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Electrical Engineering (EE) Exam  >  Electrical Engineering (EE) Tests  >  GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Electrical Engineering (EE) MCQ

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test - GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23)

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) below.
Solutions of GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) questions in English are available as part of our course for Electrical Engineering (EE) & GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) solutions in Hindi for Electrical Engineering (EE) course. Download more important topics, notes, lectures and mock test series for Electrical Engineering (EE) Exam by signing up for free. Attempt GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) | 65 questions in 180 minutes | Mock test for Electrical Engineering (EE) preparation | Free important questions MCQ to study for Electrical Engineering (EE) Exam | Download free PDF with solutions
GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 1
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Directions: In the following question five figures are given. Four of them are similar in some way but one figure is not like the other four. Point out which figure does not belong to the group.

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 1
The black ball is moving in clockwise direction, so it should be on the right side of the figure in option (e).
GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 2
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Choose the appropriate word/phrase to complete the sentence :

The _________ of skirt made it easy for the children to understand that Aunt Polly was in the corridor.

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 2
The correct word that goes with skirt is “Swishing” which is the sound made by the cloth. The other words are not sounds and hence will not tell if the person is in the corridor.Thus ‘B’ is the correct answer.
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GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 3
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The number lock of a suitcase has 4 wheels, each labelled with ten digits, i.e. from 0 to 9. The lock opens with a sequence of four digits with no repeats. What is the probability of a person getting the sequence to open the suitcase?

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 3
There are 10C4 x 4! = 5040 sequences of 4 distinct digits, out of which there is only one sequence in which the lock opens.

Required probability = 1/5040

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 4
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In DPS Noida the number of students who are opting for various subjects are given below.

In which of the two subjects, number of students is more than 35% and less than 40% of total students in Noida?
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 4
Total number of students in the college

= 847+1158+734+781+1102+923 = 5545

Total number of students opting for Maths and Chemistry = 1158+1102 = 2260

Hence, required % =

Total number of students opting for Hindi and Physics = 781+847=1628

Hence, required % =

Total number of students opting for History and Maths = 1158 + 923 = 2081

Hence, required % =

Total number of students opting for Hindi and Art = 781 + 734 =1515

Hence, required % =

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 5
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Directions: The table lists the size of building lots in the Orange Grove subdivision and the people who are planning to build on those lots. For each lot, installation of utilities costs $12,516. The city charges impact fees of $3,879 per lot. There are also development fees of 16.15 cents per square foot of land.

What approximate percentage is the area of the smallest lot listed, as compared to the area of the largest lot?
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 5
The smallest lot is 6,699 square feet and the largest lot is 9,004 square feet.

Required percentage = (6699/9004) × 100 = 0.744 × 100 ≈ 75%

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 6
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Which of the following combination is incorrect?
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 6
Option A has words which are opposite in meaning. ‘imbecile’ means a stupid person and ‘insightful’ means someone who is wise and all the rest options are having similar meaning.
GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 7
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A reservoir will be filled in 12 hours if two pipes function simultaneously. The second pipe fills the reservoir 10 hours faster than the first. How many hours will the second pipe take to fill the reservoir?
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 7
Let the first pipe fill the reservoir in x hours.

Then, the second pipe can fill the reservoir in x – 10 hours.

Now, according to the question,

1/x + 1/(x-10) = 1/12

⇒ 12(x – 10 + x) = x2 – 10x

⇒24x – 120 = x2 – 10x

⇒x2 – 34x + 120 = 0

⇒x2 – 30x - 4x + 120 = 0

⇒(x – 30)(x – 4) = 0

⇒x = 30, 4

But, x ≠ 4 (∵ x – 10 should be positive)

∴ x = 30 hours

x – 10 = 20 hours

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 8
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If x+y = 45.6, then the value of (x-25.2)3 + (y-20.4)3 is
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 8
Given

x+y = 45.6

Method 1 :

We can write the above given equation as

x + y = 25.2 + 20.4

So, (x - 25.2) + (y - 20.4) = 0…...(i)

Cubing both the sides and using the formula

We have,(x - 25.2)3 + (y - 20.4)3 + 3(x - 25.2)(y - 20.4){x - 25.2) + (y - 20.4)} = 0

Substituting value from (i), we have

(x - 25.2)3 + (y - 20.4)3 + 0 = 0

(x - 25.2)3 + (y - 20.4)3 = 0

Hence, the value of is(x - 25.2)3 + (y - 20.4)3 is 0

Method 2 :

x+y = 45.6

To find(x - 25.2)3 + (y - 20.4)3

Putting value of x from equation (i),

(20.4 - y)3 + (y - 20.4)3

= (y - 20.4)3 - (y - 20.4)3 = 0

Thus, the value of(x - 25.2)3 + (y - 20.4)3 is 0.

