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Practice Test: Electronics Engineering (ECE)- 2 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test - Practice Test: Electronics Engineering (ECE)- 2

Practice Test: Electronics Engineering (ECE)- 2 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The Practice Test: Electronics Engineering (ECE)- 2 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Practice Test: Electronics Engineering (ECE)- 2 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)- 2 below.
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Practice Test: Electronics Engineering (ECE)- 2 - Question 1

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 Practice Test: Electronics Engineering (ECE)- 2 - Question 1
Let the rate % = R

According to the question

100R + 120 R = 2200

2200 R = 2200

R = 10%

Hence required rate % = 10%

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

Amit rows a boat 9 kilometres in 2 hours down-stream and returns upstream in 6 hours. The speed of the boat (in kmph) is:

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 2

Down-stream rate = 9/2 = 4.5 kmph

Upstream rate = 9/6= 1.5 kmph

The speed of the boat = (4.5 – 1.5) kmph = 3 kmph

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Practice Test: Electronics Engineering (ECE)- 2 - Question 3

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 Practice Test: Electronics Engineering (ECE)- 2 - 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.

Hence the correct answer is option D.

Practice Test: Electronics Engineering (ECE)- 2 - Question 4

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

Turgid

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 4
Turgid = swollen and distended or congested; pompous

Bloated = inflated

Humble = plain, simple

Puffy = inflated

Tumescent = swollen

Hence, humble is the correct answer.

Practice Test: Electronics Engineering (ECE)- 2 - Question 5

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 chart 2 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 Practice Test: Electronics Engineering (ECE)- 2 - Question 5
Required ratio = 13% of 7000: 23% of 4500 = 182:207
Practice Test: Electronics Engineering (ECE)- 2 - Question 6

Directions: In each of the questions below, some statements are given followed by some conclusions. You have to consider the statements to be true even if they seem to be at variance with commonly known facts. You have to decide which of the following conclusions logically follows from the given statements. Give answer.

Statements:

Some jacket is shirt.

Some shirts are trouser.

No shoes are t-shirt.

All trousers are shoes.

All pants are t-shirt.

Conclusions:

I. All shirts being t-shirt is a possibility.

II. Some shoes are trouser.

III. Some jackets are t-shirt.

IV. Some shirts being t-shirt is a possibility.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 6

Practice Test: Electronics Engineering (ECE)- 2 - Question 7

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 Practice Test: Electronics Engineering (ECE)- 2 - 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 and option A is the correct answer.

Practice Test: Electronics Engineering (ECE)- 2 - Question 8

Direction: Which of the following is the MOST SIMILAR in meaning to the given word?

Remnant

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 8

‘Remnant’ means ‘remainder’ which is same as ‘residue’.

Practice Test: Electronics Engineering (ECE)- 2 - Question 9

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 Practice Test: Electronics Engineering (ECE)- 2 - 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%

Practice Test: Electronics Engineering (ECE)- 2 - Question 10

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

Seven people, P, Q, R, S, T, W and X sitting in a straight line facing north, not necessarily in the same order. R sits at one of the extreme ends of the line. T has as many people sitting on his right, as on his left. S sits third to the left of X. Q sits on the immediate left of W. Q does not sit at any of the extreme ends of the line.

If all the people are made to sit in alphabetical order from right to left, the positions of how many people will remain unchanged?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 10

From the given conditions, we can conclude

After arranging in alphabetical order,

Hence, only Q's position will remain unchanged.

Practice Test: Electronics Engineering (ECE)- 2 - Question 11

The figure shows the circuit of a gate in the resistor transistor logic (RTL) Then the circuit represent which gate?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 11

Hence gate is OR Gate.

Practice Test: Electronics Engineering (ECE)- 2 - Question 12

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 12

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 2 - Question 13

For the given p+ -n junction diode the minority carrier life time in the n -region is ΥP = 1.1 μsec. If the excess minority charges present in the n -region at any time t is given as QP(t) then QP(t) at t = 1.5 μsec is _____ μC?


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 13
For t < />

I0 = 2A

Diode in FB has diffusion capacitance, as current increases the charge inside increase. Hence voltage increase.

Now

Here

Y = life time of holes in N-region

= 1.1μsec

Multiplying equation (1) by C gives

And t = 1.5 \musec

Hence

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 2 - Question 14

For the given network, the power given to the load is ____ kW? [Assume R =1 Ω and ideal diode]

V = 100. sinωt


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 14
(3,4)

Using mirror symmetry

The resistive network between AB can be simplified as.

Further simplifying

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

A current source with an internal resistance of RS, supplied power to load RL. The plot of power delivered as a function of load RL is?

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

Hence option B is correct.

Practice Test: Electronics Engineering (ECE)- 2 - Question 16

where (x) denotes greatest integer

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 16

Practice Test: Electronics Engineering (ECE)- 2 - Question 17

For given system

Angle of arrival is?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 17
Zeros:- s= -3 ± j

Poles:- s = -1 ± j

ϕ’ = 180 - tan-11 = 135°

ϕA = 180° + ϕ

Where ϕ = SP - SZ

= 180 + 135 - 90

= 225°

∴ ϕA = 180° + ϕ = 405°

= 45°

Hence ϕA = ±45°

Practice Test: Electronics Engineering (ECE)- 2 - Question 18

A system is shown in figure as follows

Where, hk (n) = δ [n – k.(1/2)k]

Then value of overall impulse response of above system h(n) at h = 2 is ______?


