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Gate Practice Test: Electrical Engineering(EE)- 13 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test - Gate Practice Test: Electrical Engineering(EE)- 13

Gate Practice Test: Electrical Engineering(EE)- 13 for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The Gate Practice Test: Electrical Engineering(EE)- 13 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The Gate Practice Test: Electrical Engineering(EE)- 13 MCQs are made for Electrical Engineering (EE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Gate Practice Test: Electrical Engineering(EE)- 13 below.
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Gate Practice Test: Electrical Engineering(EE)- 13 - Question 1

There are two lines made by joining points A, B, C. B lies between the line joining A and C. Is the distance between the A and C passes through the B more then 7 Km.

I. The distance between A and B is 6 Km

II. Distance between B to C is 7 Km long,

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 1
From Statement I

We know the distance between Points B and C but we do not know about the distance between A and B

Statement II

This statement tells us about the distance between A and B and by using Both statement we can calculate the distance between A and C that is longer than the 7 Km.

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 2

Comprehension:

Direction: In the given question, a word/phrase is given followed by three statements; I, II and III. Choose the pair of sentences which can be combined using the given word/ phrase when used at the beginning of the new sentence.

And

I: When Lionel Messi, Ronaldo and Neymar kick and dribble the football, they don’t remain confined to Argentina, Portugal or Brazil, respectively.

II: Let the entire world realise that through sports, especially football, we all can cement our bonds, wash away our bitterness and prejudices.

III: Live with a greater sense of closeness and joy with each other.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 2
‘And’ is used as a conjunction here which is used to introduce an additional comment or interjection.

Here sentences II and III are displaying same sense that is realizing the benefit of football in our lives whereas sentence I is providing entirely different aspect of football as a sport.

Hence option (B) will be the most appropriate choice.

New sentence: Let the entire world realise that through sports, especially football, we all can cement our bonds, wash away our bitterness and prejudices, and live with a greater sense of closeness and joy with each other.

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Gate Practice Test: Electrical Engineering(EE)- 13 - Question 3

Which of the following is the MOST SIMILAR in meaning to Accreditation?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 3
Accreditation = an acknowledgement of a person's responsibility for or achievement of something.

Certification = an official document attesting to a status or level of achievement.

Meticulous = showing great attention to detail; very careful and precise.

Lurid = very vividly shocking.

Suppressive = tending or acting to suppress.

Agreement = harmony or accordance in opinion or feeling.

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 4

In how many ways can you place 2 white bishops on an empty chess board?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 4

One white bishop can be placed on any of the 32 white boxes and the other white bishop can be placed on any of the 32 black boxes.

Ways of doing that = 32C1 × 32C1

= 32 x 32 = 1024

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 5

Comprehension:

Direction: In the question below are given statements followed by some conclusions. You have to take the given statements to be true even if they seem to be at variance with commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts.

Question:

Statement:

No physics is maths.

Some chemistry is maths.

All sciences are chemistry.

Conclusion:

I. No science is physics.

II. Some physics are science.

III. Some physics are chemistry.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 5
The least possible Venn diagram for the given statements is as follows.

Conclusions:

I. No science is physics → False (It is possible but not definite).

II. Some physics are science → False (It is possible but not definite)

III. Some physics are chemistry → False (It is possible but not definite)

Conclusion I and II form complementary pair.

Hence, either conclusion I or II follows.

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 6

Comprehension:

Direction: Read the information carefully and give the answer of the following questions-

(This pie chart shows the percentage of students appear in six different exams in 2016)

Question:

This bar graph shows the percentage of failed students which appear in these six different exams in 2016)

If in 2016, total number of failed students in exam F was 4080, then how many passed students appears in the exam B?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 6
14280 ( Let total number of students be = 100z.

Total number of students appears in exam F = 16% of 100z = 16z;

Total number of failed students in exam F = 30% of 16z = 4.8z = 4080;

=> z = 850;

Total number of passed students appears in the exam B = (100 – 30)% of 24% of 100z = 16.8z = 16.8 × 850 = 14280 ;

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 7

Comprehension:

Direction: Two sentences with two blanks in each, followed by five alternatives with two words in each, are given. Choose that option as the answer which can fill both the blanks of both the sentences.

Question:

i. A sinking feeling of panic ________ over them and a temporary paralyzing fear engulfed them ________.

ii. Being a cleanliness freak, she ________ the floor and went down to the market only after the house was _________ clean.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 7
‘Sweep’ is to clean (an area) by brushing away dirt or litter. It fits the first blank of the second sentence, as the subject is cleaning the floor. Further, ‘sweep over’ means to overcome or overwhelm, thus fits in the first blank of the first sentence. The second blank of both the sentences can be filled by "completely", thus conveying an appropriate sense.

Thus, option D is the correct answer.

