Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  GATE ECE (Electronics) Mock Test Series 2025  >  Practice Test: Electronics Engineering (ECE)- 5 - Electronics and Communication Engineering (ECE) MCQ

Practice Test: Electronics Engineering (ECE)- 5 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test GATE ECE (Electronics) Mock Test Series 2025 - Practice Test: Electronics Engineering (ECE)- 5

Practice Test: Electronics Engineering (ECE)- 5 for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) Mock Test Series 2025 preparation. The Practice Test: Electronics Engineering (ECE)- 5 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Practice Test: Electronics Engineering (ECE)- 5 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Practice Test: Electronics Engineering (ECE)- 5 below.
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Practice Test: Electronics Engineering (ECE)- 5 - Question 1

Choose the word or phrase which is nearest in meaning to the key word:

        Pilfer

Practice Test: Electronics Engineering (ECE)- 5 - Question 2

The following question comprises two words that have a certain relationship between them followed by four pairs of words. Select the pair that has same relationship as the original pair of words

            Fox : Cunning

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 2

Fox is cunning animal similarly
Ants are Industrious.
The second word shows the quality of the first.

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Practice Test: Electronics Engineering (ECE)- 5 - Question 3

Replace the phrase printed in bold to make it grammatically correct ?

    A twenty-first century economy "cannot be held" hostage by power cuts nor travel on nineteenth century roads.

Practice Test: Electronics Engineering (ECE)- 5 - Question 4

Improve the sentence with suitable options by replacing the underlined word.

 He "lay" on the grass enjoying the sunshine.

Practice Test: Electronics Engineering (ECE)- 5 - Question 5

Fill in the blanks :

 At times, we are all ______ to be mistaken.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 5

At times, we are all likely to be mistaken.

Practice Test: Electronics Engineering (ECE)- 5 - Question 6

There are two solutions of wine and water, the concentration of wine being 0.4 and 0.6 respectively. If five liters of the first solution be mixed with fifteen liters of the second, find the concentration of wine in the resultant solution.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 6

Resultant concentration of wine

Practice Test: Electronics Engineering (ECE)- 5 - Question 7

Rajiv bought some apples at the rate of 25 for Rs. 400 and an equal number at the rate of 30 for Rs. 270. He then sold all at the rate of 10 for Rs. 180. Find his profit/loss percentage

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 7

Let Rajiv bought x apples of each kind.

Total cost price of Rajiv =

Total selling price of Rajiv =

SP > CP , Hence he makes profit;

 Profit percentage =

Practice Test: Electronics Engineering (ECE)- 5 - Question 8

Out of 10 persons working on a project, 4 are graduates. If 3 are selected, what is the probability that there is alteast one graduate among them ?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 8

P (at least one graduate) = 1 – P (no graduate)

Practice Test: Electronics Engineering (ECE)- 5 - Question 9

2 men and 3 boys can do a piece of work in 10 days, while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work ?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 9

Let work done by 1 man in 1 day = x units

Let work done by 1 boy in 1 day = y units

2 men and 1 boy can finish work

No. of days taken by 2 men and 1 boy = = 12.5 days

Practice Test: Electronics Engineering (ECE)- 5 - Question 10

In an exam, 60% of the candidates passed in Maths and 70% candidates passed in English and 10% of the candidates failed in both the subjects, 300 candidates passed in both the subjects. Find the total number of candidates appeared in the exam, if they took test in only two subjects that is Math and English.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 10

Let total number of students appear be X.