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 18
We have properties

Therefore h(n) can be drawn as

∴ at n = 2h(n) = 1

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 2 - Question 19

A square wave with frequency = 5 kHz

This signal is passed through an ideal low-pass filter with 0 dB of passband gain and 10 kHz of cut-off frequency. The filtered signal is subsequently buried additively into a zero-mean noise processed with one-sided power spectral density (PSD) of 20 nW Hz–1. Upto a frequency of 2.5 MHz. The PSD of noise is zero beyond 2.5 MHz. The signal to noise ratio of the output is _______dB.


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 19
We have

Now, Noise power = Area

= One sided PSD × Frequency range

Practice Test: Electronics Engineering (ECE)- 2 - Question 20

The output of filter is a

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 20
x(t) can be drawn like that

Hence impulse response of matched filter h(t) = x(T – t) is

∴ output y(t) = x(t) * h(t)

∴ Hence option D.

Practice Test: Electronics Engineering (ECE)- 2 - Question 21

For a MOD -30 ripple counter, the maximum operating frequency is 8 MHz. The propagation delay of each flip flop must be _________nsec?


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 21
Mod-30 counter required

∴ Propagation delay pe flip flops

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 2 - Question 22

For a successive approximation type of ADC shown.

If Va = 8V and resolution of D/A = 0.25 volt

Then the error voltage between Va and SAR output at 5th clock is_______volt?


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 22
Resolution = 0.25

Va = 8v

VDAC = Resolution × Decimal equivalent of binary o/p of SAR

∴at 5th clock max o/p = 7.75 volt

∴ error = 8 – 7.75 = 0.25 volt.

Practice Test: Electronics Engineering (ECE)- 2 - Question 23

For the transfer function G(jω) = 6 + jω. The corresponding Nyquist plot for positive frequency has the form

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 23
As real part σ = 6

Only imaginary port increases with ω.

Hence option D is correct.

Practice Test: Electronics Engineering (ECE)- 2 - Question 24

Consider a random process as z(t) = x(t) + y(t)

Where x(t) and y(t) are two random processes with zero mean and they are individually WSS. The PSD of x(t) is Sx(f) and y(t) is Sy(f). Then PSD of z(t) is?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 24
We know

Rz(T) = E[z(t).z(t − T)]

Here z(t) = x(t) + y(t)

∴ Rz(T) = E[(x(t) + y(t))⋅(x(t − T) + y(t − T))]

= E[x(t)⋅x(t − T)] + E[y(t)⋅x(t − T)] + E[x(t)⋅y(t − T)] + E[y(t)⋅y(t − T)]

∴ Rz(T) = Rx(T) + Ry(T) + Rxy(T) + Ryx(T)

We know

Rz(T) ⟶ PSD Sz(f)

∴ S2(f) = Sx(f) + Sy(f) + S(f) + Sxy(f)

Hence option D is correct.

Practice Test: Electronics Engineering (ECE)- 2 - Question 25

A digital circuit is designed in a way that it accept 3-bit number and generate it’s square. The circuit is to be designed using ROM with minimum hardware. Then the size of ROM is _______?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 25
Truth table is

It can be seen that

D1 = 0 and D0 = z

Hence ROM size = 23 × 4 = 8 × 4

The design is

Practice Test: Electronics Engineering (ECE)- 2 - Question 26

Eight voice signals, each limited to 4 kHz and sampled at Nyquist rate, are converted into binary PCM signal using 256 quantization levels. The bit transmission rate for the time-division multiplexed signal will be (in kbps)____


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 26
Nyquist rate = 4×103×2 = 8×103 samples/sec

256 levels = 8 bits.

So one sample is represented by 8 -bit. So total = 8×8×103 bits/sec

For 8 voice signal = 8×8×8×103 bits/sec

= 512 kbps

Practice Test: Electronics Engineering (ECE)- 2 - 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 Practice Test: Electronics Engineering (ECE)- 2 - 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. Thus, option B is the correct answer.
Practice Test: Electronics Engineering (ECE)- 2 - Question 28

A signal of bandwidth 100 MHz and strength 10mW is transmitted through a transmitter through a cable that has 40 dB loss. If N0/2 = 0.5 × 10−20ω/Hz then the signal to noise ratio at the input of the receiver is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 28

Practice Test: Electronics Engineering (ECE)- 2 - Question 29

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

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - 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.
Practice Test: Electronics Engineering (ECE)- 2 - Question 30

Consider a medium such that μ=μ0,ε = 81ε0,σ = 20S/m. The frequency at which the conduction current density is 10 times the displacement current density in magnitude is ____(GHz)


Detailed Solution for Practice Test: Electronics Engineering (ECE)- 2 - Question 30

on putting all these values.

= 36 GHz

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