GATE Mock Test Electronics Engineering (ECE)- 10 - Electronics and Communication Engineering (ECE) MCQ

GATE Mock Test Electronics Engineering (ECE)- 10 - Electronics and Communication Engineering (ECE) MCQ

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65 Questions MCQ Test GATE ECE (Electronics) Mock Test Series 2025 - GATE Mock Test Electronics Engineering (ECE)- 10

GATE Mock Test Electronics Engineering (ECE)- 10 for Electronics and Communication Engineering (ECE) 2024 is part of GATE ECE (Electronics) Mock Test Series 2025 preparation. The GATE Mock Test Electronics Engineering (ECE)- 10 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The GATE Mock Test Electronics Engineering (ECE)- 10 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)- 10 below.
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GATE Mock Test Electronics Engineering (ECE)- 10 - Question 1

Rohit and Ronit start together from a point in opposite direction on bike. Rohit’s speed is 21Km/h and Ronit’s speed is 15km/h. What will be the distance between them after 20 mintues?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 1

Rohit’s speed = 21 km/h

Ronit’s speed = 15 Km/h

As we know that they start from opposite direction.

Thus, the relative speed = 21 Km/h + 15 Km/h = 36 Km/h.

The distance between Rohit and Ronit after 20 minutes = 36/3

= 12 km

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 2

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)- 10 - Question 2
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)- 10 - Question 3

Each of the letters in the figure below represents a unique integer from 1 to 9. The letters are positioned in the figure such that each of (A+B+C),(C+D+E),(E+F+G) and (G+H+K) is equal to 13 each. Which integer does E represent?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 3
A + B + C = C + D + E = E + F + G

=G + H + K = 13

If we add all, we will get = 4×13 = 52

But sum of all natural number 1 to 9 =45=(9×10)/2

A + B + C + C + D + E + E + F + G + G + H + K

= 52

A + B + C + D + E + F + G + H + K =45

Hence, C + E + G=7

Also, C + D + E =13

D - G = 6

E = 4

Alternative Method

By checking other equations

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 4

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)- 10 - Question 4
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

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 5

Following bar chart provides the percentage of Adult Males, Adult Females and Children out of total population in five colonies A, B, C, D and E

If total number of residents in colonies A, B and C are 1250, 2050 and 1800 respectively, then what is the total number of adult females in colonies A, B and C together?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 5

The total number of adult females in colonies A, B and C together

= 1250 x 36/1000 + 2050 x 30/100 + 1800 x 42/100

= 450 + 615 + 756

= 1821

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 6

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)- 10 - Question 6
The context mentions that wars have been fought for 'voting' and 'privilege' means 'a special right'.
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 7

In the basement of an apartment, the total number of wheels of all cars and bikes is 70 more than twice the number of parked vehicles. The number of cars parked in the basement is:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 7
Let the total number of cars and bikes parked in the basement be ‘a’ and ‘b’ respectively.

As we know that a car and a bike has four and two wheels respectively.

Therefore, total number of wheels = 4a + 2b

Also, it is given that the total number of wheels is 70 more than twice the number of parked vehicles.

Accordingly,

4a + 2b = 2(a + b) + 70

On simplifying we get,

a = 35.

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 8

Two trains start at the same time from two stations and proceed towards each other at speeds of 90 km/h and 70 km/h, respectively. When they meet, it is found that one train has travelled 140 km more than the other. What is the distance (in km) between the two stations?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 8

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 9

Direction: In the given question, an idiom/phrase has been printed in bold in the sentence. Choose the alternative that best expresses the meaning of the idiom/phrase

The family of the victim said that the killer had got his just deserts when he was jailed for life.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 9
To get (or receive) one's just deserts receive what one deserves, especially appropriate punishment.
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 10

A tree of the graph shown below is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 10

Following figure shown that a d f is a tree.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 11

If Eigen values of A2×2Matrix are 2, 3 then Eigen values of A2 – 3A + 4I will be 'g' and 'h'. What will be the sum of 'g' and 'h'.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 11
Eigen values of f(A) = A2−3A+4Iare f(2), f(3)

∴ 2, 4 are Eigen values

g + h = 6

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 12

Directions: Select the most appropriate word/phrase among the choices to complete the sentence.