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 9
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If prices are reduced by 20% and sales are increased by 15%, what is the net effect on gross receipts?
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 9
Let original price = p, and original sales = s.

Therefore, original gross receipts = ps. Let new price = 0.80p, and new sales = 1.15s.

Therefore, new gross receipts = 0.92ps. Gross receipts are only 92% of what they were.

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 10
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In the question below, a part of the sentence is underlined. Below are given alternatives to the underlined part which may improve the sentence. Choose the correct alternative. In case no improvement is needed choose No improvement.

Why do you think this course is necessary for the students?
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 10
The sentence does not need any improvement.
GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 11
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A segment of a circuit is shown in the figure. Vr= 5 V, Vc = 4 sin 2t .The voltage VL is given by
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 11
Applying KCL at the centre node

( current through inductor)

Voltage across inductor

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 12
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If is the Fourier transform of g(t), then g(t) is
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 12
Given

Conversion of frequency domain signal in time domain is given by

By using property,

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 13
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A 3-phase, 10,000 kVA, 11 kV alternator has a subtransient reactance of 8%. A 3-phase short-circuit occurs at its terminals. The fault MVA and fault current are
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 13

Alternator percentage reactance is based on its own voltage and kVA ratings.

Let us choose 10,000 kVA as base kVA and 11 kV as base kV

Base current IB =

Per unit fault current Isc pu =

Fault current ISC = ISC x IB = 12.5 x 525 = 6.562 kA

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 14
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Consider the following circuit shown in below figure.

If V0 = 10.3V, then Zener voltage(VZ) is __________ Volts. (rounded upto one decimal place)

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 14
Given : V0 = 10.3V

By concept of virtual ground,

Vx = Ei = -5 V

Apply KVL at output side,

5 - VZ + V0 = 0

VZ = Vo +5

Vz = 15.3 V

Hence, the Zener voltage is 15.3 V.

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 15
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Two generators G1 and G2 are to be operated in parallel to deliver a total current of 60 A. Generator G1 has a terminal voltage of 280 V on no-load and 240 V when supplying 40 A current. Similarly, the second generator has a voltage of 300 V, when on no-load and 240 V, when supplying 50 A of current.

The output voltage of each generator will be
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 15
Generator G1

Voltage drop per ampere = (280-240)/40 = 1 A

Let output current is IG1, then output voltage will be

VG1 = 280 - 1 x IG1

Generator G2

Voltage drop per ampere = (300-240)/50 = 1.2 A

Let output current is IG2, then output voltage will be

VG2 = 300 - 1.2 x IG2

For parallel operation, output of each generator would be same i.e.

So, Total current delivered

Solving equation (i) and (ii) , we get

So the output voltage

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 16
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The order of differential equation is __________. (in integer)

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 16
Given

Order of the given differential equation is 2

Hence, the correct answer is 2

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 17
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A three phase bridge inverter is fed from a 500 V D.C. source. The inverter is operated at 180°C conduction mode and it is supplying a purely resistive, star-connected load. The RMS value (in V) of the output (line) voltage is (Answer up to the nearest integer)

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 17
For a three phase bridge inverter, rms value of output line voltage

Vline(rms) = √2/3 Vdc

Vdc = 500 V

= 0.816 × 500 = 408 V

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 18
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Consider the root locus plot of the unity negative feedback system shown in figure below, the range of K for the system to be stable is
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 18
From pole zero diagram, the transfer function is of form

Characteristics equation is given by

1 + G(s) = 0

(s - 5) + K(s + 5) = 0

s - 5 + Ks + 5K = 0

(1 + K)s + (5K - 5) = 0

For the system to be stable

1+K>0 and 5K-5>0

K>-1 andK>1

Combining both the conditions

K>1

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 19
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Trigonometric Fourier series for a periodic function shown in figure below is given as

The constant k, an and bn are
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 19

n≥1(integer)

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 20
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The circuit is used to produce a square wave. If switching times for multiplexers are1μs,2μs , 2μs and 2μs as shown in figure, then the frequency (f) and duty cycle (D) of V0 are respectively
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 20
A 2x1 MUX with Io =1 and I1 = 1 will give output y = Ā as shown in figure,

