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GATE Mock Test Electronics Engineering (ECE)- 6 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test - GATE Mock Test Electronics Engineering (ECE)- 6

GATE Mock Test Electronics Engineering (ECE)- 6 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The GATE Mock Test Electronics Engineering (ECE)- 6 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The GATE Mock Test Electronics Engineering (ECE)- 6 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)- 6 below.
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GATE Mock Test Electronics Engineering (ECE)- 6 - Question 1

Direction: Study the following pie-chart and tables carefully and answer the questions given below.

Total Number of email received by the organization = 90000

Ratio of Read emails to Unread emails received by the organization

What is the ratio of the number of emails read in January to those unread in the month of April in the organization?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 1
Number of read emails in the month of January = (90000*17/100*8/15)

Number of unread emails in the month of April = (90000* 8/100 * 5/12)

Ratio = 68:25

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

The shape in the figure given below is folded to form a box. Choose from the alternatives (1), (2), (3) and (4), the boxes that are similar to the box formed from the shape below.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 2
The dot will lie opposite to one of the shaded surfaces.

Therefore, figure (2) cannot be formed.

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GATE Mock Test Electronics Engineering (ECE)- 6 - Question 3

In the following question, out of the given four alternatives, select the one which is opposite in the meaning of the given word.

Scurrilous

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 3
The meanings of the given words are as follows

Scurrilous— abusive

Coarse— rough or harsh in texture

Sophisticated— developed to a high degree of complexity.

Insolent— showing a rude and arrogant lack of respect

Complimentary— expressing a compliment; praising or approving

Clearly, option D is the antonym of the given word.

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 4

What is the chance that a leap year, selected at random, will contain 53 Sundays?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 4
There are 52 complete weeks in a calendar year i.e. 52 x 7 = 364 days

Number of days in a leap year = 366

∴ Probability of 53 Sundays = 2/7

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 5

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

P, Q, R, S, T, U, V and F are sitting around a circle facing the centre. U is third to the right of Q, who is third to the right of F. P is third to the left of F. R is fourth to the left of P. T is third to the right of S. S is not a neighbour of P.

Four of the following five are similar in a certain way based on their positions in the seating arrangement and so form a-group. Which of the following does not belong to that group?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 5

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

In a group, the ratio of the number of boys to the number of girls is P. The ratio of the number of girls to the number of boys is Q, then P + Q is always

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 6
We know that (x + 1/x) is always greater than or equal to 2.

Now, let x be the number of boys and y be the number of girls, then

x/y = P and y/x = Q

So, P + Q = (x/y) + (y/x) ≥ 2

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 7

In a school, 12th class consists of 30% male students of which 30% male students failed in the class. Total 82% students passed in 12th examination out of 900 students. Calculate the total number of female passed students?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 7
Short Trick

Total students = 900

Boys : Girls = 30 % : 70% = 3 : 7

So, exactly Boys : Girls = 270 : 630

Total failed students = 18% of 900 = 162

Failed boys = 30% of 270 = 81

Failed girls = 162 - 81 = 81

Passed female students = 630 - 81 = 549

Basic Method:

⇒ Let us consider total number of students be 100x

⇒ ∴ Males students = 30x , female students = 70x

⇒ According to condition given in the problem, of this 30x male students, 70 % male students pass the exam.

ie. Passed males = 30x × 70/100 = 21x

⇒ Total passed students = 100x × 82/100 = 82x

⇒ ∴ female students passed = 82x – 21x = 61x

⇒ ∴ Girls passed = 61% of total

∴ Answer is 61/100 x 900 = 549

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

A cube of side 10 cm is coloured red, with a 2 cm wide green strip along all the sides on all the faces. The cube is cut into 125 smaller cubes of equal sizes. How many cubes have at least two green faces each?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 8
There are 12 cubes in each of the layers 1 and 5 and 4 cubes in each of the layers 2, 3 and 4 which have two faces coloured green. Also, there are 4 cubes in each of the layers 1 and 5 which have three faces coloured green.

3 faces coloured cubes = 8

2 faces coloured cubes = 24 + 12 = 36

Total cubes = 44

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 9

Direction: In the following number series, only one number is wrong. Find out the wrong number.