You can make your quixotic experiments with someone else, I do not wish to be your ________.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 12

'Guinea pig' is term used to refer any person or thing used in an experiment. As first part of sentence refers the experiments as quixotic (impractical or daring), so perfect word suited to this blank is 'guinea pig', which is used in labs for testing on biological experiments.

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 13

One hundred identical coins, each with probability P of showing up heads are tossed once. If 0 < P < 1 and the probability of heads showing on 50 coins is equal to that of heads showing on 51 coins. Then the values of P is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 13
Given100C50q50p50

= 100 C51q49p51

51 q = 50 p

⇒ 51 ( 1 - p )

⇒ 50p, p = 51/101

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 14

If prices reduce by 20% and sales increase by 15%, what is the net effect on gross receipts?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 14
Let original price = p, and original sales = s. Therefore, original gross receipts = ps.

Let new price = 0.80p, and new sales = 1.15s. Therefore, new gross receipts = 0.92ps.

Gross receipts are only 92% of what they were. Hence, gross receipts decrease by 8%.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 15

A specimen of a Si has a resistivity of 200kΩ -cm, magnetic flux B = 0.3Wb / m2 and d = w = 6mm. if the value of hall voltages and hall currents are 60mV and 10μA. Then the value of μp is (cm2 / V - sec )

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 15
μ = σRH

σ = 1 / resistivity

=1 / 200kΩ−cm

= 5 × 10−4 mho / m

RH = VHW / BI

= 360m3 / C

Thereforerμ = σ RH

= 5 х 10 -4 х 360

= 1800 cm2 / V - sec

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 16

Kirchhoff's current law is applicable to:

1. Closed loops in a circuit
2. Junction in a circuit
3. Magnetic circuits

Which of the above is/are correct?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 16

Applying KCL at mode 'N':
I1 + I2 = I3

KCL is applicable to junction in the circuit.

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 17

In the circuit given below, what is the operating region of the transistor, if it has a β of 100?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 17

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 18

Most Reality TV shows centre on two common motivators: fame and money. The shows transform waitresses, hairdressers, investment bankers, counsellors and teachers, to name a few, from obscure figures to household names. A lucky few successfully parlay their fifteen minutes of fame into celebrity. The luckiest stars of Reality TV also reap huge financial rewards for acts, including eating large insects, marrying someone they barely know, and revealing their innermost thoughts to millions of people.

Which of the following options best supports the above paragraph?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 18
The correct answer is the first option as it is stated in the passage that most Reality TV shows centre on two common motivators: fame and money.

Choices 2 and 4 are not supported by the passage because passage is about Reality TV stars and not Reality TV.

Choice 3 is incorrect because the paragraph states that some Reality TV stars manage to parlay their fifteen minutes of fame into celebrity.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 19

Consider a circuit as shown in fig below

Equivalent resistance across the terminals A & B is_____Ω.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 19

⇒ Req = 1 + 3/4

= 7/4 Ω

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 20

Directions: Insert the missing number.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 20
Middle number = (Difference between numbers of left column) × (Sum of numbers of right column)

i.e. 80 = (15 - 5) × (2 + 6) = (10) × (8) and 65 = (9 - 4) × (7 + 6) = (5) × (13)

∴ ? = (13 - 11) × (16 + 8) = 2 × 24 = 48

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 21

The ROC of z-transform of the discrete time sequence is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 21

ROC of (1/4)n u(n−1) is | Z | > 14, the signal being right sided

ROC of 5nu(−n−1) is | Z | <5, the signal being left sided

Thus the common ROC will be 1/4 < | Z | < 5

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 22

Consider a discrete time periodic signal x [n] = sin(πn/5) . Let ak be the complex Fourier series coefficients of x[n]. The coefficients {ak}, are non-zero when k = Bm±1, where m is any integer. The value of B is _______.(Answer up to the nearest integer)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 22
Given the discrete time periodic signal,

So, we have the fourier series coefficients

Also, we have the period of the function as

Since, the fourier series coefficients are also periodic, so we have

i.e. the fourier series coefficients are non zero for

k = 10 m ± 1 where m = 0, 1, 2, ……….

