correct.In schottky defect pair of oppositely chargedions leave their lattice sites, creating vacancies.so, statement-II is correct.





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But statement II don’t explain the cause ofstatement-I.

116. Statement (I): A cache is a memory unitplaced between the CPU and main memory Mand is used to store instructions, data or both.Statement (II): The cache’s effect is to increasethe average time required to access aninstruction or data word, typically to just asingle-clock cycle.

Ans. (c)Sol. A cache is a memory unit placed between the

CPU and main memory. It is used to storeinstruction and data that are used repeatedlyin the operation of programs. The cache’s effectreduces the average time required to accessinstruction or data word.

117. Statement (I): A buffer is not an area in RAMor on the hard drive designated to hold inputand output on their way in or out of the system.Statement (II): The process of placing items ina buffer so they can be retrieved by the appropriatedevice when needed is called spooling.

Ans. (d)Sol. The process in which data is temporarily held

is he buffer, which is special area in memoryor on a disk (hard drive) where a device canaccess them when it is ready.

118. Statement (I): The power diodes are three-layer devices.Statement (II): The impurity concentrationsof power diodes vary layer to layer.

Ans. (b)Sol. In power diode, to withstand large voltage a

less doped n layer is inserted between twoheavily doped P and N layers which is knownas drift region.

Anode (+)

Cathode (–)

P+

N–

N+

PN JunctionDrift Region

So, power diodes are three-layer devices andthe impurity concentrations of power diodesvary layer to layer. So, statement (II) is not anexplanation of statement (I).

119. Statement (I): Registers are used for storageof small data in the microprocessor.

Statement (II): All registers are accessible tothe user through instructions.

Ans. (c)Sol. In p registers are used for storage of data.

So, statement-I is true.All registers are not accessible to the userthrough instruction.Ex- W and Z are temporary register whichcan’t be access by instruction.So, statement-II is false.

120. Statement (I): In a three-phase inductionmotor, the maximum torque is directlyproportional to standstill reactance.Statement (II): In a three-phase inductionmotor, the speed or the slip at which maximumtorque occurs is determined by the rotorresistance.

Ans. (d)Sol. Maximum torque in a 3-phase induction motor

is given by

Tmax =

2

s 2

3 V2X

where, s = synchronous speed

V = supply voltage

2X = standstill leakage reactance of rotor

Maximum torque is inversely proportional tothe standstill reactance.Hence, statement-1 is incorrect. Slip for maximum torque is given by

max2 2

T2 2

R RSX X

where, R2 = rotor resistance.Therefore, statement-2 is correct.





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121. According to Gauss’s theorem, the surfaceintegral of the normal component of electricflux density D over a closed surface, containingfree charge is

(a) Q (b) 0

Q

(c) 0 Q (d)

2

0

Q

Ans. (a)Sol. Surface integral of D over a

closed surface = D.dS Q

[A/c to gauss theorem]So option (a) is the correct answer.

122. A unit magnetic pole may be defined as thatpole which when placed in vacuum at adistance of one metre from a similar and equalpole repels it with a force of

(a) 14 Newtons (b)

0

4 Newtons

(c)04 Newtons (d) 0

14 Newtons

Ans. (b)Sol. Since current loop is considered as magnetic

dipole.Since two dipole 1m apart

1m

F = BI sin

F = BI(here I = 1 given unit magnetic pole)

B =

o2

Id sin4 r 90

r = 1mId = 1 (given)

So F =

o2

Id I4 r

;

so

oF

4 Newton

So option (b) is the correct answer.

123. An analogous of magnetic circuit ‘permeability’in electrical circuit is(a) Reluctivity (b) Conductance(c) Conductivity (d) Resistivity

Ans. (c)Sol. Since permeability is ability or measure of

allowing or easiness for magnetic field inmagnetic path.So it must be analogous to conductivity inelectric circuit because it is the measure ofcasiness in electric circuit.

