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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  GATE Mock Test Electronics Engineering (ECE)- 4 - Electronics and Communication Engineering (ECE) MCQ

GATE Mock Test Electronics Engineering (ECE)- 4 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test - GATE Mock Test Electronics Engineering (ECE)- 4

GATE Mock Test Electronics Engineering (ECE)- 4 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The GATE Mock Test Electronics Engineering (ECE)- 4 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The GATE Mock Test Electronics Engineering (ECE)- 4 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)- 4 below.
Solutions of GATE Mock Test Electronics Engineering (ECE)- 4 questions in English are available as part of our course for Electronics and Communication Engineering (ECE) & GATE Mock Test Electronics Engineering (ECE)- 4 solutions in Hindi for Electronics and Communication Engineering (ECE) course. Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free. Attempt GATE Mock Test Electronics Engineering (ECE)- 4 | 65 questions in 180 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
GATE Mock Test Electronics Engineering (ECE)- 4 - Question 1
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A lent Rs. 5000 to B for 2 years and Rs. 3000 to C for 4 years on simple interest at the same rate of interest and received Rs. 2200 in total from both as interest. The rate of interest per annum is.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 1

Let the rate % = R

According to the question,

(5000×2×R)/100 + (3000×4×R)/100 = 2200

100R + 120 R = 2200

220 R = 2200

R = 10%

Hence required rate % = 10%

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 2
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Directions: In the following question, a set of three figures X, Y and Z shows a sequence in which a piece of paper is folded and finally cut from a particular section. Below these figures, a set of answer figures marked (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)- 4 - Question 2

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GATE Mock Test Electronics Engineering (ECE)- 4 - Question 3
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The sentences given with blanks are to be filled with an appropriate word(s). Four alternatives are suggested for each question. For each question, choose the correct alternative and click the button corresponding to it.

They abandoned their comrades ______the wolves.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 3
When something is left in between more than 2 things or persons, ‘among’ is used. There are a number of wolves in the sentence.
GATE Mock Test Electronics Engineering (ECE)- 4 - Question 4
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One of the four words given in the four options does not fit the set of words. The odd word from the group is:
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 4
'Hiatus' means 'a pause or break in continuity of a sequence or an activity'. So, options 1, 2 and 3 refer to gap or rest. But, 'end' means 'a termination of a state or situation', thus the odd one out.
GATE Mock Test Electronics Engineering (ECE)- 4 - Question 5
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Direction: Study the following information carefully and answer the given questions:

The following pie chart shows the percentage distribution of total number of Apples (Dry + Wet) sold in different days of a week :

The pie chart2 shows the percentage distribution of total number of Apples (Wet) sold in different days of a week.

Find the ratio of Apples (Dry + Wet) sold on Tuesday to that of the Apples (Wet) sold on Thursday?
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 5
Required ratio = 13% of 7000: 23% of 4500 = 182:207
GATE Mock Test Electronics Engineering (ECE)- 4 - Question 6
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Directions: In the following question, four positions of the same dice having different characters are shown.

Which character is on the surface opposite to '+'?
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 6
Four characters will be adjacent to each character. As shown, *, ?, # and $ are adjacent to character +. So, the only character that will be opposite to '+' is '@'.
GATE Mock Test Electronics Engineering (ECE)- 4 - Question 7
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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: July 1969 was to see a transformed Indira Gandhi.

Q: Quite a few people contributed with ideas.

R: She sounded the bugle through her historic ‘Note on Economic Policy and Programme’ that was circulated among delegates at Bangalore on July 9, 1969.

S: But the pivot was P.N.Haksar who gave shape, structure and substance with the help of some of his colleagues in the Prime Minister’s Secretariat.
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 7
P is clearly the first sentence as it introduces the main person of the paragraph, i.e. Indira Gandhi. The pronoun ‘she’ in R refers to Indira Gandhi mention in P. Thus, R forms the second statement. Now, in the sentence R, ‘delegates’ is mentioned which says about the people, who are given in the sentence Q. So, sentence Q must follow R.

Thus, the sequence after rearrangement is PRQS

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 8
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Directions: Read the given statement carefully and answer the question that follows.