Given that the system coefficients are non-zero for

k = Bm ± 1

We get the value

B = 10

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 23

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 23

The transmission line generates capacitive reactive volt-amperes in its shunt capacitance and absorbing reactive volt-amperes in its series inductance. The load at which the inductive and capacitive reactive volt-amperes are equal and opposite, such load is called surge impedance load i.e the receiving end voltage is equal to the sending end voltage.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 24

A discrete random variable X has possible values xi = i2, i = 1, 2, 3, 4 which occur with probabilities 0.4, 0.25, 0.15, 0.1. The mean value = E [X] of X is ___. (Answer up to two decimal places)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 24

= 1.0 (0.4) + 4 (0.25) + 9 (0.15) + 16 (0.1) = 4.35

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 25

In a transistor amplifier, change of 0.025V in signal voltage causes the base current to change by 15 µA and collector current by 1.2 mA. If collector and load resistances are of 6kΩ and 12kΩ, determine input resistance in kΩ.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 25
Input resistance = change in input voltage / change in input current

= 0.025/15 x 10 ­-6

= 1.67kΩ

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 26

A source has an alphabet {a1, a2, a3, a4, a5, a6} with corresponding probabilities {0.1, 0.2, 0.3, 0.5, 0.15, 0.2}. The entropy (in bits / symbol) of this source is ______. (Answer up to one decimal place)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 26

= -(0.1 log20.1 + 0.2 log20.2) + 0.3 log20.3 + 0.05 log20.05 + 0.15 log20.15 + 0.2 log20.2

= 2.4087 bits/symbol

or H(x) = 2.4 bits/symbol

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 27

In the figure shown below, voltage V1 and V2 are

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 27
KCL at node o: 30 = I2 + 6 + 9

I2 = 15 A

KCL at node b: 10 + 6 = I1

I1 = 16 A

KCL at node d: 9 + I3 = 40

I3 = 31 A

Applying KVL around the loop a o c d a

V2 – 3I2 – 4I3 – 40 × 1 = 0

V2 – 3 (15) – 4(31) – 40 = 0

V2 = 209 V

Applying KVL around the loop o b c o

V1 – 2I1 + 3I2 = 0

V1 = – 13V

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 28

The Fourier transform of signal is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 28

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 29

A white noise signal having two sided spectral density of 5*10-6 V2/Hz is applied to a simple RC low pass filter whose 3 dB cutoff frequency is 5 kHz. The mean squared value of noise obtained at the output of filter would be

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 29
White noise with 2 sided PSD 5*10-6 V2/Hz.

RC low pass filter has 3 dB cutoff 5 kHz.

Mean Squared value of the output power can be found using following way,

E(y2)=Ry(0)

This will give the Mean squared output power to be 39.27*10-3 V2

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 30

Consider the given circuit and waveform for the input voltage. The diode in circuit has cutting voltage Vy = 0.

The waveform of output voltage vo is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 30
For vi < 4 V the diode is ON and output v0 = 4 V.

