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

Practice Test: Electronics Engineering (ECE)- 14 for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) Mock Test Series 2025 preparation. The Practice Test: Electronics Engineering (ECE)- 14 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Practice Test: Electronics Engineering (ECE)- 14 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)- 14 below.
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Practice Test: Electronics Engineering (ECE)- 14 - Question 1

A invest 1/3 part of the capital for 1/6 of the time, B invest 1/4 part of the capital for 1/2 of the time and C invest rest of the capital for rest of the time. Out of a profit of Rs. 23000, B’s share is?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 1
Raio of their Investment

Practice Test: Electronics Engineering (ECE)- 14 - Question 2

Pipe A can fill a tank three times as fast as pipe B. If together two pipes can fill the tank in 48 min, the slower pipe alone will be able to fill the tank in:

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 2
A = 3B

Ratio of efficiency, A:B = 3:1

Ratio of times, A:B = 1:3

Total capacity = Total efficiency × Total time = 4 × 48 = 192 unit

Time taken by slower pipe

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Practice Test: Electronics Engineering (ECE)- 14 - Question 3

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

Gratuitous

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 3
The word “gratuitous” means given or done free of charge. Thus, the word “costly” would be the correct antonym of the given word.

Gratis means without charge; free. Thus, option B is the correct answer.

Practice Test: Electronics Engineering (ECE)- 14 - Question 4

Find the area bounded between parabola and the line y2 = x,y = 2.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 4

y = 2, y2 = x

⇒ x = 22 = 4

∴ Parabola and intersect at point (4,2)

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

The bar graph shows the number of employees working under the six different Departments (A, B, C, D, E, F) of a certain company. Study the diagram and answer the following questions.

If departments F and D are merged to create a new department G, then which department will have the least number of employees?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 5
If departments F and D are merged to create a new department G, then

Employees in department A = 25

Employees in department B = 6

Employees in department C = 10

Employees in department E = 15

Employees in department G = 8

∴ Department B has the least number of employees

Practice Test: Electronics Engineering (ECE)- 14 - Question 6

In the following question, a sentence is given with a blank to be filled in with an appropriate word. Select the correct alternative out of the four and indicate it by selecting the appropriate option.

Confusion prevails in madrasas in Uttar Pradesh over the distribution of free NCERT textbooks as the academic session ____________ from August.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 6

The answer is ‘has begun’ because we use Present Perfect Tense, if the action is important and not the time of action or an action that has recently finished. Thus, option D is the correct answer.

Practice Test: Electronics Engineering (ECE)- 14 - Question 7

A sum of Rs.400 amounts to Rs.480 in 4 years. What will it amount to if the rate of interest is increased by 2 % for the same time?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 7

We know that,

Now rate is increased by 2 %

So, new rate is 7%

New Amount = S.I + P = 112 + 400 = Rs.512

Practice Test: Electronics Engineering (ECE)- 14 - Question 8

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: The Information and Broadcasting Ministry plans to conduct an independent study to gauge the impact of government advertisements on people.

Q: The advertisements are carried on various platforms, including print and visual media.

R: The Directorate of Advertising and Visual Publicity (DAVP) is the nodal agency of the government for advertising on behalf of the various ministries.

S: The initiative comes ahead of the Lok Sabha election in 2019 for which the government is expected to reach out to the people and highlight the works done by it in the past 4 years.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 8

The paragraph talks about the plans and advertisements of The Information and Broadcasting Ministry, which is given in sentence P.

P is the first statement. The word ‘initiative’ given in the sentence S is talking about the plans.

Hence, S must follow P. Now, the introduction the advertising agency is given in the sentence R, which must be the next statement.

Thus, the sequence after rearrangement is PSRQ and option B is the correct answer.

Practice Test: Electronics Engineering (ECE)- 14 - Question 9

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 the sentence is free from error, select 'No error'.

The gold foil used liberal (1)/ in Thanjavur paintings serves (2)/ many objectives that makes the painting more attractive. (3)/ No error

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 9

The error is in part (1) of the sentence. Change ‘liberal’ to ‘liberally’ because in this sentence it is in adjective form while the proper usage of liberal is in its adverb form i.e. ‘liberally’ as it qualifies the gold foil here.