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 8

Comprehension:

Direction: Five statements are given below, labelled 1, 2, 3, 4 and 5 which are supposed to be in a logical order. A statement labelled P is given thereafter. P can replace one of the five statements such that the four statements along with P would make a coherent paragraph. You have to identify which statement should P replace and then the find out the correct sequence from the options. If the five options are in logical order and form a coherent paragraph/passage, choose the fifth option “12345”.

Question:

1) The future of Germany’s coalition government is hanging in the balance after the country’s interior minister reportedly announced his intention to resign over a migration showdown with Angela Merkel.

2) Horst Seehofer, who is also leader of the Christian Social Union, on Sunday night offered to step down from his ministerial role and party leadership in a closed-door meeting in which he and fellow CSU leaders had debated the merits of the migration deal Merkel hammered out with fellow European Union leaders in Brussels.

3) But with CSU hardliners believed to have tried to talk the combative interior minister into staying, a press conference was postponed until Monday, with Seehofer seeking to go back to Merkel in search of a final compromise.

4) At a 2am media conference, Seehofer said he had agreed to meet again with Merkel’s party before he made his decision final.

5) “We’ll have more talks today with the CDU in Berlin with the hope that we can come to an agreement,” Seehofer said.

P. Merkel said on Sunday she wanted the CDU and its Bavarian allies to continue working together.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 8

The paragraph talks about the issues pertaining to the state of coalition government in Germany. The first sentence introduces the issue to us. The second sentence explains the problem. The third sentence then goes on to elaborate upon the meeting, and the fourth and fifth sentence talk about the current state and measures being taken. Thus, they are in perfect order. Sentence P is completely out of place here because CDU is not the subject of this paragraph and it does not matter what Merkel thinks they should do.

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 9

4 identical solid spheres are melted and re-formed into a solid hemisphere. Then, the ratio of the curved surface area of the hemisphere to half of the surface area of a single sphere is -

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 9
Let radius of sphere be ‘r’ and radius of hemisphere be ‘R’.

Then, ATQ,

Volume of Spheres = Volume of Hemisphere

4 x (4/3)πr3 = (2/3)πR3

Or, R / r = (8)1/3 = 2 …(1)

C.S.A of hemisphere = 2πR2

And, Surface area of sphere = 4πr2

ATQ,

2πR2 :( ½) of 4πr2

= 2πR2 :2πr2 = R2 :r2 {using (1)}

= 4 :1

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 10

A study of people who reduced the calories they consumed has found the strongest evidence yet that such restrictions slow down metabolism, raising hopes that a low calories lifestyle or treatments stimulating biological effects of restricted eating, could prolong health in old age. The report provides the most robust evidence to date that everything we have learnt in other animals can be applied to humans.

Which of the following argument will prove that the above conclusion is flawed?

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 11

A single core cable has a conductor diameter of 2 cm and an overall diameter of 4 cm. If the resistivity of the insulating material is 8×108 megohm-cm while the length of the cable is 2km then calculate the insulation resistance of this cable?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 11
L= 2 km= 2×105 cm

r1= d1/2 = 1 cm

r2 = 2 cm

ρ= 8×108 megohm-cm = 8×1014 ohm-cm

(r2/r1) = 2/1 = 2

Rinsulation = (2.3ρ/2πl)× log(r2/r1)

= (2.3×8×1014/2π×2×105) × log 2

= 440.7097 MΩ ≈ 440.71 MΩ

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 12

Which of the following conditions are correct for the convergence of the fourier transform of the function x(t).

1) x(t) is absolutely integrable over the range of time period.

2) x(t) should have finite number of discontinuities within any finite interval.

3) x(t) should have finite number of maxima and minima within any finite interval.

4) x(t) must be periodic.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 12
Dirichlet’s Conditions

The conditions under which a periodic function f(t) can be expanded in a convergent Fourier Series, are known as Dirichlet’s conditions. These are as follows:

i) f(t) is a single-valued function.

ii) f(t) has a finite number of discontinuities in each period T.

iii) f(t) has a finite number of Maxima & Minima in each period T.

iv) The Integral exists and is finite or in other way,

1, 2 & 3 are the Dirichlet conditions.

Periodicity is not the necessary condition for the convergence of Fourier transform. Since an a periodic signal may be looked at as a periodic signal with an infinite period. Note what follows is not a mathematically rigorous exercise, but will help develop an intuition for the Fourier Transform for a periodic signals.

*Answer can only contain numeric values
Gate Practice Test: Electrical Engineering(EE)- 13 - Question 13

Find the series equivalent inductance of the network that causes an opposite angle (Hays bridge) to null with the following bridge arms.

ω = 3000 rad/sec, R2 = 10 kΩ

R1 = 2 kΩ, C1 = 1 μF, R3 = 1 kΩ

The value of Lx in Henry will be


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 13

From bridge balance condition

Rad . Rbc = Rab . Rcd

Separating real and imaginary part, we have

By putting the values in above equation.

