Practice Test: Electronics Engineering (ECE)- 15 - Electronics and Communication Engineering (ECE) MCQ

# Practice Test: Electronics Engineering (ECE)- 15 - 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)- 15

Practice Test: Electronics Engineering (ECE)- 15 for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) Mock Test Series 2025 preparation. The Practice Test: Electronics Engineering (ECE)- 15 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Practice Test: Electronics Engineering (ECE)- 15 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)- 15 below.
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Practice Test: Electronics Engineering (ECE)- 15 - Question 1

### There are two lines made by joining points A, B,C. B lies between the line joining A and C. Is the distance between the A and C passes through the B more then 7 Km.I. The distance between A and B is 6 KmII. Distance between B to C is 7 Km long,

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 1
From Statement I

We know the distance between Points B and C but we do not know about the distance between A and B

Statement II

This statement tells us about the distance between A and B and by using Both statement we can calculate the distance between A and C that is longer than the 7 Km.

Practice Test: Electronics Engineering (ECE)- 15 - Question 2

### Direction: In the given question, a word/phrase is given followed by three statements; I, II and III. Choose the pair of sentences which can be combined using the given word/ phrase when used at the beginning of the new sentence.AndI: When Lionel Messi, Ronaldo and Neymar kick and dribble the football, they don’t remain confined to Argentina, Portugal or Brazil, respectively.II: Let the entire world realise that through sports, especially football, we all can cement our bonds, wash away our bitterness and prejudices.III: Live with a greater sense of closeness and joy with each other.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 2
‘And’ is used as a conjunction here which is used to introduce an additional comment or interjection.

Here sentences II and III are displaying same sense that is realizing the benefit of football in our lives whereas sentence I is providing entirely different aspect of football as a sport.

Hence option (B) will be the most appropriate choice.

New sentence: Let the entire world realise that through sports, especially football, we all can cement our bonds, wash away our bitterness and prejudices, and live with a greater sense of closeness and joy with each other.

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Practice Test: Electronics Engineering (ECE)- 15 - Question 3

### Which of the following is the MOST SIMILAR in meaning to Accreditation?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 3
Accreditation = an acknowledgement of a person's responsibility for or achievement of something.

Certification = an official document attesting to a status or level of achievement.

Meticulous = showing great attention to detail; very careful and precise.

Lurid = very vividly shocking.

Suppressive = tending or acting to suppress.

Agreement = harmony or accordance in opinion or feeling.

Thus, option B is the correct answer.

Practice Test: Electronics Engineering (ECE)- 15 - Question 4

In how many ways can you place 2 white bishops on an empty chess board?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 4
One white bishop can be placed on any of the 32 white boxes and the other white bishop can be placed on any of the 32 black boxes.

Ways of doing that = 32C1 * 32C1

= 32 × 32 = 1024

Practice Test: Electronics Engineering (ECE)- 15 - Question 5

Direction: In the question below are given statements followed by some conclusions. You have to take the given statements to be true even if they seem to be at variance with commonly known facts. Read all the conclusions and then decide which of the given conclusions logically follows from the given statements disregarding commonly known facts.

Statement:

No physics is maths.

Some chemistry is maths.

All sciences are chemistry.

Conclusion:

I. No science is physics.

II. Some physics are science.

III. Some physics are chemistry.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 5
The least possible Venn diagram for the given statements is as follows.

Conclusions:

I. No science is physics → False (It is possible but not definite).

II. Some physics are science → False (It is possible but not definite)

III. Some physics are chemistry → False (It is possible but not definite)

Conclusion I and II form complementary pair.

Hence, either conclusion I or II follows.

Practice Test: Electronics Engineering (ECE)- 15 - Question 6

Direction: Read the information carefully and give the answer of the following questions-

(This pie chart shows the percentage of students appear in six different exams in 2016)

(This bar graph shows the percentage of failed students which appear in these six different exams in 2016)

If in 2016, total number of failed students in exam F was 4080, then how many passed students appears in the exam B?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 6

14280 (Let total number of students be = 100z.

Total number of students appears in exam F

= 16% of 100z

= 16z;

Total number of failed students in exam F

= 30% of 16z

= 4.8z = 4080;

⇒ z = 850;

Total number of passed students appears in the exam B

= (100 – 30)% of 24% of 100z

= 16.8z = 16.8 × 850 = 14280

Practice Test: Electronics Engineering (ECE)- 15 - Question 7

Direction: Two sentences with two blanks in each, followed by five alternatives with two words in each, are given. Choose that option as the answer which can fill both the blanks of both the sentences.

i. A sinking feeling of panic ________ over them and a temporary paralyzing fear engulfed them ________.

ii. Being a cleanliness freak, she ________ the floor and went down to the market only after the house was _________ clean.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 7

‘Sweep’ is to clean (an area) by brushing away dirt or litter. It fits the first blank of the second sentence, as the subject is cleaning the floor. Further, ‘sweep over’ means to overcome or overwhelm, thus fits in the first blank of the first sentence. The second blank of both the sentences can be filled by "completely", thus conveying an appropriate sense. Thus, option D is the correct answer.