So, all the MUX in the given circuit are working as NOT gate, in which only first three MUX will decide the time period as shown below,

T = 10μs

f = 106/10 = 100kHz

D = 5/10 = 0.5

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 21
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A 20-MVA, 6.6-kV, 3-phase alternator is connected to a 3-phase transmission line. The per unit positive-sequence, negative-sequence and zero-sequence impedances of the alternator are j0.1, j0.1 and j0.04, respectively. The neutral of the alternator is connected to ground through an inductive reactor of j0.05 p.u. The per unit positive, negative and zero-sequence impedances of the transmission line are j0.1, j0.1 and j0.3, respectively. All the per unit values are based on the machine ratings. A solid ground fault occurs at one phase of the far end of the transmission line. The voltage (in V) of the alternator neutral with respect to ground during the fault is (Answer up to one decimal place)

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 21
Total zero sequence impedance, +ve sequence impedance and -ve sequence

Impedance

Fault current

Neutral voltage

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 22
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Consider a function , then the value of where c is circle defined by

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 22
Given

and

Poles of f(z) are given by, z2 + 2z + 5 = 0

Both poles lie outside the circle

Hence, by Cauchy’s theorem .

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 23
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The graph of a function f(x) is shown in the figure.

For f(x) to be a valid probability density function, the value of h is x/3. Find the value of x. (Answer up to the nearest integer)

Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 23
Since f(x) is a valid probability density function

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 24
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Consider a combinational circuit shown in below figure.

The circuit behaves as
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 24
If V=0 then B8, B4, B2 and B1 will appear at the outputs of XOR gates and circuit will act as adder.

If V=1 then output of XOR gates will be B8, B4, B2, B1 then the circuit will act as subtractor.

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 25
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Solution of the differential equation is
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 25

Put tan y = v

[Linear equation]

v.I.F =

I.F =

⇒ v.x2 =

v.x2 = x5/5 + C

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 26
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A control system is represented by differential equation where u(t) is the input signal. The order of the state and output matrices are
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 26
Order of differential equation number of state variables = n=2

Number of input = m = 1(only u(t))

Number of output = p = 1(only y(t))

The state model is written as

Hence, order of state matrix = n✖n = 2x2

Order of output matrix = p✖n = 1✖2

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 27
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Stability of a power system can be improved by:

1. Using series compensators

2. Using parallel transmission lines

3. Reducing voltage of transmission

Which of the above statements is/are correct?
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 27
Stability of a power system can be improved by

(i) higher system voltage

(ii) the use of parallel lines to reduce the series reactance

(iii) the use of high speed circuit breakers and auto-reclosing breakers

(iv) reducing the series reactance thereby increasing Pm which increases the transient stability limit of a system

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 28
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The rotor of a 3-phase induction motor has a resistance per phase of 0.04 ohm. This motor has a 0.2-ohm standstill reactance per phase. Neglect stator resistance. If the external resistance 0.02Ω is added to the circuit then what is the percentage improvement in p.f. during starting?
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 28
Given: R2 = 0.04Ω

R2(external) = 0.02Ω, X2 = 0.2Ω

At starting power factor is given by, p.f.

Without external resistance p.f.

with external resistance p.f.

So, percentage improvement in p.f.

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 29
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Which of the following conditions must be satisfied for a transistor to be in saturation?

1. Its collector to base junction should be under forward bias.

2. Its collector to base junction should be under reverse bias.

3. Its emitter to base junction should be under reverse bias.

4. Its emitter to base junction should be under forward bias.

Which of the above conditions are true?
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 29
For a transistor to remain under saturation, both collector-to-base junction and emitter-to-base junction should be forward biased.

For a transistor to remain under cutoff, both junctions should be under reverse bias.

For a transistor to remain under active region, emitter-base junction should be forward biased and collector-base junction should be reverse biased.

GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 30
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The type of feedback used in collector to base bias circuit is
Detailed Solution for GATE Mock Test: Electrical Engineering (EE)- 7 (31-01-23) - Question 30
Collector to base bias circuit is shown below.

Sampling → Voltage (shunt)

Mixing → Current (Shunt)

Hence, shunt-shunt topology.

Remark :

When feedback is directly connected to output → voltage (shunt) sampling.

The feedback element is connected in shunted form to the input → current (shunt) mixing.

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