Practice Test: Electronics Engineering (ECE)- 13 - Electronics and Communication Engineering (ECE) MCQ

# Practice Test: Electronics Engineering (ECE)- 13 - 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)- 13

Practice Test: Electronics Engineering (ECE)- 13 for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) Mock Test Series 2025 preparation. The Practice Test: Electronics Engineering (ECE)- 13 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The Practice Test: Electronics Engineering (ECE)- 13 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)- 13 below.
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Practice Test: Electronics Engineering (ECE)- 13 - Question 1

### The average of 8 numbers is 27. If each of the numbers is multiplied by 8, find the average of new set of numbers.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 1
Sum of 8 no = 27 × 8 = 216

if each no is multiplied by 8 then

Practice Test: Electronics Engineering (ECE)- 13 - Question 2

### The product of two numbers is 2028 and their H.C.F is 13. The numbers of such pair is:

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 2
Let the numbers be 13a and 13b.

Then, 13a × 13b = 2028

ab = 2028/169 = 12

Now, the co-primes with product 12 are (1, 12) and (3, 4).

So, the required numbers are (13 × 1, 13 × 12) and (13 × 3, 13 × 4).

Clearly, there are 2 such pairs.

Hence, option B is correct.

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Practice Test: Electronics Engineering (ECE)- 13 - Question 3

### Which pair of numbers will come in place of the missing numbers.4,10,__,___, 244,730,2188

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 3
On careful observation, we find 10 = 4 x 3 – 2; 730 = 244 x 3-2; 2188=730 x 3-2

Using the same logic, we try to work out the 2 missing numbers as 10 x 3- 2=28

and 28 x 3-2=82. To verify the correctness, we check 244=82 x 3-2

Practice Test: Electronics Engineering (ECE)- 13 - Question 4

Refer the below data table and answer the following Question.

If the GDP of the country was \$8 trillion at the end of 2011, what was it at the beginning of 2013?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 4
GDP of the country at the end of 2011 = \$8 trillion

GDP growth of 2012 = 7%

Therefore at the beginning of 2013, GDP growth = 107% of 8 trillion

= \$8.56 trillion

Practice Test: Electronics Engineering (ECE)- 13 - Question 5

Select the most appropriate synonym of the given word.

Solicit

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 5
The meaning of 'solicit' is to ask for or try to obtain something from someone. The word “sanction,” which means official permission or approval for action is similar to this.

Sanction means official permission or approval for action.

Scum means a layer of dirt or froth on the surface of a liquid.

Disguise means to give someone or oneself a different appearance in order to conceal one's identity.

Practice Test: Electronics Engineering (ECE)- 13 - 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.

The little boy ran ____ fast that he was ____ for breath.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 6

Option B has the correct fillers. Running too fast is less likely to result in a fight for breath but rather a gasp for breath. Hence, ‘so, gasping’ is the most suitable response.

Practice Test: Electronics Engineering (ECE)- 13 - Question 7

Select the most appropriate option to fill in the blank.

Although his brother is blind, he is very fast _________ calculations.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 7

Let us understand the meaning of the given words :-

"In" = used for expressing the situation of something that is or appears to be enclosed or surrounded by something else.

"With" = accompanied by another person or thing.

"At" = used to show the activity in which someone's ability is being judged.

Since doing fast calculations is an ability, so the preposition "at" is used here.

Hence, option C is the correct answer.

Practice Test: Electronics Engineering (ECE)- 13 - Question 8

The monthly salaries of A. Band C are in the ratio 2 :3 :5. If C's monthly salary is Rs. 12,000 more than that of A, then B's annual salary is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 8

Let the salaries of A, B and C be 2x, 3x and 5x respectively.

According to the question,

5x - 2x = 12000

⇒ x = 4000

So, monthly Salary of B = 3×4000 = 12000

Therefore Annual Salary = 1212000 = 144000

Practice Test: Electronics Engineering (ECE)- 13 - Question 9

Mathew told his friend Sham, pointing to a photograph, “Her father is the only son of my mother.” The photograph is of whom?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 9

From the given question, Her father is the only son of my mother indicates that the photograph is of Mathew’s daughter.

Practice Test: Electronics Engineering (ECE)- 13 - Question 10

Round-trip tickets to a tourist destination are eligible for a discount of 10% on the total fare. In addition, groups of 4 or more get a discount of 5% on the total fare. If the one-way ticket fare of the trip for single person is Rs 100, a group of 5 tourists purchasing round-trip tickets will be charged Rs____.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 10

Total round trip fare for group of 5 tourist without discount = 5x200 = 1000

Discount for Round Trip = 10 % of total fare = (10/100)x1000 = 100 Rs.

Discount for having a group of 5 Tourist = 5 % of total fare

= (5/100) x 1000 = 50 Rs.

Total Discount = Discount for Round + Discount for having a group of 5 Tourist

= 100+50

= 150 Rs.

Thus the net round trip fare for group of 5 tourist after discount is

Net Fare = Total Fare - Total Discount

= 1000-150

= 850 Rs.