17 25 34 98 121 339

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 9
The pattern followed here is,

⇒17 + 23 = 17 + 8 = 25

⇒25 + 32 = 25 + 9 = 34

⇒34 + 43 = 34 + 64 = 98

⇒98 + 52 = 98 + 25 = 123

⇒123 + 63 = 123 + 216 = 339

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 10

Directions: In the question given below, a statement is followed by three courses of action labelled (A), (B) and (C). A course of action is a step or administrative decision to be taken for improvement, follow-up or further action in regard to the problem, policy, etc. On the basis of the information given in the statement, you have to assume everything in the statement to be true and then decide which of the suggested courses of action logically follow(s) for pursuing.

Statement: Many political activists have decided to stage demonstrations and block traffic movement in the city during peak hours to protest against the steep rise in prices of essential commodities.

Courses of action:

(A) The Govt. should immediately ban all forms of agitations in the country.

(B) The police authority of the city should deploy additional forces all over the city to help traffic movement in the city.

(C) The state administration should carry out preventive arrests of the known criminals staying in the city.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 10
(A) is not feasible in a democracy. (C) is not followed because the problem is not concerned with criminals. (B) is the only course that authorities can resort to.
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 11

In the following figure, the J and K inputs of all the four Flip-Flops are made high. The frequency of the signal at output Y is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 11
Output of NAND is zero when Q3Q2Q1Q0 have state 1010=(10), Dec. Therefore, the given figure represents mod- 10 up counter. And, frequency of the signal will be

f = 10KHz/10 = 1KHz

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 12

For the function e-x, the linear approximation around x = 2 is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 12
We have f(x) = e-x = e-(x - 2) - 2 = e-(x - 2)e-2

= [1 - (x - 2)]e-2 Neglecting higher powers

= (3 - x)e-2

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 13

Consider the lossless transmission line circuit shown in the figure, The voltage standing wave ratio on 50 Ω line is given by

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 13
Zin at junction is given by

Zin = Zin(λ/4) ||Zin(150Ω)

Zin = Z02/ZL = 225 Ω

VSWR on 50 Ω is 225/50 = 4.5

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 14

The Laplace transform of a function is(S + 1)/S(S + 2) . The initial and final values, respectively, of the function are

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 14
For initial value, finding

For final value, Finding

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 15

Given x(t) = e−tu(t). Find the inverse laplace transform of e−3s X(2s).

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 15

x(t) = e−tu(t)

Taking laplace transfarm

X(s) = 1/s+1

X(2s) = 1/2s+1 = 1/2[1/(s+1/2)]

Taking inverse laplace transfarm

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 16

Under low level injection assumption, the injected minority carrier current for an extrinsic semiconductor is essentially the

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 16
Under low level injection, the minority current for an extrinsic semiconductor is by diffusion as there is concentration gradient due to low level injection.
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 17

Find the differential equation of the system described by the transfer function given as:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 17
We have,

Y(s)/X(s) = s + 3/s(2s + 5)

⇒Y(s) [2s2 + 5s] = X(s)⋅(s + 3)

⇒ 2s2⋅Y(s) + 5⋅Y(s) = s⋅X(s) + 3⋅X(s)

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 18

Choose proper substitutes for X and Y to make the following statement correct.

Tunnel diode and avalanche photo diode are operated in X bias and Y bias, respectively.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 18
Tunnel diode shows the negative characteristics in forward bias. It is used in forward bias. Avalanche photo diode is used in reverse bias.
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 19

The built-in potential of an abrupt p-n junction is 0.75 V. If its junction capacitance (Q) at a reverse bias (VR) of 1.25 V is 5 pF, the value of CJ (in pF) when VR = 7.25 V is________.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 19

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 20

Consider the state equation for a system given below.

