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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  GATE Mock Test Electronics Engineering (ECE)- 9 - Electronics and Communication Engineering (ECE) MCQ

GATE Mock Test Electronics Engineering (ECE)- 9 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test - GATE Mock Test Electronics Engineering (ECE)- 9

GATE Mock Test Electronics Engineering (ECE)- 9 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The GATE Mock Test Electronics Engineering (ECE)- 9 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The GATE Mock Test Electronics Engineering (ECE)- 9 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)- 9 below.
Solutions of GATE Mock Test Electronics Engineering (ECE)- 9 questions in English are available as part of our course for Electronics and Communication Engineering (ECE) & GATE Mock Test Electronics Engineering (ECE)- 9 solutions in Hindi for Electronics and Communication Engineering (ECE) course. Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free. Attempt GATE Mock Test Electronics Engineering (ECE)- 9 | 65 questions in 180 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
GATE Mock Test Electronics Engineering (ECE)- 9 - Question 1
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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 GATE Mock Test Electronics Engineering (ECE)- 9 - Question 1
Raio of their Investment

= 4 : 9 : 10

B’s share = 23000 x 9/23 = Rs.9000

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 2
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Pick the odd one out.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 2
Swimming is done in a large water body. All the other words relate to activities involving smaller volume of water.
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GATE Mock Test Electronics Engineering (ECE)- 9 - Question 3
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In the following question, out of the four alternatives, select the word opposite in meaning to the given word.

Gratuitous

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - 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.

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 4
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Pick the odd one out.
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 4
Break, hiatus and pause, all mean a temporary halt. End is the final halt.
GATE Mock Test Electronics Engineering (ECE)- 9 - Question 5
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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 GATE Mock Test Electronics Engineering (ECE)- 9 - 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

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 6
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Fill in the missing number in the series:

7, 13, 21, ____, 43, 57
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 6
In this series, each subsequent term is obtained by adding 6, 8, 10, 12, 14,... respectively in the previous term. Hence, the missing term is 21 + 10 = 31.
GATE Mock Test Electronics Engineering (ECE)- 9 - Question 7
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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 GATE Mock Test Electronics Engineering (ECE)- 9 - Question 7

We know that,

A = S.I + P

480 = S.I + 400

⟹ S.I = 480 – 400 = 80

⟹ S.I =

⟹ 80 =

⟹ R = 5 %

Now rate is increased by 2 %

So, new rate is 7%

New S.I = = Rs. 112

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

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 8
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How many times do the hands of a clock overlap in 24 hours?
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 8
After 11 o'clock, the minute hand has to travel all the way and by the time they meet it is has to be 12 o'clock again, since we know what the clock looks like at that time. So the two hands overlap 11 times in a 12 hour period. So, in a 24 hour period, they would overlap 22 times.
GATE Mock Test Electronics Engineering (ECE)- 9 - Question 9
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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 GATE Mock Test Electronics Engineering (ECE)- 9 - 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.
GATE Mock Test Electronics Engineering (ECE)- 9 - Question 10
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In the following question, a group of four figures following certain sequence is given as problem figures. Problem figures are followed by another group of four figures known as answer figures marked (1), (2), (3), (4). Find out the figure from the answer figures which when placed next to the problem figures will continue the sequence of the problem figures.
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 10
In going from one problem figure to the next, the arrows are going from vertical to horizontal and one arrow decreases in number.

Then they again go back to vertical.

Answer figure (3) will continue the sequence of the problem figures.

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 11
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During transmitting over a binary communication channel, bit error occurs independently with a probability of p. The probability of at most 2 bits in error in a block of ‘n’ transmitted bits would be
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 11
Probability of no error = (1 - p)n

Probability of single bit in error = nC1p(1 - p)n-1

Probability of 2 bits in error = nC2p2(1 - p)n-2

Thus, total probability= (1 - p)n

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 12
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There are 60 students in a class. The students are divided into three groups A, B and C of 15, 20 and 25 students, respectively. The groups A and C are combined to form group D. What is the average weight of the students in group D?
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 12
There is no indication of weights of the students in the question. Therefore, it is not possible to find the relation between the weights of students in groups A, B, C, and D.
GATE Mock Test Electronics Engineering (ECE)- 9 - Question 13
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Transfer function of an LTI system is given as G(S)= (s-1)/(s+1)

Find the gain of the system at a frequency of 1 rad/s.
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 13

We have, putting s=jω G(jω)=

= 1

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 14
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The probability that a man will live 10 more years is 1/4 and the probability that his wife will live 10 more years is 1/3. The probability that neither of them will be alive in 10 years is
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 14
Probability that a man will live 10 more years = 1/4

Probability that man will not live 10 more years = 1 - 1/4 = 3/4

Probability that his wife will live 10 more years = 1/3

Probability that his wife will not live 10 more years = 1 - 1/3 = 2/3

Then, probability that neither will be alive in 10 years = (3/4) x (2/3) = 1/2

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 9 - Question 15
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Consider the Schmitt trigger circuit shown below the Hysteresis width in Volts is……?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 15
Apply KCL at Va

When V0 = +VsatV2 = VUTP

When V0 = +VsatVa = VLTP

Here Vref = -10

R1 = 10k,

R2 = 50k,

R3 = 10k

VUTP =

= - 3.4545V

Similarly VLTP = -5.6363

So hyseresis width

= UTP - LTP

= (-3.4545) - (5.6363)

= 2.19V

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 16
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Voting is the privilege for which wars have been fought, protests have been organized, and editorials have been written. "No taxation without representation," was a battle cry of the American Revolution. Women struggled for suffrage, as did many minorities. Eighteen year olds clamored for the right to vote, saying that if they were old enough to fight in the war, then they should be allowed to vote. Yet Americans have a deplorable voting history, and many will tell you that they have never voted.

