GATE Mock Test Electronics Engineering (ECE)- 8 - Electronics and Communication Engineering (ECE) MCQ

# GATE Mock Test Electronics Engineering (ECE)- 8 - Electronics and Communication Engineering (ECE) MCQ

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## 30 Questions MCQ Test GATE ECE (Electronics) Mock Test Series 2025 - GATE Mock Test Electronics Engineering (ECE)- 8

GATE Mock Test Electronics Engineering (ECE)- 8 for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) Mock Test Series 2025 preparation. The GATE Mock Test Electronics Engineering (ECE)- 8 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The GATE Mock Test Electronics Engineering (ECE)- 8 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)- 8 below.
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GATE Mock Test Electronics Engineering (ECE)- 8 - Question 1

### A die is thrown twice. What is the probability of getting a sum 7 from both the throws ?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 1
(1,6),(6,1) = 2

(2,5),(5,2) = 2

(3,4),(4,3) = 2

Desirable outcomes = 6

Total Outcomes in two dices = 36

Probability of sum 7 = 6/36

= 1/6

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

### Find the sentence that has a mistake in grammar or usage. If you find no mistake, mark choice 4.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 2
The correct verb form is 'has broken'.
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GATE Mock Test Electronics Engineering (ECE)- 8 - Question 3

### In the following question, some part of the sentence may have errors. Find out which part of the sentence has an error and select the appropriate option. If a sentence is free from error, select ‘No Error’.I suppose no place is more (1)/ better than (2)/ home on Christmas. (3)/ No error (4)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 3
The error is in part (1) of the sentence. The use of "better" signifies that the sentence is of comparative degree.

So, the use of "more" is incorrect here as it is redundant to be used with "better". So, remove "more".

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 4

Directions: The following verbal analogy has two parts. One part is complete while the other one is incomplete. Complete the portion that is incomplete by selecting the right choice from the given options.

Son : Nuclear : : ______ : Extended

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 4
A son is part of a nuclear family, and a cousin is part of an extended family.
GATE Mock Test Electronics Engineering (ECE)- 8 - Question 5

Direction: In the following question, a sentence, with a part of it missing and represented by a blank is given. Select the best out of the five answer choices given, to make the sentence complete and coherent (coherent means logically complete and sound).

__________ or else they would not keep electing him year after year.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 5
This type of questions needs to be solved on the basis of the contextual appropriateness of the phrase and the statement formed after the placement of the phrase into the statement.

Option A is inaccurate because a mayor is not elected by the leaders but by the people. The use of pronoun 'they' is ambiguous, as it the subject 'the party leader' is singular.

Option B makes no sense in the context of the statement, as having scandals, told to resign and getting elected repeatedly is not at all factually connected.

Option C is incorrect in the context of the statement, because the mayor can't threaten the residents who would be involved in his election year after year.

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 6

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

The federal government has diversity of jobs and geographic locations. The federal government offers flexibility in job opportunities that is unmatched in the private sector.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 6
The subordinator 'because' in choice 3 establishes the logical causal relationship between the subordinate and main clause. Choices 1 and 2 do not make sense. Choice 4 has faulty construction.
GATE Mock Test Electronics Engineering (ECE)- 8 - Question 7

Direction: Study the information carefully and answer the questions given below.

P is standing north of Q at a distance of 6m. R is standing toward east of P at a distance of 6m. R is to the north of S and distance between them is 6m. P is facing north direction, R is facing east, S is facing North and Q is facing towards west direction. All of them walk straight for 3m.

What is the distance between the final position of P and Q?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 7

Let T be the final position of P and L be the final position of Q.

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

The first term of an AP is equal to the sum of the common ratio of a GP and the first term of the GP, which is equal to the common difference of the AP. If the sum of the first two terms of the GP is equal to the sum of the first 2 terms of the AP, then the ratio of the first term of the GP to the first term of the AP is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 8
We know that the three term of the AP are α - β, α, α + β and those of the GP are a, ar, ar2.

So, according to the quesition,

α - β = r + a = β

a + ar = 2α - β

a + ar = 2α - 2β + β

a+ar = 2

a(1 + r) = 2 (r + a) + r + a

a(1+ r) = 3 (r + a)

a + r = a(1 + r)/3

Now, the required ratio is a/α - β = a/(a + r) = (a × 3)/(a (1 + r)) = 3/(1 + r)

This is independent of the first term of the GP.