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 14

The Taylor series expansion of 3 sin x + 2 cos x is

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 14

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 15

The above circuit shown above will behave as:

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 15
For positive half cycle.

When (vi > 10V)

D will be ON

∴ V0 = 10V

So, the circuit will behave as a peak clipper.

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 16

The loop transfer function of a system is given by The total number of root loci is:

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 16
Total no of root loci = max(no of poles, no of zeros)=max(4,5)=5.

Thus, option D is the correct answer.

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 17

Given below is arrangement of different counters the input frequency of clock is f0 and the output frequency is f0/n. The value of n will be

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 17
Frequency at output of counter =

For n bit ring counter no. of states = 2n

For n bit synchronous counter no. of states =2n

For n bit Johnson counter no. of state =2n

∴ Frequency at output

N = 256

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 18

The no load voltage of a transformer is 440 V and its load current is 10 A. If the equivalent resistance & reactance are 0.12 Ω and 2.42 Ω and is operating at a power factor of 0.8 lagging, then find the % voltage regulation of the transformer.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 18
ϕ=cos−1⁡0.8=36.86

Full load alg. can be calculated as-

V2=E2−I2∠⟶×(R2+jX2)=440−10∠−36.86(12+j2.42)=V2=424.93∠−2.51∘V

Required equation

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 19

A p-channel JFET has VP=4V and IDSS=12mA. It is used in the circuit of figure given below with VDD=12V. Determine RD and RSso that ID=4mA and VDS = 6V

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 19

Given, VGS=ISRS=IORS...........(1)

Substituting value in equations (1) and (2)

VGS = 4Rs

Solving we get Rs =1.58kΩ or 0.42kΩ

(smaller value selected)

KVL for OD loop−12−6+(RD+RS)=0

R0=6.08KΩ

Rs = 0.22KΩ

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 20

Which of the following plot represents the magnitude of transfer function versus frequency of the circuit shown in figure.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 20

Transfer function of the circuit is given as following

This is high pass filter, so the response is shown in figure below

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 21

For the flip flop configuration as shown above for clock input to T flip flop. The mod of the above flip flop configuration will be _______.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 21

For D flip flops. Q2 and Q1 are outputs and clock Q0 from T flip flop is applied.

∴ Truth table for D flip flop:

So mod of D flip flops will be 4

The mod of T flip flop is 2.

So equivalent mod = 2 × 4 = -8= 8

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 22

A compound DC generator is supplying a load and working as cumulative generator. If now machine behave as a motor, it will behave as:

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 22
Cumulative compound DC generator

As it is cumulative compound gen, that means series flux is in phase with shunt flux.

Now it runs as a motor.

Shunt field is same, but series flux opposes.

Hence, series flux opposes shunt flux.

It will run as differentially compound motor.

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 23

Which one is increased when of negative feedback voltage is applied to an amplifier?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 23

Feedback reduces the overall gain of the system negative feedback increases the input impedance and decreases the output impedance.

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 24

A circular disc of radius 5 m with a charge density of ρs = 12 sinϕ μC/m2 is enclosed by a surface S. The net flux crossing the S is.

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 24

The net flux crossing,

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 25

For IOL (max) = 16mA, IIL(max) = 0.0016 A. The fan-out (low) will be

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 25

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 26

Two continuous time system is described as x1 (t)=te-t u(t) and x2 (t)=te-2t u(t), then the convolution of x1(t) & x2(t) will be?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 26

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 27

What will be the time constant of the given circuit:

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 27
Time constant =

The circuit can be redraw as

Leq = [(2||2) + 2] H = 3 H

To find Req,

• Open circuit current source

• Short circuit voltage source

Find out Req across Leq:

Req=10Ω

Time constant =

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 28

Which feedback topology is employed in the amplifier circuit below?

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 28
In the given amplifier circuit, RE’ provides the negative feedback. RE’ is not directly connected to output node, hence current sampling and it is not directly connected to input node, hence series mixing.

Thus, option C is the correct answer.

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 29

Boolean expression , its maxterm form is

Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 29
Min term expression

Using De Morgan's law

Gate Practice Test: Electrical Engineering(EE)- 13 - Question 30

For a unity negative feedback system with the forward transfer function Find the sum of maximum and minimum value of K to make the system stable.


Detailed Solution for Gate Practice Test: Electrical Engineering(EE)- 13 - Question 30
The transfer function in the following case is defined as

On substituting the value of G(s), we have T(s) = On simplifying we have

Applying R-H criteria,

First of all k > 0 for R - H criteria to be valid

Clearly, 6 − 3k > 0 in order for the system, to be stable. K < />

Thus, sum of maximum and minimum value of k is = 2

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