For vi > 4 V diode is off and output v0 = vi.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 31

An electric field strength is to be measured at an observation point

θ = π/2, 500 km from a half-wave (resonant) dipole antenna operating in air at 50 MHz . The measurement of dipole would be ____(in m)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 31
The wavelength λ = c/f = 3 × 10850 × 106 = 6m

the length of half-dipole is I = λ/2 = 3m

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 32

Input impedance of an instrumentation amplifier compared with a difference amplifier is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 32
Input impedance of instrumentation amplifier is very large. Instrumentation amplifiers (in-amps) are very high gain differential amplifiers which have a high input impedance and a single ended output.
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 33

Let x(t) be a signal band width to 1 kHz. Amplitude modulation is performed to produce signal g(t) = x(t) sin 2000ϖt. A proposed demodulation technique to illustrated in figure. The ideal low pass filter has cutoff frequency 1 kHz and pass band gain 2. The y(t) would be

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 33
x1 (t) = y (t) cos ⁡(2000wt)

=x (t) sin⁡ (2000wt) cos⁡ (2000wt)

=1/2x (t) sin⁡ (4000πt)

X1( jω ) = 1/4j {x[ j (ω − 4000π) ] −x [ j (ω−4000π)] }

This implies that

X1( jω ) is zero for |ω| ≤ 2000 ϖ

Because ω < 2ϖfm = 2ϖ1000

When x1 (t) is passed through LPF with cutoff frequency 20000 , the output will be zero

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 34

In the following circuit, the respective values of iN and RN are

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 34
The short circuit current across the terminal is

isc = (6 + 4)/(4 + 2) = 4 A = iN

RN = 6||3 = 2 Ω

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 35

An 16-bit microprocessor has an external RAM which is having accessible range from A800 – AFFF H. The no. of address lines in the RAM is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 35
Solution: starting address is 1010 1000 0000 0000

Ending address is 1010 1111 1111 1111

From the given data there are 11 address lines

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 36

In a simple RC high pass filter, the desired roll-off frequency is 15 Hz and C = 10 μF. The value of R (in kΩ) would be ____. (Answer up to two decimal places)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 36

We = 2πfe = 1/RC

R = (1) / (2π x 15 x 10 x 10-6) = 1.06kΩ

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 37

If the function

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 37

Z is homogeneous function of x, y of order = 2

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 38

If is equal to

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 38

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 39

The system shown in the given figure, is subjected to a unit ramp input

By closing the switch ‘S’

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 39
Before closing the switch

s → 0

Hence both ′ξ′ and ess will increase.

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 40

The open-loop transfer function with ufb are given below for different systems. The unstable system is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 40
In characteristic equation s3 + 2s2 + K, then term s is missing, hence unstable.
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 41

For the closed loop control system shown

The values of α& β, such that the system has settling time of 0.2 sec. and steady state error for unit ramp input is 6% are (approx.)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 41
Let the damping factor ρ& natural frequency is ωn

ts = 4/ρωn

= 0.2,ess

= 2pn = 0.06

ρωn = 20 &

ρωn = 20 × 0.03

ρωn × ρ/ωn = 20 × 0.03

ρ2 = 0.6

ωn = 25.82

Characteristic equation

1 + α/s(s+2) × (1+sβ) = 0

s2 + (2 + αβ)s + α = 0

α = 0

Compare with s2 + 2ρωns + ωn2 = 0

α =ωn2 (25.82)2 = 666.7

2ρωn = 2 + αβ

2 × 20 = 2 + 666.7 × β

β = 0.057

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 42

If r = xux + yuy + zuz, then (r. ∇) r2 is equal to

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 42

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 43

If the effective mass of an electron is equal to thrice the effective mass of a hole , then the distance of fermi level in an intrinsic semiconductor from the centre of the forbidden band at room temperature is ____ eV.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 43

Formula for fermi level is:

Also we know,

= 0.0213 eV

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 44

If C is a closed curve enclosing a surface S, the magnetic field intensity the current density and the electric flux density are related as:

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 44

Maxwell Equations

Integral form

Stokes Theorem

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 45

In the circuit shown, the switch S is closed at t = 0 after long time. If (0+) is____ (in Amp).