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 9

In a class test, A scored 51 marks. The test had 4 sections. In the second section, A scored 7 marks more than in the first section, and in the third section, A scored 7 marks more than in the second section. If A did not score at all in the fourth section, how much did A score in the second section?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 9
If A scored x marks in section 2, he scored (x-7) marks in section 1 and (x+7) marks in section 3.

(x-7) + x + (x+7) + 0 = 51

⇒ 3x = 51

⇒ x = 17

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 10

Ramu is planning to get wireless internet service at his house. Two service providers A and B offer different rates as shown in the table below.

If Ramu plans on using 25 hours of internet service per month, which of the following statements is true?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 10
When used for 25 hours per month, provider A will cost \$20 + 7.5 x \$1 (for the hourly charge above the free hours).

This equals \$27.50.

Provider B will cost \$20 + 5 x \$1.50 (for the hourly charge above the free hours).

This equals \$20 + \$7.50 = \$27.50 as well.

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 11

Find the unit normal vector of the cone of revolution z2 = 9x2 + 15y2 −  z2 at a point (1,0,3)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 11

The unit normal vector is the grad of f = 9x2 + 15y2 − z2

Unit normal vector,

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 8 - Question 12

Let Q√γ be the BER of a BPSK system over an AWGN channel with twosided noise power spectral density N0/2. The parameter γ is a function of bit energy and noise power spectral density.

A system with two independent and identical AWGN channels with noise power spectral density N0/2 is shown in the figure. The BPSK demodulator receives the sum of outputs of both the channels.

BER of this system is Q(b√γ ), then the value of b is ________.(Answer up to three decimal places)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 12

Bit error rate for BPSK

⇒ Y = 2E/No

Function of bit energy and noise

Counterllation diagram of BPSK

Channel is AWGN which implies noise sample as independent

Let 2x +n1+ n2 = x1 + n1

where x1 = 2x

n1 = n1 + n2

E1 is energy in x1

No1 is PSD of h1

E1 = 4E [as amplitudes are getting doubled]

No1 = No [independent and identical channel]

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 13

The transfer function of a system is 10/(s2+2s+10). Find the nature of the given system.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 13
On comparing the given characteristic equation with the standard second order transfer function,

ωn2/(s2 + 2ζωns + ωn2).

Hence, ωn2 = 10 ; 2 ζωn = 2⇒ ζ = 0.316

For 0 < ζ < 1, the system is said to be Under-damped.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 8 - Question 14

The surface ρ = 2, ρ = 4, Φ = 45°, Φ = 135°, z = 3 and z = 4 define a closed surface. The total area of the enclosing surface is _____. (Answer up to two decimal places)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 14

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 15

There are two energy levels E1 and E2 . E1 is E eV above the fermi level and E2 is E eV below the fermi level.

P1 = the probability of E1 being occupied by an electron.

P2= the probability of E2 being empty.

Then, which of the following is correct?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 15
The Fermi- Dirac probability function also called Fermi function, provides the probability of occupancy of energy levels. It is given by

where EF is Fermi level or Characteristics energy, for the crystal in eV. The fermi level represents the energy state with 50 percent probability of being filled if no forbidden band exists.

Therefore, P1 And P2 can be calculated by using Fermi- Dirac probability function.

For P1

For P2

therefore,

P1 < />2

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 8 - Question 16

For a Hertz dipole antenna, the half power beam width (in degrees) in the E-plane is _____.(Answer up to the nearest integer)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 16
The beam-width of Hertzian dipole is 180° and its half power beam-width is 90°.
*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 8 - Question 17

Consider the circuit shown in the figure below

v(t) = 80e−1000tV ;t>0

i(t) = 5e−1000tmA ;t>0

the energy dissipated (in μ) by the resistor for 0 < t < 0.6 ms will be approximately

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 17

The energy dissipated in the resistance is

E = 200(1−e−1.2)μJ

E = 139.76 pJ

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 8 - Question 18

A die is thrown 100 times. Getting an even number is considered a success. The variance of the number of successes is ______ (Answer up to the nearest integer)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 18
Here p = 3/6 = 1/2, q = 1 - 1/2 = 1/2 & n = 100

Thus variance = npq = 100 × 1/2 × 1/2 = 25

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 19

While analysing a two-port network, which of the following parameters are obtained when one-port variables are expressed in terms of the other port variables?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 19
Expressing one port variables in terms of the other port variables,

(V1,I1) = f(V2,−I2)

The parameters so obtained are the T-parameters or the ABCD-parameters.