Statement: There were different streams of freedom movements in colonial India carried out by the moderates, liberals, radicals, socialists and so on.

Which of the following is the best inference from the above statement?
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 8
Hetero means different. It is clearly mentioned that the movement comprised of moderates, liberals, radicals, socialists and so on.
GATE Mock Test Electronics Engineering (ECE)- 4 - Question 9
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Direction: Read the information carefully and give the answer of the following questions:

Total number of children = 2000

If two-ninths of the children who play football are females, then the number of male football children is approximately what percent of the total number of children who play cricket?
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 9
57.4% (If 2/9 th of children play football are female , then male children = 1 - 2/9 = 7/9th of children play football;

Let's take z% of male football children equal to cricket children , then

=> z% of cricket children = 7/9 th of football;

=> z% of (23% of 2000) = 7/9 th of (17% of 2000);

=> z% of 23 = 7/9 th of 17

=> z = 7× 17 × 100 = 57.4%

Short-cut :

Required percentage = (7/9)×17×100/ 23 = 57.4%

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 10
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A bottle of whisky contains 3/4th whisky and the rest is water. What quantity of the mixture must be taken away and substituted by equal quantity of water to have half whisky and half water?
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 10
Percentage of whisky in mixture = 75%

Percentage of whisky in water = 0%

Therefore, mixture and water should be mixed in the ratio of 2 : 1.

or,

We need to replace 1/3rd or of the mixture with water.

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 11
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Determine Rth ands Vth for this network?
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 11
According to Millman’s theorem the expressions for total voltage and total resistance is given by,

Vth = V1G1+V2G2+V3G3/(G1+G2+G3)

Rth = 1/ (G1+G2+G3)

G1 = 1/4, G2 = 1/5, G3 = 1/6

V1 = 3V, V2 = 5V, V3 = 7V

Vth = 4.729V, Rth = 1.62Ω

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 4 - Question 12
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A card is drawn at random from an ordinary deck of 52 playing cards. If we are told that card is heart, then information, in bits, is(Answer up to the nearest integer)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 12
P (heart) = 13/52 =1/4

I = log24 = 2 bits

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 13
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Which of the following is CORRECT with respect to LAG-LEAD compensator?
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 13
The lag-lead compensation pole is more dominating than the zero and because of this lag-lead network may introduces positive phase angle to the system when connected in series.

Lead-leg compensators have different frequencies for the zero and the pole. Based on the difference of the two frequencies we get different static gains below the lower frequency and above the higher frequency.

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 14
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A 60 Ω coaxial cable feeds a 75 + j 25Ω + dipole antenna. The voltage reflection coefficient Γ and standing wave ratio s are respectively
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 14

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 15
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Find the differential equation whose solution represents the family c(y + c)2 = x3
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 15
c(y+c)2 = x3

Differentiating, we get

2c(y + c)dx/dx = 3x2

Pitting this value of 'c' in equation

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 16
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If A is a 3-rowed square matrix such that |A| = 2, then |adj(adj (adj A2))| is equal to
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 16
Let B = adj(adjA2), then B is also 3 by 3 matrix

Now,

|adj(B)| = |B|3 -1 = |B|2 = |adj(adjA2)|2 = ((|adjA2|)3 - 1)2 = |adjA2|4 = |adjA|8 = (|A|)3 - 1)8 = |A|16 = 216

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 17
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The transfer function of a phase lead controller is given as

Gc(s)=(1+αTs)/(1+Ts) where a > 1 and T is a constant depending on circuit parameter Determine the maximum value of phase lead which can be obtained from the controller.
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 17
For phase lead controller

⇒ puts = jω

Gc(jω) = 1+jaTω/1+jTω

ϕ = ∠Gc(jω)

= tan−1⁡aωT − tan−1⁡ωT

tan⁡ϕ = ωT(a−1)/1+ω2T2…………………(i)

For getting maximum value of phase angle

dϕ/dω = 0

We get ωm = 1/T√a

Putting ωminq

tan⁡ϕm = 0−1/2√a or ϕm

=tan−1⁡[a−1/2√a]

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 4 - Question 18
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A rectangular waveguide of internal dimensions (a = 4 cm and b = 3 cm) is to be operated in TE11 mode. The minimum operating frequency is (Answer up to two decimal places)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 18
Cut -off Frequency is

For TE11 mode,

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 19
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In the circuit shown in fig the op-amp is ideal. If βF= 60, then the total current supplied by the 15 V source is;
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 19
v+ = 5v

= v− = vE

i+15V = iz + ic

= iz + aF + iE

The input current to the op-amp is zero.