Which of the following words is the best synonym for the word 'privilege' as used in the passage?
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 16
The context mentions that wars have been fought for 'voting' and 'privilege' means 'a special right'.
GATE Mock Test Electronics Engineering (ECE)- 9 - Question 17
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A non-ideal diode is used in a clamper circuit. The forward and reverse resistances of the diode are 200 ohms and 250 kilo ohms, respectively. The circuit designer has designed the circuit using a 10 kilo ohm resistor. The value of resistor to be connected in parallel to correct the circuit would be approximately
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 17
We have a formula to get the value of resistor to be used in a clamper circuit and it is given by R =

This gives value of R as 7.07 kΩ.

Since 10 kΩ resistor is being used, to make effective resistance as 7.07 kΩ, a resistor of value 24.18 kΩ has to be connected in parallel.

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 18
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Two trains start at the same time from two stations A and B towards each other.

They arrive at B and A respectively in 5 hours and 20 hours after they passed each other.

If the speed of the train that started from A is 56 kmph, then what is the speed of the second train (in kmph)?
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 18
Let the speeds of two trains starting from station A and B be S1 and S2 respectively and the time taken by them after meeting be t1 and t2 respectively.

Then, we can use this equation to get the answer:

S2 = 28 kmph

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 19
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If a characteristics equation has roots S = -3 ± J4 then the values of natural frequency and damping ratio will be respectively.
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 19

We know that S =

Thus, option c is correct answer.

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 20
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A man starts at a certain point, walks one km east, then two km north, one km east, one km north, one km east and finally one km north to arrive at the destination. What is the shortest distance (in km) from the starting point to the destination?
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 20

Let us assume that man starts from point O and reaches his destination at point A.

The shortest distance between O and his destination is OA = = 5 km

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 21
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Calculate value of RL so that maximum power is transmitted through RL Also calculate maximum power generated
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 21

To find RL

For maximum power transfer RL = RS

All voltage sources should be short circuited and current sources should be open circuited and RL should be removed.

Consider the figure:

RS = 5|| 10 = 50/15 =3.33 Ω

Therefore RL = 3.33 Ω

To calculate power generated:

R= 5 + 3.33||10 = 7.5 Ω

I=V/R =10/7.5 = 1.33 A

Iab= 1.33×10/(10+3.33) = 1A

P = I2 RL = (1.33)2 × 3.33 =5.8 W

Thus, option D is the correct answer.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 9 - Question 22
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If u = log , find the value of x .(Answer up to the nearest integer)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 22

On solving, we get

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 23
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Find the transfer function for the given block diagram?
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 23

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 24
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If the closed-loop transfer function of a control system is given as T(s) = then it is
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 24
In a minimum phase system, all the poles as well as zeros are on the left half of the s-plane. In given system, as there is right half zero (s = 5), the system is a non-minimum phase system.
GATE Mock Test Electronics Engineering (ECE)- 9 - Question 25
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For the n-channel enhancement MOSFET shown in the given figure, Threshold voltage Vtn=2V,The drain current ID of the MOSFET is 4mA when drain resistance RD is 1kΩ, If value of RD is increase to 4kΩ, then drain current ID will become
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 25
ID=K(VGS-V(E(th)2

4mA=K(VGS-2)2

Here VGS=10-4×1=6

∴4=k(6-2)2

K= 4/16=¼

When RD is increased to 4kΩ.

VGS=10-4ID

Now ID= 4/16(10-4ID-2)2

Solving the above equation for ID we will get

ID =2.8 mA and ID =1.4 mA and

but ID can not be 2.8 mA as VGS becomes negative so,

ID =1.4 mA

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 26
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The root locus of the system G(s) H(s) = has the break-away point located at
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 26

Since, breakpoint must lie on root locus. So, s = -0.784 is possible.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 9 - Question 27
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As it comes in the form of 0/0 so applying LH rule QUESTION IS DISPLACED

Putting t=1/4W

P (t) = = 0.5

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 27
fm=3.4kHz

fs=8kHz

Qs=64

⇒2n=64

⇒∩=no.of bits=6

m=10

Total bits in a frame =nm+5/2

(As extra bit is added after every alternate frame)

=6×10+2.5

=62.5bits

Bit rate, rb total bits in a frame × sampling frequency

=62.5×8000

=5×105bits/sec

∴Minimum channel bandwidth,

BW = rb/2 = 2.5 x 105

= 0.25 х 105

= 0.25 x MHz

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 28
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A conducting circular loop of radius 20 cm lies in the z = 0 plane in a field B = 20 cos 377t az mWb/m2. The induced voltage in the loop is
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 28

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 9 - Question 29
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The transfer function of a network can be written as

(1+s)/(1+0.5s) . The maximum phase angle occurs at a frequency of _______ rad/sec.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 29
Comparing with standard transfer function

Here, τ=1

and aτ=0.5

or a=0.5

∵ α< />

∴ Lead network

The maximum phase occurs at a frequency of

ωm=

= 2rad/sec

GATE Mock Test Electronics Engineering (ECE)- 9 - Question 30
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If v = 2xy, then the analytic function f (z) = u + iv is
Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 9 - Question 30

By Milne's Method, f'(z) = g(z,0) + ih(z,0) = 2z + i0 = 2z

On intergrating, f(z) = z2 + c

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