These are used for the analysis of the power transmission line. The input and output ports are the sending and the receiving ends respectively.

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 20

By simplifying the given Boolean function, Y = ABC + A'B + ABC' + AC, the minimum number of literals is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 20
Y = ABC + A'B + ABC' + AC

= AB(C + C') + A'B + AC

= AB + A'B + AC = (A + A')B + AC = B + AC

So, the number of literals is 3.

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 21

The impulse response of a second order under damped system starting from rest is given by C(t) = 12.5e-8tcos6t

What are the responsive value of natural frequency and damping ratio of the system?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 21
C(t) = 12.5e8tcos⁡6t,t > 0

T.F = L{C(t} = 12.5(s−(−8))/(s−(−8))2+62)

TfF = (12.5s+100)/(s2+16s+100)

Comparing with s2+2ξωnn2

ωn2 = 100→ωn = 10

2ξωn = 16→ξ = 0.8

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 22

Which of the following conversions is performed by the circuit shown in figure?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 22
y3 = x3 (MSB)

y2 = x3 ⊕x2,y1 = y2 ⊕ x1, y0 = y1 ⊕ x0

So, the conversion is gray code to Binary code

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 8 - Question 23

For the CE amplifier shown, calculate value of lower cutoff frequency FL (in Hz) due to C2

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 23
Rc = 200Ω,Cb = ∞

C2 = Cc = 4.2μF and Rs = 1KΩ

Now Re⁡q = Ro, + RL = 1K + 200Ω = 1.2KΩ

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 24

The phenomenon known as "Early Effect" in a bipolar transistor refers to a reduction of the effective base-width caused by

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 24
In BJT, as the B-C reverse bias voltage increases, the B-C space charge region width increases, where xB (i.e. neutral base width) > A. Change in neutral base width will change the collector current. A reduction in base width will cause the gradient in minority carrier concentration to increase, which in turn will cause an increase in the diffusion current. This effect is known as "Early Effect".
*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 8 - Question 25

Calculate the hysteresis width (in Volts) for an inverting Schmitt trigger with feedback fraction 0.5. Assume the supply voltage to be 10V.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 25
For a inverting Schmitt trigger.

Given: β = 0.5, Vcc = 10V

Hysteresis Voltage VH = VUT - VLT

VUT = +β x Vcc

VLT = -β x Vcc

VH = 2 x β x Vcc

VH = 2 x 0.5 x 10

VH = 10V

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 26

Consider the following graphs:

Which of the these graphs is/are non-planar?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 26

Other three circuits can be drawn on plane without crossing as shown below.

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 27

When a relatively small d.c. voltage is applied across a thin slice of Gallium Arsenide with thickness of order of few tenths of a micrometer and the voltage gradient across the slice is in excess of about 3300 V/cm, then the negative resistance will manifest itself. Which of the following devices operate based on the above phenomenon?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 27
The above-mentioned phenomenon occurs in Gunn diode and it is called the GUNN EFFECT. Gunn Effect is exhibited by semiconductor materials like gallium arsenide, indium phosphide, cadmium telluride and indium arsenide.
GATE Mock Test Electronics Engineering (ECE)- 8 - Question 28

The current flowing through the resistance R in the given circuit has the form P cos 4t, where P is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 28
Here the value of inductance is not given we can ignore it

I1 = 2cos 4t/3 = 0.67 cos 4 t i.e. ω = 4

= (0.23 - 0.35j)Cos 4 t

GATE Mock Test Electronics Engineering (ECE)- 8 - Question 29

In a superheterodyne AM receiver ,the image channel selectivity is determined by :

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 29
Image frequency rejection can be improved by placing more no of tank circuits, in between Antenna and the IF amplifier and by increasing their selectivity against image frequency.
GATE Mock Test Electronics Engineering (ECE)- 8 - Question 30

The approximate Bode magnitude plot of a minimum-phase system is shown in the figure. The transfer function of the system is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 8 - Question 30
The given bode plot is shown below

At ω = 0.1 change in slope is + 60 dB → 3 zeroes at ω = 0.1

At ω = 10 change in slope is - 40 dB → 2 poles at ω = 10

At ω = 100 change in slope is -20 dB → 1 poles at ω = 100

Now 20 log10 K = 20

or K = 10

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