124. The magnetizing force at the centre of a circularcoil varies1. Directly as the number of its turns.2. Directly as the current.3. Directly as its radius.4. Inversely as its radius.Which of the above statements are correct?(a) 1, 2 and 3 only (b) 1 and 4 only(c) 1, 2 and 4 only (d) 2 and 3 only

Ans. (c)Sol. Magnetising force at the centre of coil flux

density at centre flux intensity (H) at thecentre

I

r

From ampere’s lawH.dl = NIH.2 r = NI

H =

NI2 r

force H N I

1r

So option (c) is the correct

125. An uncharged capacitor of 0.01 F is chargedfirst by a current of 2 mA for 30 s and then





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by a current of 4 mA for 30s. The final voltagein it will be(a) 12V (b) 18V(c) 24V (d) 30V

Ans. (b)Sol. First charging

vCO = initial capacitor voltage

v1 = 310 2 10 300.01

V1 = 100 × 2 × 10–3 × 30 = 6VV1 = Voltage after first charge that

becomes initial voltage forsecond charging

Secondary chargingSo V2(0) = 6V

v2 = 2 11 1V 0 idt V 0 i tC C

= 316 4 10 300.01

= 6 + 100 × 4 × 10–3 × 30

2V 18V

So option (b) is the correct answer

126. A capacitor of 10 pF is connected to a voltagesource of 100 V. If the distance between thecapacitor plates is reduced to 50%, while itremains connected to the 100 V supply, thevalue of potential gradient in the second casewill be(a) Half of earlier value(b) Same as earlier value(c) Twice of earlier value(d) One-fourth of earlier value

Ans. (c)Sol. A/c to the question

d = d2

Potential gradient = dV Vdx x

=V

d / 2

=2Vd

= 2×(Initial gradient)= Twice of earlier value

Assumption:Because earlier voltage V and distance d so

initial voltage gradient = Vd

So option (c) is the correct answer

127. Which of the following statements are correct?1. Accuracy is the closeness with which an

instrument approaches the true value ofthe quantity being measured.

2. Precision is a measure of thereproducibility of the measurement.

3. Precision of an instrument can be improvedupon by calibration.

4. Accuracy may be specified in terms of limitsof errors.


Ans. (c)Sol. Accuracy:- It is the closeness with which on

instrument reading approaches the true valueof the quantity being measured. It may bespecified in terms of limits of errors.Precision:- It is a measure of reproducibilityof the measurement i.e it is a degree ofagreement within a group of measured value.Calibration is the process of configuring oninstrument to provide a measured value withinan acceptable range. Accuracy can be improvedby calibration of instrument but Precisioncannot be improved by calibration ofinstrument.

128. An electrodynamometer instrument can beused as1. Wattmeter and VAR meter.2. Power factor meter and Frequency meter.3. Transfer instrument.Which of the above statements are correct?(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3

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Ans. (d)Sol. An electrodynamometer instrument can be

used as1. Wattmeter and VAR (reactive power)

meter.2. Power factor meter3. Frequency meter4. Transfer instrument.Transfer Instrument may be calibrated withdc source and then used without modificationto measure ac with same accuracy.Electrodynamometer instrument can be usedas a transfer instrument.

129. The moving iron instruments when measuringvoltages or currents.(a) Indicate the same values of the

measurement for both ascending anddescending values of current

(b) Indicate higher values of the measurementfor ascending value of current

(c) Indicate higher values of the measurementfor descending values of current

(d) Indicate lower values of the measurementfor both ascending and descending valuesof current

Ans. (c)Sol. The moving iron (MI) instruments when

measuring voltages or, currents indicatedifferent values of the measurement forascending and descending values of current.This error occurs because of hysteresis. Due tohysteresis property, the flux density is higherfor descending values of current and therefore,the instrument tends to read higher fordescending values.Hysteresis error can be minimized by makingthe iron parts small so that they demagnetizethemselves quick. It can also be reduced byoperating the iron parts of instruments atrelatively low values of flux density so that thehysteresis effects are small.

130. True RMS-reading voltmeter1. Measures the RMS value of voltage

accurately.2. Eliminates the error due to waveform.

3. Uses the thermocouple for heating.Which of the above statements are correct?(a) 1 and 2 only (b) 1 and 3 only(c) 2 and 3 only (d) 1, 2 and 3

Ans. (a)Sol. True RMS-reading volumeter uses some

property of a circuit element which is heatedby flow of current. Such type of voltmeter correctly measure

the rms value of voltage or, currentirrespective of the waveform and henceeliminates the error due to waveform.