= 49.4 mA

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 20
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A random process has the power density Pxx(ω) = 6ω2/[1 + ω2]3 . The average power in process is
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 20

= 3/8

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 21
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Let A be a 3 x 3 matrix with eigen values 1, -1 and 3. Then
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 21
Singular Matrices have Zero Eigenvalues. Suppose A is a square matrix. Then A is singular if and only if λ=0 is an eigenvalue of A.

Eigen values ofA2+ A are 2,0,12

Eigen values of A2 - A are 0,2,6,

Eigen values of A2 +3A are 4,-2,18

Eigen values of A2 - 3A are -2,4,0.

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 22
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Consider the ufb system shown below:

The root - loci, as α is varied, will be
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 22

Thus an equivalent system has G(s) = 1/(s2 + 1) and H(s) = αs. For this function root-locus is (2).

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 23
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For the lattice circuit shown in Figure,Za = j2 and Zb = 2. The values of the open circuit impedance parameters

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 23

We know that

V1 = z11I1 + z12I2

V2 = z11I1 + z22I2

where

Consider the given lattice network, when I2=0. There is two similar path in the circuit for the current I1

So I = 1/2I1

For z11 applying KVL at input port, we get

For this circuit z11

= z22 and z12

= z21. Thus

Here Za = 2jΩ&Zb

= 2Ω

Hence Z11

= (2+2j)/2

= 1 + j;Z22 = 1 + j

Z21 = (2j − 2)/2

= −1 + j

= Z12

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 24
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The Fourier Series coefficient for the periodic signal shown below is
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 24

T = 2π, ωo = 2π/2π = 1,

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 25
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For the circuit in fig let β = 100. The values of Q-point (ICQ .VCEQ) is;
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 25
R1 = 12kΩ, R2 = 2kΩ

RTH = R1 || R2 = 12 || 2 = 1.71 kΩ

3.57 = 1.71k × IBQ + VBE + (β + 1)IBQ × 0.5k − 5

5 − 3.57 − 0.7

= (1.71 + 50.5)IBQ

IBQ = 14μA

IEQ = (100 + 1)IBQ

= 1.412mA

ICQ = 100IBQ

= 1.4mA

VCEQ = 5 − RCICQ − REIBQ + 5

= 5 − (5)(1.4) − (0.5)(1.412) + 5

= 2.3V

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 26
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Consider two signals x [n] = {1, 2, - 1} and h [n] = x [n]. The convolution y [n] = x [n] x h [n] is
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 26
y[n] = {1, 4, 2, -4, 1}

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 27
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For the AM envelope shown below, determine the total power transmitted is _____W.
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 27
From the envelope shown

Vmax = 10…(1)

Vmin = 2V…(2)

But Vmax = Ac(1 + μ)…(3)

Vmin = Ac(1 − μ)…(4)

From (1),(2),(3),(4)

1 + μ/1 − μ = 10/2 = 5

⇒ μ = 2/3 = 0.67

⇒ Ac = 6V

Peak amplitude of Carrier = AC = 6V

Modulation coefficient = μ = 2/3 = 0.67

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 4 - Question 28
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At 20 GHz, the approximate gain of a parabolic dish antenna of 1 metre and 70% efficiency is(Answer up to two decimal places)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 28
λ = c/f = (3 x 108)/(20 x 109) = 3/200

= 44.87 dB

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 29
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The inverse z-transform of the signal X(z)= In (α/α - z-1); ROC : |z| > 1/α
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 29
Given X(z) = In (α/α - z-1); ROC : |z| > 1/α

= - in(1 - (αz)-1)

now expanding it by Taylor series, we get

Taking the inverse z - transform, we get

GATE Mock Test Electronics Engineering (ECE)- 4 - Question 30
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The simplified form of a logic function is
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 4 - Question 30

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