# GATE Mock Test Electronics Engineering (ECE)- 12

## 65 Questions MCQ Test GATE ECE (Electronics) 2023 Mock Test Series | GATE Mock Test Electronics Engineering (ECE)- 12

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Attempt GATE Mock Test Electronics Engineering (ECE)- 12 | 65 questions in 180 minutes | Mock test for GATE preparation | Free important questions MCQ to study GATE ECE (Electronics) 2023 Mock Test Series for GATE Exam | Download free PDF with solutions
QUESTION: 1

### Which of the following states the meaning of the idiom ‘Into the bargain’?

Solution: ‘Into the bargain’ has the meaning‘ in addition to what has already been mentioned or was expected’
QUESTION: 2

### Direction: Find the correct relationship between the given quantities-Quantity I: Let E be the set of all integers with 6 in their unit place. The probability that a number chosen from the set {10, 11, 12, ........59} is a member of E is?Quantity II: A boy speaks truth in 70% of cases while another boy speaks truth in 65% cases. The probability that they will contradict each other on stating the same fact is?

Solution: Quantity I: The sample space is n(s) = 50

The favourable outcomes are {16, 26, 36, 46, 56} i.e. 5

Probability = 5/50 = 0.1

Quantity II: Boy 1 speaks truth in 70% cases so 70/100 =7/10 and

Boy 2: 65/100 = 13/20

Boy 2 does not speak truth in 35/100 cases i.e. 7/20

Probability= p (A')*p (B) +p (A)*p (B’)

[7*7/ (10*20)]+ [3*13/ (10*20)]

= 0.44

QUESTION: 3

### A square metallic sheet of an area of 1296 square cm is used for making the toys. For taking the metallic sheet from that square – four equal sized of the circular plates have been cut –off. If a fencing with the Silver thread is required to make the base of toys, how many rolls of the thread is required, if a roll contains 30 cm of the thread only?

Solution: Side * Side = 1296

⇒ Side= 36

Radius of each circular plate = ¼ of 36 = 9 cm

Circumference of each plate = 2 π 9 = 56.57 cm

Total Circumference = 4 *56.57 = 226.29 cm

No. of rolls = 226.29/30= 7.54 = 8 rolls

QUESTION: 4

The wrong (misfit) number of the sequence 36, 81, 144, 225, 256, 441 is

Solution: The pattern in the above sequence is the squares of consecutive multiples of 3;

62 = 36

92 = 81

122 = 144

152 = 225

182 = 324 ≠ 256

212 = 441

Thus the wrong number is 256

Hence Option D is correct

QUESTION: 5

Direction: Read the pie chart given below and answer the following questions-

The pie chart given shows the population of India in the 7 states considering there is no other states apart from these. How much percent more is the population of UP than that of Odisha?

Solution: Population of UP more than Odisha = 63⁰ - 36⁰ = 27⁰,

Now for percentage we have to calculate how much more population is in UP with respect to Odisha. So,

% more= (27⁰×100)/ 36⁰

= 75%

QUESTION: 6

Direction: In the following question, choose the word opposite in meaning to the given word as your answer.

Drought

Solution:

Drought (noun) means a prolonged period of abnormally low rainfall, leading to a shortage of water. Its opposite is ‘flood’, which means an overflow of a large amount of water beyond its normal limits.

QUESTION: 7

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

P is standing north of Q at a distance of 6m. R is standing toward east of P at a distance of 6m. R is to the north of S and distance between them is 6m. P is facing north direction, R is facing east, S is facing North and Q is facing towards west direction. All of them walk straight for 3m.

What is the distance between the final position of P and Q?

Solution:

3√10m Let T be the final position of P and L be the final position of Q.

Now LT = √(32 + 92) = √(9 + 81) = √ 90 = 3√10m

QUESTION: 8

In the following question, a statement has been given followed by two arguments numbered I and II. Based on the argument, you have to decide which of the following argument is strong and which is a weak argument.

Statement: Shiva trilogy written by Amish Tripathi is the best fiction book.

Arguments

I. Yes, None other fiction books have three parts.

II. No, the best fiction book is subjective and it always differs from person to person.

Solution:

Best fiction book or novel is a subjective thing and differs from person to person. Hence, the second argument is strong.

QUESTION: 9

A year after the deadliest Ebola outbreak ever in Africa, the virus has resurfaced in Congo this time around. It is not clear yet how the first victim—a man who has since died—caught the virus. But this outbreak is not the first in Congo; the country has suffered seven prior epidemics of the virus since 1976.

Which of the following is a conclusion that can be drawn from the above paragraph?

Solution:

It has been said that the biggest ever Ebola epidemic in Africa struck last year. Hence, (A) is false.

The paragraph says that the first victim of the most recent Ebola outbreak in Congo is dead. Hence, (B) can be concluded from the paragraph.

It has nowhere been mentioned that Congo was not affected by last year’s epidemic. Hence (C) cannot be concluded.

Scientists have not yet figured out how the epidemic resurfaced this time, since the first victim is dead. Hence (D) cannot be concluded.

QUESTION: 10

Direction: Following bar graphs shows time taken by different pipes to fill that particular percentage of tank which is described in tabular data. For Ex- Pipe 1 fills the 10 % of tank in 6 mins.  A Tank has three pipes; two of them are used to fil the tank while another one used to empty the tank namely: pipe 3, pipe 5 to fill the tank and a new third pipe for making the tank empty. When all three pipes are open, 7/18th part of the tank is filled in 1 hours. What is the ratio of the efficiency of Pipe-3: Pipe 5: all three pipes working together.