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 45
t < />

S→open for long time i.e. Steady state inductor→Short circuit

At t = 0+ S is closed, inductor acts as a current source

- 2 + 1. (i + 1) + i.1 = 0

2i = 1 ⇒ i = 0.5 A

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 46

If Eb, the energy per bit of a binary digital signal, is 10-5 watt-sec and the one sided power spectral density of the white noise, N0 = 10-6 W/Hz, then the absolute output SNR (in dB) of the matched filter is (Answer up to the nearest integer)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 46

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 47

Consider the following network

The value of ZL (in Ω) for maximum power transfer to it is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 47
Let us use the method of injection of source in order to find Zth. Replace all independent source by their internal resistances.

For maximum power transfer ZL = Z*th =

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 48

For the diode circuit shown in figure,

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 48

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 49

The maximum amplitude of cross correlated sequence of x(n)=(2,5,0,4) and h(n)=(3,1,4) is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 49
x(n) correlation h(n) = x(n)*h(-n)

h(-n)= (4,1,3)

=(8,22,11,31,4,12)

Maximum value = 31

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 50

For the following circuit, what will be the current (mA) flowing through collector of the transistor? (Answer up to one decimal place)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 50
VB = - 10 + 3.2 = - 6.8 V

VE = VB - 0.7 = 6.8 - 0.7 = - 7.5 V

IE = (-7.5 + 10)/(1k) = 2.5 mA = I

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 51

In the below circuit, the input offset voltage and input offset current are Vio=4mV and Iio=150nA The total output voltage is _______mV

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 51
The offset due to

Vio = 404mV = (1+RF/R,)Vio

(1+500/5) 4m = 404mV

Due to Iio,Vo=RFIio = 500k × 150n = 75mV

∴ Total off-set voltage Vo 404 + 75 = 479mV

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 52

Directions: For the regulator circuit shown below, various parameters are

Vz = 6.3 V

IzT = 40 mA

Rz = 2 Ω

Vs = 12 V to 18 V

Minimum load current = 0 mA

Minimum zener diode current = 1 mA

Maximum zener diode current = 119 mA

The value of RS that limits the zener current Iz(max) (in Ω) will be (Answer up to one decimal place)

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 52
Zener current iz becomes maximum when supply voltage is maximum and the lead current is minimum

VS = 18 V, iz(max) = 119 mA

RS =

= = 96.3 Ω

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 53

A full wave rectifier uses a capacitance of 60 μF in parallel with a load of 1 kΩ to regulate the voltage obtained through a centre tapped transformer. If the centre tapped transformer has 40 V (RMS) as voltage in secondary coil during operation at frequency of 50 Hz, what will be the regulation provided(in %)?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 53
We have value of VRMS given here, so

=40

Thus, Vm = 56.56 V

Voltage Regulation is given by,

%Regulation = R/RL∗100%

where R is the resistance provided by transformer and the diode circuit, and RL is the load connected.

For a Full Wave rectifier with Capacitor filter, R is given as,

R = 1/4f0c = 83.33Ωand RL = 1000 Ω

Thus, Regulation provided by the configuration is 8.33 %.

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 54

A unity feedback system is given as:

G (s) =

Indicate the correct root locus diagram.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 54
Any point on real axis of s-is part of root locus if number of OL poles and zeros to right of that point is even. Thus (2) and (3) are possible option.

The characterstics equation is

For break away & break in point

Which gives = 3, -1

Here -1 must be the break away point and 3 must be the break in point.

*Answer can only contain numeric values
GATE Mock Test Electronics Engineering (ECE)- 10 - Question 55

AWGN channel having B.W of 4 kHz, 2-sided noise PSD is given by 1012Watts / Hz The Channel capacity( in kbps) required to get the signal power of 0.1 MW at channel o/p is __

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 55
Given B=4 kHz

S = 0.1MW = 10-4W

N = N0B = 8 × 10-9 Watts

Therefore,C = Blog2

= 4 log2

= 54.4 kbps

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 56

The general solution of - 2y = 10 cos x is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 56
The given equation is

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 57

In a spectrum analyser a signal with 100% amplitude modulation is observed, what type of display should be observed?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 57
Modulation index (μ) = 1

Therefore,

AMwaves(t) =

Power of carrier component, Pc=

Power of LSB component = power of USB component

=

Therefore, ratio of the power of carrier and power of side band

PC/PSB=14

SO, USB and LSB will be 6 dB down compare to carrier.

