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GATE Practice Test: Electrical Engineering (EE)- 1 - Electrical Engineering (EE) MCQ


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30 Questions MCQ Test - GATE Practice Test: Electrical Engineering (EE)- 1

GATE Practice Test: Electrical Engineering (EE)- 1 for Electrical Engineering (EE) 2024 is part of Electrical Engineering (EE) preparation. The GATE Practice Test: Electrical Engineering (EE)- 1 questions and answers have been prepared according to the Electrical Engineering (EE) exam syllabus.The GATE Practice Test: Electrical Engineering (EE)- 1 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)- 1 below.
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GATE Practice Test: Electrical Engineering (EE)- 1 - Question 1

Choose the most approximate word from the options given below to complete the following sentence.

If I had known that you were coming, I _______ you at the airport

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 1

The sentence talks about a past incident and a condition has been mentioned. Thus option 2 fits here correctly. 'Would have' is the correct tense to be used here. Option 1 is illogical. 'Would' must be used here and not 'will' as it is in the past tense. Past perfect tense is incorrect here.

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 2

Choose the most approximate word from the options given below to complete the following sentence.

I believe in the _____ of positive thinking thus always recommend these books to my clients.

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 2

The word that fits here is a noun thus options 3 and 4 are eliminated. The form should be singular as for abstract nouns we do not use the plural form. Option 1 is thus the correct answer.

The word powerful is an adjective and 'empowering' is a verb.

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GATE Practice Test: Electrical Engineering (EE)- 1 - Question 3

Consider a circle of radius r. Fit the largest possible square inside it and the largest possible circle inside the square. What is the radius of the innermost circle?

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 3

The radius of outer circle = r

∴ Diagonal of square = 2r

Side of square (a) = 

Now, the radius of the inner circle = 

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 4

A bird files along the three sides of a field in the shape of an equilateral triangle at speeds of 3, 6, 8 km/hr respectively. The average speed of the bird is

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 4

Average speed = Total distance/Total time

Total distance = a + a + a = 3a

Total time = 

 

Average speed  = 

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 5

Choose the most appropriate words from the options given below to fill in the blanks.

She was ______ to travel abroad and _____ in the field of commerce as per her wishes.

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 5

The sentence mentions a person doing something as per her wishes thus the first word must be a positive reaction. 'Restive' which means 'restless' and 'jinxed' which means 'cursed' are incorrect here. Between options 1 and 3,the word 'elated' which means 'too happy' and the word 'venture' which means 'undertake a risky or daring journey or course of action' fit here correctly. The word 'cease' means 'stop' and does not convey a proper meaning. Thus option 3 is the correct answer.

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 6

In a 1600 m race around a circular track of length 400 m, the faster runner and the slowest runner meet at the end of the sixth minute, for the first time after the start of the race. All the runners maintain uniform speed throughout the race. If the faster runner runs at twice the speed of the slowest runner. Find the time taken by the faster runner to finish the race.

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 6

As, the faster runner is twice as fast as the slowest runner, the faster runner would have complete two rounds by the time the slowest runner completes one round.

Their first meeting takes place after the fastest runner takes 6 min to complete two rounds and the slowest runner completes one round.

Time taken to complete one round by fastest runner = 6/2 = 3 minutes

Rounds needed to complete the race = 1600/400 = 4

Time taken to complete the race = 4 × 3 = 12 min

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 7

Select the pair that best expresses a relationship similar to that expressed in the pair:

Horse: Foal

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 7

The young one of a horse is called a foal. Similarly, the young one of a goose is known as gosling. Thus, option 2 contains the pair that best expresses a relationship similar to that expressed in the given pair. In all the other pairs, the first word expresses the male form of the latter.

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 8

Read the following passage and find out the inference stated through passage.

Juvenile delinquency is also termed as Teenage Crime. Basically, juvenile delinquency refers to the crimes committed by minors. These crimes are committed by teenagers without any prior knowledge of how it affects the society. These kind of crimes are committed when children do not know much about outside world.

Which of the following Inferences is correct with respect to above passage? 

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 8

Since teenagers here have tender age and commit crimes unknown to its consequences, clearly they should not be treated as any criminal. They need to be taught so that they can make difference between right and wrong and do not commit these crimes in future. Hence inference in option 1 is correct.

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 9

If n and y are positive integers and 450 y = n³, which of the following must be an integer? 

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 9

450 y = n³ implies that 450 y is the cube of an integer. 

When we prime-factorize the cube of an integer, we get 3 (or a multiple of 3) of every prime factor.

8 is the cube of an integer because 8 = 2³ = 2*2*2.

Thus, when we prime-factorize 450 y, we need to get at least 3 of every prime factor. 

Here's the prime factorization of 450 y:

450 y = 2 * 3² * 5² * y 

Since 450 provides only one 2, two 3's, and two 5's, and we need at least 3 of every prime factor, the missing prime factors must be provided by y. Thus, y must provide at at least two more 2's, one more 3, and one more 5. 

Smallest possible case: 

y = 2² * 3 * 5

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 10

The graph shows cumulative frequency % of research scholars and the number of papers published by them. Which of the following statements is true?

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 10

In this problem, cumulative frequency is given, so converting into frequency

From the table, it is clear that option b, i.e.

60% of the scholars published at least 2 papers is correct.

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 11

For which value of x, the following matrix A is a singular matrix.

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 11

A matrix is singular if and only if its determinant is zero.

 

x(3+1)−2(x2+1) = 0−2x2+4x−2 = 0

x2−2x+1 = 0

(x−1)2 = 0

x = 1

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 12

Find the laplace transform of t2sin (2t).