Practice Test: Electronics Engineering (ECE)- 14 - Question 10

A, B and C can do a job in 6 days, 12 days and 15 days respectively. C works till 1/8 of the work is completed and then leaves. Rest of the work is done by A and B together. Time taken to finish the remaining work by A & B together is how much?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 10

Remaining work

∴ Time taken in doing 7/8 part of work

Practice Test: Electronics Engineering (ECE)- 14 - Question 11

Given a network with values of components depicted in the figure.

Find the sum of current through 5Ω and 4Ω resistor.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 11
Assume the current in the first loop be I1 and in the second loop be I2.

Then, we can apply mesh analysis to solve it.

Mesh 1: 15I1 - 10I2 = 5

Mesh 2: -10I1 + 20I2 = 10

On solving them, we have

I1 = 1A

I2 = 1A

Thus, sum = 2A

Practice Test: Electronics Engineering (ECE)- 14 - Question 12

The number of possible distinct Boolean expression of 3 variables will be ___________.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 12
The number of distinct Boolean expression of n variables is 22n, where n is number of variables

Practice Test: Electronics Engineering (ECE)- 14 - Question 13

Given the following system

T{X[n]} = X[n] + 3u[n+1]

Where u[x] represents unit step functions-

Which of the following is a correct representation of the system?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 13
We have,

T{X2[n] + X1[n]} = X1[n] + X2[n] +3u[n+1]

And

T{X1[n]} = X1[n] + 3u[n+1]

T{X2[n]} = X2[n] + 3u[n+1]

Since,

T{X2[n] + X1[n]} ≠ T{X1[n]} + T{X2[n]}

Thus, system is non linear.

T{X[n-no]} = X[n-no] + u[n+1]

≠ y[n-no]

Thus, system is Time Variant.

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

Consider the following sentences regarding a constant signal.

1) A constant signal is a periodic signal because it has no fundamental period.

2) A constant signal is an anti periodic signal because it never repeats itself.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 14
1 is true because a constant signal is an special type of periodic signal whose fundamental period is not defined because it repeats itself at each and every small increment of time and in continuous time domain small increment can’t be calculated.
Practice Test: Electronics Engineering (ECE)- 14 - Question 15

List-1 (pole location) with list-2(shown constant amplitude with impulse response).

Select the correct answer using the codes given below.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 15
For A

If we plot A then it is similar to the 4 which is followed by equation K1+K2e-at+K3eat

For B

If we plot B then it is similar to the 1 which is followed by equation (sinat + sinbt) u(t)

For C

If we plot c then it is similar to the 3 which is followed by equation eatsinbt u(t)

For D

If we plot D then it is similar to the 2 which is followed by equation sinat u(t)

Practice Test: Electronics Engineering (ECE)- 14 - Question 16

In the root locus for open loop transfer function is the ‘break away/in points are located at

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 16
1 +GH = 0

s2 + 8s + 15 + K(s+6) = 0

Practice Test: Electronics Engineering (ECE)- 14 - Question 17

In a PCM system, if the code word length is increased from 6 to 8 bits, the signal to quantization noise ratio improves by the factor

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 17
When word length is 6

When word length is 8

Thus it improves by a factor of 16.

Practice Test: Electronics Engineering (ECE)- 14 - Question 18

The op-amp configuration shown below has following transfer function The feedback resistance used has 1.5 MΩ , the value of capacitance will be ___________ µf.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 18

Practice Test: Electronics Engineering (ECE)- 14 - Question 19

A semiconductor sample at room temperature has intrinsic concentration of 2.5 X 1017 /m3. After doping what will be the minority carrier concentration if the majority carrier concentration is given as 5.5 X 1021 /m3.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 19
In a pure Semiconductor (Intrinsic Semiconductor), the electron and hole concentrations are n1p1 respectively. By doping impurity atoms the SC becomes extrinsic then the electrons and hole concentrations n2p2 respectively, then the following equations are acceptable

n1p1 = n2p2 = ni2For Intrinsic Semiconductor, n = p = ni2 and as per questions before doping n1p1 = ni2

Therefore,

Practice Test: Electronics Engineering (ECE)- 14 - Question 20

An LED is connected as shown in figure. It should glow when V1 is at logic ‘0’ state (0.2V). Calculate R( in Ω). (Assume in active state current as 15 mA)


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 20
V1 is at logic ‘0’ ⇒ V1 = 0.2

By KVL

5 - 1.7 - 0.2 - (15 + R)15mA = 0

3.1/ (15 × 10-3) = 15 + R

R = 191.67 Ω

Practice Test: Electronics Engineering (ECE)- 14 - Question 21

For an ideal p-channel MOSFET, μp = 300cm2/v-s, W = 15μm, L = 1.5μm, tox = 300A, Vt = -0.7V. If the transistor is non-saturation region at VSD = 0.5V, then what is the Transconductance gm?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 21
ID = (μpCox/2)(W/L)(2(VSG+VT)VSD-VSD2)2,

where

Cox = (3.9 x 85 x 1014)/(300 x 1010).

