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GATE Mock Test Electronics Engineering (ECE)- 2 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test - GATE Mock Test Electronics Engineering (ECE)- 2

GATE Mock Test Electronics Engineering (ECE)- 2 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The GATE Mock Test Electronics Engineering (ECE)- 2 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The GATE Mock Test Electronics Engineering (ECE)- 2 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for GATE Mock Test Electronics Engineering (ECE)- 2 below.
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GATE Mock Test Electronics Engineering (ECE)- 2 - Question 1

Direction: In the following question, there are three sentences. Each sentence has a pair of words which is underlined. From the underlined words, select the most appropriate word (1) or (2) to form correct sentences. Choose the correct options to form correct sentences.

The report submitted by the committee does not quite jibe(1)/gibe(2) with the observations.

Ryan tried to flesh(1)/flush (2) out his novel with incidents borrowed from other writers.

Anger is always a bad thing, even when it is a condign (1)/ condescend(2) response to events.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 1
The word ‘jibe’ means ‘be in harmony’ and ‘gibe’ means ‘to mock’. Thus 1 fits here best.

The word ‘flesh out’ means ‘to expand’ and ‘flush out’ means ‘to clean’. Thus 1 conveys the correct meaning.

The word ‘condign’ means ‘appropriate’ and ‘condescend’ means ‘to lower to a less dignified level’. Thus 1 fits here correctly.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 2

Directions: In the following question, a set of three figures X, Y and Z shows a sequence in which a paper is folded and finally cut from a particular section. Below these figures, a set of answer figures (a), (b), (c) and (d) shows the design which the paper actually acquires when it is unfolded. You have to select the answer figure which most closely resembles the unfolded piece of paper.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 2

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GATE Mock Test Electronics Engineering (ECE)- 2 - Question 3

If the rate of change in the circumference of a circle is 0.5 cm/s, then find the rate of change in the area of the circle when the radius is 8 cm.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 3
C = 2πr

⇒ dC/dt = 2πdr/dt

⇒0.5/2π = dr/dt

∴ A = πr2

⇒dA/dt = 2πrdr/dt = 0.5 x 8 = 4 sq.units

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 4

Which of the following is an antonym of the word PROFESSIONAL?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 4
Professional is the person who does something as a part of his job. Amateur is a person who does something because he loves doing it.
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 5

Find the wrong number in the series

2, 9,5, 36, 125, 648

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 5

5 is wrong number, starting new series with 5, we get

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 6

The ratio of cost price and marked price of a cell phone is 2 : 3 and ratio of profit percentage and discount percentage is 3 : 2. What is the discount percentage?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 6

CP:MP = 2:3

Let CP = 200, MP = 300

(%) profit : (%) discount = 3:2

⇒ 3x/100 x 200 + 2x/100 x 300 = 100

⇒ x = 8.33%

Discount = 2x = 16.66%

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 7

The line graph shows Net exports in million USD of a certain country (Net exports = Exports - Imports). Study the diagram and answer the following questions. 

Cumulative net exports (in million USD) from the beginning of 2011 to the end of 2014 was?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 7

Cumulative net exports (in million USD) from the beginning of 2011 to the end of 2014 
= 20 - 10 - 20 + 30 = 20

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 8

Directions: Choose the sentence that best combines the given sentences.

The airport is called the Glynco Jetport. The airline reservations and travel systems refer to the location as Brunswick, Georgia.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 8
Option 3 is the most logical subordinating phrase, showing a contrast. The other choices are not only illogical but also grammatically incorrect.
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 9

The question below consists of a set of labelled sentences. These sentences, when properly sequenced form a coherent paragraph. Select the most logical order of sentences from among the options.

P. This partial eclipse began at 7:18 IST on July 13, reached its maximum magnitude of 0.3367 at 8:31 IST and ended at 09:43 IST.

Q. The year's second solar eclipse occurred on July 13, 2018.

R. The partial eclipse was seen only in parts of Australia and New Zealand as it took place almost entirely over open waters of Antartica.