It uses the thermocouple not for heatingbut to measure the temperature. There isa heater whose temperature is directlyproportional to the square of current beingmeasured.

I

Heater PMMC

ThermocoupleHeat developed, H = I2R.tWhere I = Current being

measured.R = Resistance of heating

elementt = time.

131. Instrument transformers are(a) used to extend the range of the AC

measuring instruments only(b) Used to isolate the measuring instruments

from the high voltage only(c) Used to extend the range and isolate the

measuring instruments(d) Not used at generating stations and

transformer stations.Ans. (c)Sol. Instrument transformers are used:

1. To extend the range of instrument, sothat current, voltage, power and energycan be measured with moderate size ofinstruments.





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In power system, currents and voltagesare very high, therefore, directmeasurement is not possible. In suchtype of measurement, we useinstrument trnasformer.

2. To isolate the measuring circuit fromthe power circuit. Since, secondarycircuit of an instrument transformer areisolated electrically from the primaryside, and hence safe to measure currentor, voltage at secondary side ofinstrument transformer.

Instrument transformers are used formetering as well as protection purpose inpower system.

132. The power in a 3-phase circuit is measuredwith the help of 2-wattmeters; the readings ofone of the wattmeters is positive and that ofthe other is negative. The magnitude ofreadings is different. It can be concluded thatthe power factor of the circuit will be(a) Unity (b) Zero(c) 0.5 (d) Less than 0.5

Ans. (d)Sol. The reading of the two wattmeters are:

P1 = 3VIcos 30° –

and P2 = 3VIcos 30° +Since, readings of one of the wattmeter ispositive and that of the other is negative,but magnitude are differentLet P1 reading is positive and P2 readingis negative.

P2 = 3VIcos 30° + <0

cos 30° + < 0

30° + > 90° > 60°

cos < 0.5

i.e., power factor of the circuit will be lessthan 0.5.

133. In a Q-meter, distributed capacitance of a coilis measured by changing the capacitance of

the tuning capacitor. The value of tuningcapacitor are C1 and C2 for resonant frequenciesf1 and 2f1 respectively. The value of distributedcapacitance will be

(a)1 2C C2 (b)

1 2C 2C3

(c)1 2C 4C3 (d)

1 2C 3C2

Ans. (c)Sol. For Q-meter, tunning capacitance is C1 at

resonant frequency f1 and that of C2 at resonantfrequency 2f1.

i.e., f1 = 1 d

12 L C + C ... (i)

and , 2f1 = 2 d

12 L C + C ... (ii)

From equation (i) & (ii), we get,

2. 1 d

12 L C + C = 2 d

12 L C + C

1 d

4C + C =

2 d

1C + C

4C2 + 4Cd = C1 + Cd 3Cd = C1 – 4C2

1 2d

C – 4CC =3 ;distributed

capacitance.

134. In a digital voltmeter, during start ofconversion, zero indication is displayed and iscalled auto zeroing. This is achieved by(a) Using a positive reference voltage(b) Using a negative reference voltage(c) Properly charging the differentiator circuit

capacitance to ground(d) Properly discharging the integrator circuit

capacitance to groundAns. (d)Sol. Digital voltmeter has three distinct features:

1. Auto-ranging2. Auto-zeroing3. Polarity detection





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Auto-zeroing: During start of conversionin a digital voltmeter, zero indication isdisplayed. This is called auto-zeroing. Thisis achieved by properly discharging theintegrator circuit capacitance to ground.

Vin

SR

C

During auto-zeroing, switch S is moved fromVin to ground so that the capacitance of thedual slope converter discharge to the ground.