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 58

A speaks truth in 75% and B in 80% of the cases. In what percentage of cases are they likely to contradict each other narrating the same incident?

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 58
Let E = event that A speaks the truth.

F = event that B speaks the truth

P (A and B contradict each other)

= P [( A speaks truth and B tells a lie ) or ( A tells a lie and B speaks the truth)]

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 59

The input signal Vin shown in the figure is a 1 KHz square wave voltage that alternates between +7V and -7V with a 50% duty cycle. Both transistors have the same current gain, which is large. The circuit delivers power to the load resistor RL. What is the efficiency of this circuit for the given input? Choose the closest answer.

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 59
This is a class-B amplifier whose efficiency is given as

η = πVp/4VcC

Where Vp→ peak value of input signal

VCC→ Supply voltage

Here VP = 7 volt, VCC = 10 volt

So, η= π/4 × 7/10 × 100 = 54.95% ≈ 55%

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 60

All transistors in the N output mirror shown below are matched with a finite gain â and early voltage VA = . The expression for each load current is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 60

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 61

Consider an excess-3 to BCD code converter as shown below which accepts excess-3 code as input and outputs corresponding BCD code

The simplified Boolean expression for the output bit Y is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 61
For the given circuit, the minterms m0, m1, m2, m13, m14 and m15 correspond to invalid excess-3 codes. Hence these input combinations are don’t cares. The truth table is as shown below (for valid excess-3 codes only)

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 62

In the following circuit, transistors Q1 and Q2 has following parameters:

The voltages V1, V2 and V3 respectively are

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 62

Assume transistors are in saturation

For saturation region

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 63

The contents of some memory location of an 8085 microprocessor based system are as shown below

The program is as follows

LHLD 3000H

MOV E, M

INX H

MOV D, M

LDAX D

MOV L, A

INX D

LDAX D

MOV H, A

The contents of HL pair after the execution of the program will be

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 63
LHLD 3000H ; (L)←(3000), (H)←(3001)

(HL) = 3002H

MOV E, M ; (E)←(3002) (E) = 00H

INX H ; (HL)←(HL) +1 (HL) = 3003H

MOV D, M ; (D)←(3003H), (D) = 30H

LDAX D ; (A)←DE

i.e (A)←(3000H), (A) = 02H

MOV L, A ; (A)←(L), (L) = 02H

INX D ;( DE)←DE)+1. (DE) = 3001H

LDAX D ; (A)←DE

i.e (A)←(3001H), (A) = 30H

MOV H. A ; (H)←(A) (H) = 30H

Hence (H) = 30H

(L) = 02H

(HL) = 3002H

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 64

Two random variable X and Y have the density function

The X and Y are

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 64

Form

From marginal densities fx(x)

GATE Mock Test Electronics Engineering (ECE)- 10 - Question 65

Parameters of transistors are given,

VTN = 0.8V, Kn′ = 40μA / V2. The width to-length ratio of M2 is

(w/L)2= 1. if V0 = 0.10V when Vi=5V, then

(W/L)1 for M1 is

Detailed Solution for GATE Mock Test Electronics Engineering (ECE)- 10 - Question 65
Given, VT = 0.8V

V0 = 0.10V When Vi = 5V

M2 : VGS2 = 5 - V0 = 5 -0.1 = 4.9V

M1 : VGS1 = VG - VS = Vi - 0 = 5V

VDS1 = VD - VS = V0 - 0 = 0.1V

VGS1 - VT = 5 - 0.8 = 4.2V

VDS1 < />GS1 - VT ⇒ Linear

= 20.25

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