Practice Test: Electronics Engineering (ECE)- 15 - Question 8

Direction: Five statements are given below, labelled 1, 2, 3, 4 and 5 which are supposed to be in a logical order. A statement labelled P is given thereafter. P can replace one of the five statements such that the four statements along with P would make a coherent paragraph. You have to identify which statement should P replace and then the find out the correct sequence from the options. If the five options are in logical order and form a coherent paragraph/passage, choose the fifth option “12345”.

1) The future of Germany’s coalition government is hanging in the balance after the country’s interior minister reportedly announced his intention to resign over a migration showdown with Angela Merkel.

2) Horst Seehofer, who is also leader of the Christian Social Union, on Sunday night offered to step down from his ministerial role and party leadership in a closed-door meeting in which he and fellow CSU leaders had debated the merits of the migration deal Merkel hammered out with fellow European Union leaders in Brussels.

3) But with CSU hardliners believed to have tried to talk the combative interior minister into staying, a press conference was postponed until Monday, with Seehofer seeking to go back to Merkel in search of a final compromise.

4) At a 2am media conference, Seehofer said he had agreed to meet again with Merkel’s party before he made his decision final.

5) “We’ll have more talks today with the CDU in Berlin with the hope that we can come to an agreement,” Seehofer said.

P. Merkel said on Sunday she wanted the CDU and its Bavarian allies to continue working together.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 8

The paragraph talks about the issues pertaining to the state of coalition government in Germany. The first sentence introduces the issue to us. The second sentence explains the problem. The third sentence then goes on to elaborate upon the meeting, and the fourth and fifth sentence talk about the current state and measures being taken. Thus, they are in perfect order. Sentence P is completely out of place here because CDU is not the subject of this paragraph and it does not matter what Merkel thinks they should do. Thus, D is the correct answer.

Practice Test: Electronics Engineering (ECE)- 15 - Question 9

4 identical solid spheres are melted and re-formed into a solid hemisphere. Then, the ratio of the curved surface area of the hemisphere to half of the surface area of a single sphere is -

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 9

Let radius of sphere be ‘r’ and radius of hemisphere be ‘R’.

Then,

ATQ,

Volume of Spheres = Volume of Hemisphere

4 x (4/3)πr3 = (2/3)πR3

Or, R / r = (8)1/3 = 2 …(1)

C.S.A of hemisphere = 2πR2

And, Surface area of sphere = 4πr2

ATQ,

2πR2 :( ½) of 4πr2

= 2πR2 :2πr2 = R2 :r2 {using (1)}

= 4 :1

Practice Test: Electronics Engineering (ECE)- 15 - Question 10

A study of people who reduced the calories they consumed has found the strongest evidence yet that such restrictions slow down metabolism, raising hopes that a low calories lifestyle or treatments stimulating biological effects of restricted eating, could prolong health in old age. The report provides the most robust evidence to date that everything we have learnt in other animals can be applied to humans.

Which of the following argument will prove that the above conclusion is flawed?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 10

The given passage gives the benefits of reducing calories to attain a healthier lifestyle and out of the given options, option B seems to give a contradictory fact stating that the lack of required calories may cause problems to the body. Hence, option B is correct.

Practice Test: Electronics Engineering (ECE)- 15 - Question 11

The value of k such that the system of equation

x + ky + z = 0, x + 3y + kz = 0, 2x + y + z = 0 has non-trivial solution.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 11

To find the value of K, we have to make rank of matrix less than unknown of the equation.

Apply R2 = R2 - R1

R3 = R3 - 2R1

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 12

The frequency at the output of the following cascaded circuit is …………. Hz

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 12
Modulus of 10 -bit ring counter = 10

Modulus of 4 -bit ripple counter = 2N = 24 = 16

Modulus of 5-bit Johnson counter = 2N = 2 × 5 = 10

Frequency at the output

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 13

A PCM (Pulse Code Modulation) system uses 8-bit encoder and uniform quantization. What is the maximum bandwidth(in MHz) of the low-pass input message signal for which the bit rate of the system is 20Mbps?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 13
Given system is PCM, n = 8bits, Rb = 20. Mbps

We have,

Again, by Nyquist Criterion, fz ≥ 2f

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 14

A super heterodyne receiver having RF amplifier is tuned to 1200 kHz, the intermediate frequency is 450 kHz. The quality factor of the tuned circuit at RF amplifier and at the mixer are same and is equal to 65. The image rejection ratio is

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

a = Image rejection ratio

[Turned circuit of same quality factor used ar two stages]

Therefore,

Practice Test: Electronics Engineering (ECE)- 15 - Question 15

Simplify the given Boolean expression

A+AB'+AB'C'+AB'C'D'+AB'C'D'E'+…………………

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 15
We have A (1 + B'+B'C'+B'C'D'+B'C'D'E'+…………………)

We know A + 1 = 1 (Rule)