135. A CRT has an anode voltage of 2000 V andparallel deflecting plates 2 cm long and 5 mmapart. The screen is 30 cm from the centre ofthe plates. If the input voltage is applied to thedeflecting plates through amplifiers having anoverall gain of 100, the input voltage requiredto deflect the beam through 3 cm will be(a) 1 V (b) 3 V(c) 5 V (d) 7 V

Ans. (a)Sol. Given,

Anode voltage, Ea = 2000 VLength of deflecting plates, ld = 2cm = 2 × 10–2 mDistance between plates, d = 5mm = 5 × 10–3mDistance of screen, L = 30 cm = 30 × 10–2m

Gain = 100D = 3 cm = 3 × 10–2

Voltage applied to deflecting plates,

Ed = a

d

2d E DL

=

3 2

2 22 5 10 2000 3 10

30 10 2 10So, input voltage required for a deflection,

= dEGain

=100100

= 1V

136. An aquadag is used in a CRO to collect(a) Primary electrons only(b) Secondary emission electrons only(c) Both primary electrons and secondary

emission electrons(d) heat emission electrons

Ans. (b)Sol. Secondary emission electrons:

The bombarding of electrons strike the screenwhich releases secondary emission electrons.These secondary electrons are collected by anaqueous solution of graphite called “Aquadag”,which is connected to the second anode. Thecollection of secondary electrons is necessaryto keep the CRT screen in a state of electricalequilibrium.

137. A resistance wire strain gauge with a gaugefactor of 2 is bonded to a steel structural membersubjected to a stress of 100 MN/m2. The modulusof elasticity of steel is 200 GN/m2. Thepercentage change in the value of the gaugeresistance due to the applied stress will be(a) 0.1% (b) 0.3%(c) 0.5% (d) 0.7%

Ans. (a)Sol. Given,

Gauge factor, Gf = 2Stress = 100 MN/m2 = 100 × 106 N/m2

= 200 × 109 N/m2

Since, Gauge factor

Gf =

R / R R / R

/ strainWe know,Modulus of elasticity

=stressstrain

strain,

= stress

=

6

9100 10200 10

= 0.5 × 10–3

So,RR = fG /

= 2 × 0.5 × 10–3 = 10–3

So, percentage change in the value of the gaugeresistance,





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R 100R = 10–3 × 100 = 0.1%

138. Capacitive transducers can be used for themeasurement of liquid level. The principle ofoperation used in this case is the change ofcapacitance with change of(a) Distance between plates(b) Area of plates(c) Dielectric(d) Resonance

Ans. (c)Sol. Capacitance transducers can be used for the

measurement of non-conducting liquid level.Capacitive transducers using the principle ofchange of capacitance with cahnge of dielectricare normally used for measurement of non-conducting liquid level.

139. The hexadecimal of the binary number(11010011)2 is(a) D316 (b) D416

(c) C316 (d) C416

Ans. (a)Sol. In Question hexadecimal equivalent of binary

number has been asked.Table shows the hexadecimal binary number:

Binary Hexadecimal0000 00001 10010 20011 30100 40101 50110 60111 71000 81001 91010 A1011 B1100 C1101 D1110 E1111 F

Given binary member is (1101 0011)2 fromtable hexadecimal of (1101 0011)2

3D

will be (D3)16.

option (a) is correct

140. Which one of the following relations from theBoolean algebra pertaining to ‘AND’ operationcannot be verified when A and B can take ononly the value 0 1 ?(a) AB = BA (b) AA = A(c) A1 = 1 (d) A0 = 0

Ans. (c)Sol. Boolean Agebra

AB = BA ...(i) LHS = AB = BA = RHSIt is correct according to commutative law ofmultiplication of Boolean variable.Similarly when B = A in the equation (1)

A.A = A2 = ASo it is also correctNow A.1 = ABut option (c) A.1 = 1 which is incorrectNow AO = O is also correctSo Option (c) is the correct option because wehave to tell which is not correct.

141. Which of the following design levels of acomputer are widely used in computer design?1. Gate level 2. Processor level3. Register level 4. User level(a) 1 and 3 only (b) 2 and 4 only(c) 3 and 4 only (d) 1, 2 and 3 only

Ans. (a)Sol. Register level design is geared toward the

processing of words as the basic limits ofinformation or signals. Gate level design:individual bits are treated as basic signals.

142. Which one of the following is a powerful webplatform for web applications and web services,built-in virtualization technologies, variety ofnew security tools, enhancements andstreamlined configuration and managementtools?





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(a) Internet Explorer(b) Internet Information Services(c) Web Matrix(d) Visual Web Developer

Ans. (c)Sol. Web matrix is an integrated environment

offered by many famous web hosting providers.It enables developers to build websites usingbuilt in templates with full support for ASPNet, PHP, Node.55 etc.