S. Solar eclipse is a natural phenomenon that occurs when the moon comes in between the sun and the earth.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 9
The given passage talks about the second solar eclipse of this year. Hence, Q should be the first sentence as it introduces the topic. R should come next as it tells about the eclipse is that it was a partial eclipse. P should follow it as it uses the term ‘this partial eclipse’ (another hint that partial eclipse introduced in earlier sentence) gives further information about timing of the eclipse. In the end, sentence S gives general information about solar eclipse. Hence the correct sequence is QRPS
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 10

Leo participated in a race of motor bikes. Flags of different colours are lowered to give different signals. How many different signals can be made here, using any number of flags from 5 flags of different colours?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 10
Signals can be made by using at a time one, two, three, four and five flags.

∴ Total number of signals that can be made = 5P1 + 5P2 + 5P3 + 5P4 + 5P5

= 5!/4! + 5!/3! + 5!/2! + 5!/1! + 5!/1

= 5 + (5 × 4) + (5 × 4 × 3) + (5 × 4 × 3 × 2 × 1) + (5 × 4 × 3 × 2 × 1)

= 5 + 20 + 60 + 120 + 120

= 325

So, 325 signals can be made using five flags.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 11

If A is 3 × 3 matrix, whose elements are given by aij = i2 – j2 where 1 ≤ i, j ≤ 3 then A–1 = _____

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 11
aij = i2 – j2∀i, j

∴A3×3is singular

∴|A| = 0

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 12

The load curve is useful in deciding:

  1. The operating schedule of generating units

  2. The total installed capacity

Which of the above statements is/are correct?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 12
The load curve indicates average (KWh) energy consumption during a given period and hence, it is useful in deciding both the operating schedule of generating units and the total installed capacity.
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 13

What is the behavior of following one i/p Flipflop ‘X’ ?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 13

→ Qn+1 = X

Therefore it is D-Flipflop.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 14

Consider the circuit diagram shown below. The circuit represents

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 14

In the case of transistor AND gate, when the inputs X, Y = 0 V or when X = 0 V and Y = +5 V or when X = +5 V and Y = 0 V, both the transistors Q1 and Q2 are at OFF state. At the same time, transistor Q3 gets enough base drive from the supply through resistor R3, so transistor Q3 will be ON. Therefore, output voltage Z = Vce(sat) corresponds to 0 V. When both the inputs are equal to +5 V, transistors Q1 and Q2 will be ON; therefore, the voltage at the collector of transistor Q1 will drop. As a result, transistor Q3 does not get enough base drive and turns OFF. Therefore, no current flows through the collector resistor of Q3, hence no voltage drop across it. So, the final output voltage corresponds to +5 V.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 15

In the given figure, Zin at the generator is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 15

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 16

For static electric and magnetic fields in an homogeneous source-free medium, which of the following represents the correct form of Maxwell equations?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 16
Maxwell equations

For static electric and magnetic fields

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 17

Consider the circuit below with initial state Q0 = 1, Q1 = Q2 = 0. The state of the circuit is given by the value of 4Q2 + 2Q1 + Q0

Which one of the following is the correct state sequence of the circuit?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 17

State= 4Q2 + 2Q1 + Q0

This State Equation is Decimal Equivalent of binary State.

Ex: Qo Q1 Q2 = 100, state = 4 × 0 + 2 × 0 + 1 = 1

Hence the Answer is 1,2,5,3,7,6,4 (See the table)

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 18

Consider the following circuits:

  1. Oscillator

  2. Emitter follower

  3. Power amplifier

Which of the above circuits employ feedback?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 18
Oscillator

With the use of positive feedback or regenerative feedback, the oscillations will take place in a circuit and the circuit acts as an oscillator.

Emitter follower:

It has negative current feedback as shown in the figure below.

Power Amplifier:

A power amplifier has negative feedback. The negative feedback is used to reduce the distortion and noise and to increase the bandwidth.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 19

A sinusoidal message signal of amplitude Am and frequency fm is passed through a delta modulator whose step size in 0.628 V and sampling rate is given by 40K sample/ sec then which of the following delta modulator will be slope overloaded.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 19
Step size (s)=0.628 V

Condn for slope overload Δ/Ts < />mAm

0.628 x fs < />mAm

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 20

The switch in the figure below was closed for a long time. It is opened at t = 0. The current in the inductor of 2 H for t ≥ 0, is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 20
At t = 0-, inductor acts as short circuit,

Now, the circuit is reduced to,

Equivalent resistance Req = 10 Ω

Current, I =50/10 = 5 A

By using current division rule,

IL (0-) = 5/2 = 2.5 A

At t = 0+,

IL (0+) = IL (0-) = 2.5 A

At steady state (t =∞ )

Time constant of the circuit (τ) = L/Req

Req is the equivalent resistance across inductor.