Which of the following conditions is true for complete controllability?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 20

Controllability matrix = [B AB]

must not be equal to zero;

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 21

Assume electronic charge q = 1.6×1019C,kT/q = 25mVand electron mobility μn = 1000cm2/V−s.If the concentration gradient of electrons injected into a P-type silicon sample is 1×1021 cm4,the magnitude of electron diffusion current density (in A/cm2) is _________.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 21
Given q = 1.6×10−19; kT/q = 25mV

μn = 1000cm2/v−s

From Einstein relation, Dnn = kT/q

⇒D = 25mV × 1000cm2/v−s

⇒ 25cm2/s

Diffusion current Density

J = qDndn/dx

= 1.6 x 10-19 x 25 x 1 x 1021

= 4000 A/cm2

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 22

Without any additional circuitry, an 8 : 1 MUX can be used to obtain

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 22
A 2n : 1 MUX can implement all logic functions of (n + 1) variable without any additional circuitry.

Here, n = 3.

Thus, an 8 : 1 MUX can implement all logic functions of 4 variables.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 23

The voltage gain of an amplifier is 100. A negative feedback is applied with β=0.04. The overall gain of the amplifier is:


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 23
Initially the voltage gain of the amplifier is 100 i.e. A=100.

Then a negative feedback is applied to the amplifier where the gain of the feedback loop is given as β = 0.04

Now the overall gain of the negative feedback amplifier is given as, = A/1+Aβ

= 100/[1+(100)(0.04)]

= 100/(1+4)

= 100/5

= 20

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 24

In the circuit shown, the device connected to Y5 can have address in the range

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 24
The simple decoder is shown below

Addressing is done based on the inputs and their respective logic circuit connected to the address decoders.

Calculation:

In the given question the chip select for Y5 is 101.

A10A9A8 are used for that chip select so A10A9A8 = 101.

A7 ⋯ A0 are initially zeroes and at the final address these will be all 1's.

To get the active high output for the NAND gate all the inputs must be 1 always.

The total addressing range is shown below:

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 25

Consider two resistors of 50 kΩ and 100 kΩ at room temperature of 27 OC. These resistors are passing a signal of bandwidth 50 kHz. Let N1 and N2 be the noise voltage generated while operating in series and parallel configuration, respectively. Determine N1 and N2.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 25
Given parameters,

T = 300 K

RS = 150 kΩ

RP = 100/3 kΩ.

B = 50 kHz.

We can find out the noise voltage generated through resistors using following formula,

Vn2 = 4RkTB, where k is Boltzmann constant.

Thus, N1(Series) would be 11.15 µV and N2(Parallel) would be 5.25 µV.

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 26

The transfer function of a phase-lead compensator is given by Gc = , where T > 0. What is the maximum phase shift of the compensator?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 26
The transfer function of given compensator is

Comparing with

The maximum phase shift is

or Φmax = π/6

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 27

The amplitude of a random signal is uniformly distributed between -5 V and 5 V.

If the signal to quantization noise ratio required in uniformly quantizing the signal is 43.5 dB, the step size (in V) of the quantization is approximately

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 27
(S/Nq)0dB = 1.76 + 6.02u

⇒43.5 = 17.6 + 6.024

⇒ n ≈ 7

Δ(Step size) = 2A/2′′

=10/27 = 0.07 Volts

Which is close to 0.0667 Volts

GATE Mock Test Electronics Engineering (ECE)- 6 - Question 28

During transmission over a certain binary communication channel, bit errors occur independently with probability p. The probability of at most one bit in error in a block of n bits is given by

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 28
By Binomial distribution the probability of error is

Pe = nCr Pr (1- P) n - r

Probability of at most one error

= Probability of no error + Probability of one error

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 29

For a telephone line, β= 0.02rad/km. If frequency is 1 kHz. Calculate time taken to travel 3.14 km by the wave in µs


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 29
β = 0.02rad/km,f = 1kHz,d = 3.14km

Wavelength λ = 2π/β = 2π/0.02 = 314.159km

Velocity v = ω/β = 2πf/β = 2π∗1000/0.02 = 3.14 × 105km/sec

Time t = d/v = 3.14/(3.14 × 105) = 105Sec

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 6 - Question 30

The Laplace transform of i(t) is given by I(s) = 2/s(1 + s) . As t ⟶∞, the value of i(t) tends to be (Answer up to the nearest integer)


Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 6 - Question 30
From the Final value theorem, we have

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