X = 750

Practice Test: Electronics Engineering (ECE)- 5 - Question 11

In the figure, the value of resistor R is (I + 5) ohms, where I is the current in amperes. The value of current I is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 11

150 – IR = 0

            150 – I (I +5) = 0

            150 – I2 - 5I = 0

            I2 +5I – 150 = 0

I2 +15I – 10I - 150 = 0

I (I +15) – 10 (I +15) = 0

(I +15) (I-10) = 0

I = 10A & I = -15A

Practice Test: Electronics Engineering (ECE)- 5 - Question 12

The velocity of propagation of electromagnetic wave in an underground cable with relative permittivity of 9 is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 12

Practice Test: Electronics Engineering (ECE)- 5 - Question 13

The function shown in the figure can be represented as

Practice Test: Electronics Engineering (ECE)- 5 - Question 14

Consider an LTI system with impulse response h(t) = e-5t u(t). Determine the output of the system for the input x(t) = 5 u(t)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 14

y(t)      = h(t) * x(t)

                    = e-5t u(t) * 5u(t)

y(t)      = (1 - e-5t) u(t)

Practice Test: Electronics Engineering (ECE)- 5 - Question 15

The value of integral   is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 15

                   = e0 ×  0

                               = 0

Practice Test: Electronics Engineering (ECE)- 5 - Question 16

A rectangular pulse x(t) is shown in the figure. Determine the value of  

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 16

According to Parswell’s theorem

Practice Test: Electronics Engineering (ECE)- 5 - Question 17

If   and f(t) is     as t → ∞ . Determine the value of K

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 17

⇒K = 2

Practice Test: Electronics Engineering (ECE)- 5 - Question 18

In the signal flow graph shown in the given figure, the value of C/R ratio is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 18

There are three path

Practice Test: Electronics Engineering (ECE)- 5 - Question 19

What is the open-loop transfer function for an unity feedback system having root locus shown in the following figure ?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 19

Practice Test: Electronics Engineering (ECE)- 5 - Question 20

The digital circuit shown in the figure works as a  

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 20

For D flip-flop

            Qn 1 = D

            Here, D = X Qn

        So Qn 1 = X Qn

     So it becomes T flip-flop

Practice Test: Electronics Engineering (ECE)- 5 - Question 21

The simplified form of Boolean function F(A,B,C) implemented by the given 3 × 8 decoder is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 21

Practice Test: Electronics Engineering (ECE)- 5 - Question 22

The contents of the accumulator in an 8085 microprocessor is altered after the execution of the instruction

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 22

Compare operation does not affect the accumulator and ORAA does not change the content of accumulator.

Practice Test: Electronics Engineering (ECE)- 5 - Question 23

Assuming that all the diodes are ideal in figure, the current in the diode D2 is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 23

Assume both the diodes are ON

So, I1 = -1A

But current can not flow from n to p side in a diode so D1 will be off and equivalent circuit becomes

      Therefore,

Practice Test: Electronics Engineering (ECE)- 5 - Question 24

Determine the upper and lower threshold voltage of the schmitt trigger shown below :

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 24

Practice Test: Electronics Engineering (ECE)- 5 - Question 25

The transistor in the given circuit, should always be in saturation region. Take VCE (sat) = 0V and VBE = 0.7V. the minimum value of RC is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 25

Practice Test: Electronics Engineering (ECE)- 5 - Question 26

A differential amplifier has input at non inverting terminal is 1050 μV and at inverting terminal 950 μV. What is the error in the differential output if CMRR is 1000.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 26

V0 = AdVd AcmVc

Practice Test: Electronics Engineering (ECE)- 5 - Question 27

Which one of the following type of negative feedback increases the input resistance and decreases the output resistance of an amplifier ?

Practice Test: Electronics Engineering (ECE)- 5 - Question 28

The spectral density and auto correlation function of white noise are respectively.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 28

Auto corelation function and PSD are fourier transform pair

            Here, PSD = N0

Practice Test: Electronics Engineering (ECE)- 5 - Question 29

For a random variable x having the probability density function is shown in the figure, what are the variance of random variable.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 29

=  4/3

Practice Test: Electronics Engineering (ECE)- 5 - Question 30

An amplitude modulated signal is given as, S(t) = 100 Cos 2pfct 50 Cos 2pfmt.Cos 2pfct, where f fm. and fc is the carrier frequency and fm is the modulating signal frequency. The power efficiency is.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 5 - Question 30

S(t) = 100 Cos 2pfct 50 Cos 2pfmt.Cos 2pfct

                  = 100 [1 0.5 Cos 2pfmt] Cos 2pfct

            Here modulation index, μ = 0.5

      = 11.11%

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