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 12

According to formula, L(tf(t)) =​

Here, f(t) = sin (2t) ⇒ F(s) =

∴ L(t2sin (2t)) =  = 

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 13

Consider the following function:

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 13

ii) (4 - x)x=2 = 2

Hence at x = 2, f(x) is continuous

∴ f(x) is not differentiable.

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 14

 value of b = ?

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 14

The given limit is in 0/0 form, So

Applying L hospitals rule

⇒ log b - 1 = -1 - log b

⇒ 2 log b = 0

⇒ b = 1

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 15

The integrating factor of equation y log y dx + (x – log y) dy = 0 is

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 15

y log y dx + (x – log y) dy = 0

It is a Leibnitz’s equation in x.

Integrating factor =

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 16

Relationship between input x(t) and output y(t) of a system is given as

The transfer function of this system is

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 16

By applying Laplace transform,

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 17

For the two square inputs in figure, the PMMC meter will read, maximum when the phase difference between them is

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 17

PMMC meter reads average value if the output of the XOR gate.

The output of XOR gate will be high when both the inputs are different.

For a phase difference of T/2 both X and Y will be different as shown below.

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 18
  1. For the circuit shown the hybrid parameter matrix [h] is

 

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 18

Equations of hybrid parameter are,

When, V2 = 0

V1 = -2I2

By applying KCL at 0,

⇒ h21 = -2

V1 = -2I2 = -2 (-2I1) = 4I1

⇒ h11 = 4

When, I1 = 0

⇒ h22 = 0.5

By applying KVL,

⇒ -V1 + 2V2 – I2 = 0

⇒ -V1 + 2V2 – 0.5 V2 = 0

⇒ h12 = 1.5

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 19

H(z) is discrete rational transfer function. To ensure that both H(z) and its inverse are stable its

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 19

For H(z) to be stable the poles of H(z) must be inside the unit circle.

For the inverse of H(z) to be stable the poles inverse of H(z) must be inside the unit circle.

The poles of inverse of H(z) are the zeros of H(z)

Hence, both poles and zeros of H(z) must be inside the unit circle.

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 20

In the dc circuit shown in the below figure, the node voltage V2 at steady state is

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 20

By voltage division

By voltage division,

*Answer can only contain numeric values
GATE Practice Test: Electrical Engineering (EE)- 1 - Question 21

A DC-DC converter is used to feed the resistive load and its output is always inverted. An alternating square wave voltage is getting produced across in inductor having rms value of 200 V and frequency of 50 Hz. the duty cycle of the switch is______


Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 21

Given circuit is Buck Boost converter

Voltage across inductor would be, V= 200 V

Vo = 200 V

⇒ δ = 1 – δ

⇒ δ = 0.5

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 22

If the multiplexer is controlled such that the channels one sequenced every 5 μs as 1, 2, 1, 3, 1, 4, 1, 2, 1, 3, 1, 4, 1, … , the input connected to channel 1 will be sampled at the rate of

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 22

Sampling period, Ts = 2 × 5 = 10 μs

Sampling rate, 100k samples/sec

*Answer can only contain numeric values
GATE Practice Test: Electrical Engineering (EE)- 1 - Question 23

What percentage of current IDSS is the drain current for a JFET, if the gate to source voltage is 65% of the pinch-off voltage.


Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 23

VGS = 0.65 Vp

The drain current of JFET is given as

 

 (1−0.65)2 = 0.1225 = 12.25% 

*Answer can only contain numeric values
GATE Practice Test: Electrical Engineering (EE)- 1 - Question 24

A uniform line charge of infinite length with PL = 15 nC/m lies along the z-axis.

The electric field at (5, 12, 20) will be – v/m


Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 24

Electric field at a point P(x, y, z) is given as

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 25

A three phase, 50 Hz transmission line of length 120 km has a capacitance of  It is represented as a π – model. The shunt admittance at each end of the transmission line will be

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 25

In π model, the shunt admittance at each end of the line is 

= j300×10−6  

= 300×10−6 ∠90∘ mho

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 26

The fundamental period of the sequence x[n] = 3 sin (1.3 πn + 0.5 π) + 5 sin (1.2 πn) is

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 26

x[n] = 3 sin (1.3 πn + 0.5 π) + 5 sin (1.2 π n)

Time period = LCM (20, 5) = 20

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 27

What should be the input voltage to turn off the SCR in the circuit shown in figure below when holding current of the SCR is 4 mA and load resistance is 4 kΩ.

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 27

To turn off the SCR anode current should go less than holding current

⇒ V = 4 × 10-3 × 4 × 103

⇒ V = 16 V

Form the options, V = 15 V is correct. for the remaining options, holding current is less than anode current.

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 28

 

The inverse Fourier transform of a function given 

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 28

Fourier transform of

⇒ g(t) = δ(t) + 2.e-3|t|

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 29

Figure shows the polar plot of a system. The transfer function of the system is

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 29

From the given polar plot,

From the options, 5(1 + 0.1s) satisfies this condition.

GATE Practice Test: Electrical Engineering (EE)- 1 - Question 30

In the circuit of figure A, B, C are the inputs and P, Q are the two outputs. The circuit is a

Detailed Solution for GATE Practice Test: Electrical Engineering (EE)- 1 - Question 30

From the given circuit diagram,

P = A⊕B⊕C

Q = AB + BC + CA

The above two expressions presents sum (P) of a full adder and carry (Q) of a full adder

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