Cox = 1.15 x 10-7 F/m2. Also, gm = ∂(ID)/∂(VSG).

On substituting and solving, gm = 0.172mS.

Practice Test: Electronics Engineering (ECE)- 14 - Question 22

In the circuit shown below express the current I0 in terms of Vi

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 22
The current going in the op amp is shown above

As per virtual short,

Practice Test: Electronics Engineering (ECE)- 14 - Question 23

From the circuit given below, find out the operating region of the transistors T1 and T2

(VTH = -0.4)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 23
For T1

VSD = VS – VD = 1.5 – 0 = 1.5 V

VSD(sat) = VSG + VTH = (1.5 − 0.5) − 0.4

= 1 – 0.4 = 0.6V

Here, VSG > (VTH) & VSD > VSD(sat)

So, T1 is in Saturation region

Similarly for T2,

VSD = VS − V0 = 0.9 − 0.9 = 0

VSD(sat) = VSG + VTH = (0.9 − 0) − 0.4 = 0.5V

Here VSD < VSD(sat) & VSG > (VTH)

∴ T2 in linear region

Practice Test: Electronics Engineering (ECE)- 14 - Question 24

In the circuit shown below, assume that the voltage drop across a forward biased diode is 0.7 V. The thermal voltage V1 = kT/q = 25 mV. The small signal input Vi = Vp cos(ωt) where Vp = 100 mV.

The bias current IDC through the diodes is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 24
IDC = [12.7-(0.7 + 0.7 + 0.7 + 0.7)]/9900 = 1mA

Hence option A is correct

Practice Test: Electronics Engineering (ECE)- 14 - Question 25

A communication channel having AWGN characteristics is operating in such a way that SNR >> 1. The bandwidth of signal being transmitted is B and capacity C1. Determine the capacity of channel if a signal with half the bandwidth is transmitted through the same channel with same quality.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 25
Here same quality implies that SNR for transmitted signals is kept equal. We have a formula for channel capacity as

Now, B2 = 0.5 × B1. SNR is same, thus C2 would be,

Practice Test: Electronics Engineering (ECE)- 14 - Question 26

A three stage amplifier with identical stages with lower cut-off frequency per stage 'f1' is given overall negative feedback. Depending on the overall gain, the system may oscillate at a low frequency fc given by

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 26

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 27

Consider the differential equation 3yn(x) + 27y(x) = 0 with initial conditions y(0) = 0 and y'(0) = 2000. The value of y at x = 1 is


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 27
3y′′(x) + 27y(x) = 0, y(0) = 0, y′(0) = 2000

Auxiliary equation, 3m2 + 27 = 0 ⇒ m2 + 9 = 0

⇒ m = 0 + 3i

yc = c1 cos⁡ 3x + c2 sin⁡3x and yp = 0

∴ yc = c1 cos ⁡3x + c2 sin ⁡3x

y(0) = 0 ⇒ c1 + 0 = 0 ⇒ c1 = 0

∴ y = c2 sin⁡3x

y = 3c2 cos⁡3x

y′(0) = 2000 ⇒ 2000 = 3c2 ⇒ c2 = 2000/3

Here in sin⁡ 3, 3 is in radian as it is a value.

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 14 - Question 28

Let z be a complex variable. For a counter-clockwise integration around a unit circle C, centred at origin, the value of A is


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 28
5z - 4 = 0

z = 4/5 lies circle

Practice Test: Electronics Engineering (ECE)- 14 - Question 29

In the given circuit, the equivalent impedance of the circuit between terminals A-B is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 29
To find out Thevenin’s equivalent, we put a test source between terminals A-B.

KCL at node A

Practice Test: Electronics Engineering (ECE)- 14 - Question 30

The ROCs of different impulse responses are shown below. Which of the following impulse responses are stable?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 14 - Question 30
For stability of a system, ROC must include imaginary axis.
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