Now A.1 = A

Practice Test: Electronics Engineering (ECE)- 15 - Question 16

If the skin depth is 60 µm at 3 MHz in a certain conducting medium then what will be the skin depth if the frequency is changed to 27 MHz?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 16

From the expression we find that δ is inversely proportional to square root of frequency,

Practice Test: Electronics Engineering (ECE)- 15 - Question 17

Find the Standing Wave Ratio of a transmission line of characteristic impedance 80Ω is connected to a load of 100Ω.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 17
Given, Z0 = 80Ω, ZL = 100Ω

Voltage reflection ratio,

Standing Wave Ratio, SWR

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 18

Consider the following continuous time signal x(t)

If Ck is the complex Fourier series coefficient of signal x(t). Then the value of |C2| will be

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 18

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 19

The circuit shown in the figure below is in steady state when switch is closed at t = 0, then the value of

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 19
At t = 0, the circuit is in steady state condition during this condition inductor will acts as short circuit

iL(0-) = 10/(9 + 1) = 10/10 = 1A

After t = 0+

By Applying KVL in the loop

VL + 1(1) = 0

VL = -1

Practice Test: Electronics Engineering (ECE)- 15 - Question 20

Characteristic equation of a system having unity feedback is

(s + 1)(s - 2)+ k(s - 3)(s + 4). The system is then which of the following?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 20
Transfer Function

For minimum phase system, all finite poles and zeroes are located in the left half of the s-plane. For a non-minimum phase system, one or more zeroes are located in the right half of the splane and remaining all zeroes and poles are located on the left half of the s-plane. For All-pass system, zeroes lies on the right half of the s-plane and poles lies on the left half of the s-plane and are symmetrical about the imaginary axis. Also, magnitude should be 1 and phase should be -180

Practice Test: Electronics Engineering (ECE)- 15 - Question 21

Consider a spatial curve in 3D space given on parameters form by

X = t

Y = 1

Z = 3t/π in 0 ≤ t ≤ π. The length of the curve is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 21

Practice Test: Electronics Engineering (ECE)- 15 - Question 22

The modified work function of N-channel MOSFET is -0.8V. The effective positive charge at the Semiconductor oxide layer interface is 9.8 x 10-5 C/m2 and the oxide capacitance is 300 μF/ m2, then calculate the flat band voltage.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 22
φms = -0.8V, Q0x = 9.8 x 10-8C/m2, C0x = 300μF/m2

The flat band voltage is given by

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 23

In the voltage controller circuit shown in the figure (assume that op-amp is ideal). The BJT has VBE = 0.7V & β is very large and V2 = 4.7V. For a regulated output of 9.4V, the value of R (in kΩ) is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 23
Out voltage for given circuit

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 24

Consider a lossless antenna with a directive gain of +6db. If 1mW of power is fed to it the total power radiated by the antenna will be

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 24
Lossless antenna with directive gain of +6dB = 4 (In linear)

Input power to antenna = 1 mW

Power radiated by antenna = 4 mW

Thus,

Radiated power = 4 × 1 mW = 4 mW

Hence option (a) is correct.

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 15 - Question 25

The output voltage Vo(in mV) of the circuit shown in figure

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 25

Practice Test: Electronics Engineering (ECE)- 15 - Question 26

For a given P+ - N Si junction diode with ND = 1015 cm-3 and ni = 1010 cm-3 and e = 8.85 × 10-14 C2/N-cm3. Find the electric field at the midpoint of the depletion region on the n-side. (let reverse bias voltage = 100V)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 26

We have

Practice Test: Electronics Engineering (ECE)- 15 - Question 27

The divergence of the vector

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 27
We have

Divergence is,

Practice Test: Electronics Engineering (ECE)- 15 - Question 28

For the voltage amplifier circuit shown in the figure, Av is very high. Also RS = 1mΩ, RL = 1kΩ R1 = 1 kΩ and R2 = 1mΩ

Then what is the value of |Vo/Vs| = ?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 28
This is a negative feedback amplifier

Hence option A is correct.

Practice Test: Electronics Engineering (ECE)- 15 - Question 29

The value of d/dt x(t) at t = 1.5 for x(t) = u(t) + r(t) - 2r(t - 1) + r(t - 2) - u(t - 2) is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 29
Plotting each signal one by one

By adding all above signals we get x(t) as

by differentiation of x(t)

Practice Test: Electronics Engineering (ECE)- 15 - Question 30

Consider an angle modulated signal X(t) = 10.cos [108 πt + Sin 2π 103t]

The maximum frequency deviation is ________kHz?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 15 - Question 30
Given x(t) = 10. Cos [108 πt + 5. Sin 2π103 t]

θ(t) = 108 πt + 5. Sin 2π103t

= ωct + ϕ (t)

∴ ϕ(t) = 5. sin 2π103t

∴ ϕ’ (t) = 104.π.cos 2π 103 t

Maximum frequency deviation is

Δω = | ϕ’(t)|max

∴ ∆fmax = 5 kHz.

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