Req = 16||32||32 = 8 Ω

Time constant:

(τ) = L/Req = 2/8 = 0.25 sec

= 0.25 sec

= 0 + (2.5 - 0) e-4t = 2.5 e-4t A

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 21

In the question a circuit and a waveform for the input voltage is given. The diode in circuit has cut-in voltageVγ=0 Choose the option for the waveform of output voltage V0.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 21
During positive cycle when vs < 8V, both diode is OFF vo = vs⋅ For vs > 8,v0 = 8V,D1 is ON.

During negative cycle when |vs|<6, both diode is OFF, vo = vs. For |vs| > 6V,D2 is on = -6 V

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 22

If a, b, c are 3-input variables, then Boolean function y = ab + bc + ca represents:

  1. A 3-input majority gate

  2. A 3-input minority gate

  3. Carry output of a full adder

  4. Product circuit for a, b, and c

Which of the above statements are correct?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 22
In the given input combination, if the majority of inputs are 1, then the Boolean (SOP) function gives a majority gate. In the given 3-input a, b, c if at least two inputs are '1' then the Boolean function y = ab + bc + ca represents majority gate. Consider the truth table of full adder

By using K-map method, the carry output is obtained as follows:

Thus the given Boolean function represents carry o/p of a full adder.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 23

A gallium arsenide pn junction is operating in reverse-bias voltage VB = 5V. The doping profile are Na = Nd = 1016cm−3. The minority carrier life- time are τp0 = τn0 = τ0 = 10−8s. The reverse-biased generation current density is s (εr = 13.1, ni = 1.8 x 106)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 23

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 24

Using the residue theorem, the value of the integral (counterclockwise) around a circle with center at z = 0 and radius = 8 (where z is a complex number and i = √-1 ), is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 24

circle having centre z = 0

Radius = 8

z = 4 is a simple pole.

= -20

Using Residue theorem,

= 2πi (-20) = -40πi

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 25

Under which of the following conditions, the N-MOS transistor operates in depletion mode?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 25

For an N-MOSFET, when Vt < 0, the transistor operates in depletion mode and when Vt > 0, the transistor operates in enhancement mode.

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 26

In the following circuit, a network and its Thevenin and Norton equivalent are given.

The value of the parameters are

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 26
Thevenin voltage: (Open circuit voltage)

VTh = 4 + (2 x 2) = 4 + 4 = 8 V

Thevenin resistance:

RTh = 2 + 3 = 5Ω = RN

Norton current:

IN = VTh/RTh = 8/5A

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 27

Which of the following meters is based on the principle of Hall Effect?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 27
Voltmeter and ammeter works on the principle of electrostatic principle. Gauss meter works on the principle of Half effect.
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 28

In the Op-Amp circuit shown, assume that the diode current follows the equation I = Is exp (V/VT). For Vi = 2V, V0 = V01; and for Vi = 4 V, V0 = V02.

The relationship between V01 and V02 is:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 28
Here the inverting terminal is at virtual ground and the current in resistor and diode current is equal i.e.

IR = ID

or VD = VT In Vi/IsR

For the first condition

VD = 0 - V01 = VT In 2/IsR

For the second condition

VD = 0 - V02 = VT In 4/IsR

Subtracting above equation

V01 - V02 = VT In 4/IsR - VT In 2/IsR

or V01 - V02 = VT In 4/2 = VT In 2

GATE Mock Test Electronics Engineering (ECE)- 2 - Question 29

Let f(x) be a real, periodic function adequate the condition f (-x) = -f(x). The general form of its Fourier series representation would be

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 29

Given signal is odd function so a0=0

f (-x) = -f(x)

And the Fourier series will content sine term only

So

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 2 - Question 30

A white noise process X(t) with two-sided power spectral density 1 x 10-10 W/Hz is input to a filter whose magnitude squared response is shown below.

The power (in W) of the output process Y(t) is given by ___× 10-6

(Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 2 - Question 30
Power of the output process Y(t),

Y(t) = |H(t)|2 . X(t)

X(t) = N0/2 = 1 x 10-10W/Hz

|H(t)|2 = area under curve |u(t)|2

= 2(1/2 x 10 x 103 x 1) = 104 Hz

Then, Y(t) = 10-6 W

or Y(t) = 1 x 10-6 W

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