One way single person fare = 100

Two way fare for single person = 200

For 5 persons two way fare = 1000

Now, total discount = (10 + 5)% = 15%

Amount to be paid = (1000 - 150) = 850

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

Practice Test: Electronics Engineering (ECE)- 13 - Question 12

Determine Vo (in Volts) in the given network.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 12
Applying current-divider

Hence,

Practice Test: Electronics Engineering (ECE)- 13 - 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 Practice Test: Electronics Engineering (ECE)- 13 - Question 13
Zin at junction is given by

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

The output of a continuous-time, linear time-invariant system is represented by T{x(t)} where x(t) is the input signal. A signal z(t) is called eigen-signal of the system T, when T{z(t)} = Y z(t) , where Y is a complex number, in general, and is called an eigen-value of T. Assume the impulse response of the system T is real and even. Then which of the following statements is TRUE?

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 14
Given impulse response is real and even

Thus H(jω0) will also be real and even

So sin(t) and cos(t) are Eigen signal with same Eigen values.

Practice Test: Electronics Engineering (ECE)- 13 - Question 15

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

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 15
x(t) = e−tu(t)

Taking laplace transform

Taking inverse laplace transform

Practice Test: Electronics Engineering (ECE)- 13 - Question 16

For the given signal, x(t) = 3 cos 80πt, determine the sampling frequency(in Hz) at which aliasing will take place.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 16
According to the sampling theorem, the sampling frequency should be at least equal to twice the highest frequency component of the signal, below which overlapping, or aliasing tales place. Highest frequency component,

Hence, minimum sampling frequency to avoid aliasing, fs = 2 × 40 = 80Hz

Hence, aliasing will occur if the sampling frequency is less than 80Hz, i.e., at 40Hz.

Practice Test: Electronics Engineering (ECE)- 13 - Question 17

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

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 17
We have,

Practice Test: Electronics Engineering (ECE)- 13 - Question 18

Let the state transition matrix of a system be given as

Then which of the following is equal to φ-1(t)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 18
We have, φ−1(t) = (eAt)−1 = eA(−1) = φ(−t)

Hence, replace 't' in state transition matrix by '-t'.Thus, option A is the correct answer.

Practice Test: Electronics Engineering (ECE)- 13 - 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 Practice Test: Electronics Engineering (ECE)- 13 - Question 19

Practice Test: Electronics Engineering (ECE)- 13 - Question 20

A Si Solar cell has short-circuited current of 100 mA and open-circuit voltage of 0.7 V under full illumination. If the fill factor is 0.71 then the Maximum power delivered (in mW) to load by this cell is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 20

Practice Test: Electronics Engineering (ECE)- 13 - Question 21

Assume electronic charge q = 1.6×10−19C, kT/q = 25mV and 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 Practice Test: Electronics Engineering (ECE)- 13 - Question 21

From Einstein relation,

Practice Test: Electronics Engineering (ECE)- 13 - Question 22

In an abrupt p-n junction, the doping concentrations on the p-side and n-side are NA = 9 X 1016 /cm3 and ND = 1 X 1016 /cm3 respectively. The p-n junction is reverse biased and the n-side depletion width is 3μm. The depletion width on the p-side is

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 22
We know that

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

Practice Test: Electronics Engineering (ECE)- 13 - Question 24

For common emitter input characteristics VCE is 6V and VBE is changed from 0.32 to 0.8V also base current changes from 40μA to 60μA, then by using the given data calculate the input impedance (in KΩ) kept VCE constant.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 24
To calculate input impedance we need to calculate h – parameter.

hei = ΔVBE/ΔIB|VCE = 6V,

Constant = 0.48V /20μA = 24KΩ

Practice Test: Electronics Engineering (ECE)- 13 - Question 25

Consider two resistors of 50 kΩ and 100 kΩ at room temperature of 27ºC. 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 Practice Test: Electronics Engineering (ECE)- 13 - 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,

where k is Boltzmann constant.

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

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 13 - Question 26

A receiver has been operating at 290 K and receiving a signal of bandwidth 400 kHz. The amplifier used in the receiver has an average output resistance of 1.5 kΩ. This will lead to Johnson noise voltage of ________ μV.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 26
We have a relation between voltage generated, temperature, bandwidth and resistance offered as follows,

We have, k = 1.38 × 10-23 J/K,

T = 290 K, B = 400 × 103 Hz,

R = 1.5 × 103Ω

Thus, vn = 3.099 μV.

Practice Test: Electronics Engineering (ECE)- 13 - Question 27

An amplifier operating over the frequency range of 18 to 20 MHz has a 10kΩ input resistance. The RMS noise voltage at the input to the amplifier at ambient temperature is (assume Boltzman’s constant

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 27

Practice Test: Electronics Engineering (ECE)- 13 - Question 28

The return loss due to a 150W cable terminated by a 100W load is _______________ (in dB)

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 28

*Answer can only contain numeric values
Practice Test: Electronics Engineering (ECE)- 13 - Question 29

What is the maximum torque (in Nm) on a square loop of 200 turns in a field of uniform flux density of 1 Wb/m2. The loop has 15cm side and carries a current of 5A.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 29
Max torque, Tmax = NBIS

N-no of turns = 200

B = 1 Wb/m2

I = 5A

S-area of loop=(15 x 10-2)2 = 0.0225

Tmax = 200×1×5×0.0225 = 22.5Nm

Practice Test: Electronics Engineering (ECE)- 13 - Question 30

Consider the following sequential circuit consisting of 2 J-K flip flops and D flip flop :

The Mod value for this counter is_____________.

Detailed Solution for Practice Test: Electronics Engineering (ECE)- 13 - Question 30
The truth table can be given as follows

Hence Mod value of the counter is 8.

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