143. Which one of the following is the correctsequence of steps for executing an instrumentduring CPU’s processing ?(a) Fetch instruction, Read data, Decode

instruction, Store data and Executeinstruction

(b) Decode instruction, Read data, Executeinstruction, Fetch Next instruction andStore data

(c) Decode instruction, Decode Next operands,Fetch Next instruction, Execute instructionand Store data

(d) Fetch instruction, Decode instruction, readoperands, Execute instruction and Storedata

Ans. (d)Sol. Fetch instruction: Instruction will be fetched

from memory.Decode instruction: Instruction will bedecoded and calculate the effective address.Read operands: Fetch the operand frommemory.Execute instruction: Execute the instructionand store the result in the proper place.

144. Which one of the following is the correctcombination of registers in DMA controller?(a) Data register, Stack pointer and Data

counter(b) Data register, Address register and Data

counter(c) Data register, Stack pointer and Address

register(d) Data register, Program counter and Data

counter

Ans. (b)Sol. DMA controller consist of data register, address

register and data counter.

145. A multiprocessing technology which enablessoftware to treat a single processor as twoprocessors to utilize the processing power inthe chip that would otherwise go unused andlets the chip operate more efficiently resultingin faster processing is called(a) Systematic multiprocessing(b) Massively parallel processing(c) Co-processing(d) Hyper threading

Ans. (d)Sol. Hyper-threading is a technology used by some

intel microprocessor’s that allows a singlemicroprocessor to act like two separateprocessors to the operating system and theapplication program that use it.

146. A physical implementation of the typedeclaration in high-level programminglanguages where major information typesshould be assigned formats for identification iscalled(a) Storage order (b) Tag(c) Error correction (d) Error detection

Ans. (d)Sol. In the semantic analysis phase of compilation

the type error is detected before a program isrun. So the programmers create programs thatare more likely to run error-free in theirphysical implementation.

147. Which of the following factors are to beconsidered while selecting numberrepresentations to be used in computer?1. Number types to be represented2. Range of values to be encountered3. Cost of the hardware to store and process

the numbers4. Positional notation with fixed weight(a) 1, 2 and 4 only (b) 1, 3 and 4 only(c) 2, 3 and 4 only (d) 1, 2 and 3 only

Ans. (a)





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Sol. 1. A variables represent an area of memoryto hold a value. In a typed language suchas C, every variable must be declared witha type. The type tells the compiler aboutwhat we expect to store in a variable.

2. Range of values to be encountered.3. Positional rotation with fixed weight, value

of number is determined by multiplyingeach digit by a weight ad then summing.The weight of each digit is a power of thebase and is determined by position.

148. If a negative binary number is to be representedby n-bits, then the standard format will be(a) Sign bit ‘0’ on left and magnitude on right(b) Sign bit ‘1’ on left and magnitude on right(c) Sign bit ‘0’ on right and magnitude on left(d) Sign bit ‘1’ on right and magnitude on left

Ans. (b)Sol. In signed binary number first digit is sign

digit which is written on the left most andthen magnitude value is written on the rightof sign bit.Since, for negative number sign bit is 1 so, 1will be written first (i.e., left most) and theright of this sign bit magnitude will be written.So, option (b) is the correct answer.

149. The physical address translation in virtualmemory address with Memory ManagementUnit (MMU) is done by which one of thefollowing mechanisms?(a) Multiply virtual address by some constant(b) Translation lookaside buffer (TLB)(c) Encryption key(d) Using general purpose register in CPU

Ans. (b)Sol. The reference to the page table during address

translation from physical address to virtualaddress taken one memory cycle because pagetable resides in main memory. Translation lookaside buffer (TLB) is used to avoid this delay.

150. Which one of the following satellite system ismost often used for Global Positioning System(GPS)?(a) Geosynchronous(b) Geostationary(c) Low Earth Orbit(d) Medium Earth Orbit

Ans. (d)Sol. GPS satellite placed in medium earth orbit

(MEO)

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ELECTRICAL ENGINEERING · 2020-01-06 · (c) Shunt with inverting input (d) Shunt with non-inverting input Ans. (a) Sol. Opertational amplifier gives finite output even if input is