Solution:

part of work in one hour =

(P3+P5+Third pipE. X 1 = 7/18th part of work

⇒ 7/18th part of work = 1 hr

⇒ Total work done = 18/7 hr by all three pipes

Pipe 3 does 50% of work in 90 mins

100 % of work = 90/50 * 100 = 180 mins = 3 hr

Pipe 5 does 50% of work in 60 mins

100 % of work = 60/50 * 100 = 120 mins = 2 hr

Time Tw Efficiency

Pipe -3 = 3 hr Total work

Pipe-5 = 2 hr Total work

All three = 18/7 hr Total work

hence efficiency ratio: 6: 9: 7

QUESTION: 11

Find the unit normal vector of the cone of revolution z2 = 9x2 + 15y2 at a point (1,0,3)

Solution: The unit normal vector is the grad of f = 9x2 + 15y2 - z2 QUESTION: 12

Minimum of real value function f(x) = (x−1)2/5 occurs at x equal to

Solution: as f(x) is square of hence its min, value be 0 where at (x=1)

QUESTION: 13

The transfer function of a system is 10/(s2 + 2s + 10). Find the nature of the given system.

Solution: On comparing the given characteristic equation with the standard second-order transfer function,

ωn2/(s2+2ζωns+ωn2).

Hence, ωn2 = 10 ; 2 ζωn = 2 ⇒ ζ = 0.316

For 0 < ζ < 1, the system is said to be Under-damped.

*Answer can only contain numeric values
QUESTION: 14

In a transistor circuit , ICO = ICBO = 2μA , also it is given that α = 0.98

Then the value of ICEO is _______(μA).

Solution: QUESTION: 15

There are two energy levels E1 and E2. E1 is E eV above the fermi level and E2 is E eV below the fermi level.

P1 = the probability of E1 being occupied by an electron.

P2 = the probability of E2 being empty.

Then, which of the following is correct?

Solution: The Fermi- Dirac probability function also called Fermi function, provides the probability of occupancy of energy levels. It is given by where EF is Fermi level or Characteristics energy, for the crystal in eV. The fermi level represents the energy state with 50 percent probability of being filled if no forbidden band exists.

Therefore, P1 And P2 can be calculated by using Fermi- Dirac probability function.

For P1 QUESTION: 16

Match List I (Regions of Bipolar Transistor in a Monolithic IC) with list II (Physical Properties)

List I

a) Emitter

b) Base

c) Collector

d) Substrate

List II

(1) Moderate resistivity

(2) Very high resistivity

(3) Large size

(4) Very high Conductivity

Select the correct answers from the following.

Solution: Emitter – High Doping and very high conductivity

Collector – Large Size

Substrate- Very high resistivity

Base- Moderate resistivity

QUESTION: 17

Consider the circuit shown in the figure below v(t) = 80e−1000tV ;t>0

i(t) = 5e−1000tmA ;t>0

the energy dissipated (in μ) by the resistor for 0 < t="" />< 0.6="" ms="" will="" be="" />

Solution: The energy dissipated in the resistance is  QUESTION: 18

A practical voltage source consists of:

Solution: An ideal current source is expected to have zero losses, but practically this is not possible. The internal loss of the current source is represented as an internal resistance ‘r’ placed in series with the current source. It is placed in parallel because current remains equal in series components and divides in parallel components.
QUESTION: 19

While analysing a two-port network, which of the following parameters are obtained when one-port variables are expressed in terms of the other port variables?

Solution: Expressing one port variables in terms of the other port variables,

(V1,I1) = f(V2, - I2)

The parameters so obtained are the T-parameters or the ABCD-parameters. These are used for the analysis of the power transmission line. The input and output ports are the sending and the receiving ends respectively.

QUESTION: 20

Consider the following sentences regarding a periodic signal

a) A periodic signal always repeats itself after minimum fundamental time period T.

b) A periodic signal is exist from -∞ to +∞ and converse is also true.

c) A periodic signal exist from -∞ to +∞.

Which of the following is/are correct option?

Solution: A- Is the definition of any periodic signal.

B- A periodic signal is only exist from -∞ to +∞ and converse is not true because any signal or anything practically always take birth from origin means from it increases. If it comes from positive to negative then it is a decaying signal not a periodic signal. Thus, option A is the correct answer.

QUESTION: 21

The impulse response of a second order under damped system starting from rest is given by

C(t) = 12.5e-8tcos 6t

What are the responsive value of natural frequency and damping ratio of the system?

Solution:  QUESTION: 22

If x(t) = δ(2t/3), what will be its amplitude at t = 0?

Solution: According to the property of Unit Impulse function

δ(t) = 0, t ≠ 0

≠ 0, t = 0 (Very Large Infinite)

Since delta function represents an infinite amplitude spike at t = 0 having area = 1.

Hence, value of x(t) at t = 0 would be infinite

*Answer can only contain numeric values
QUESTION: 23

For the CE amplifier shown, calculate value of lower cutoff frequency FL (in Hz) due to C2 Solution:  QUESTION: 24

In the given figure, β = 100, the magnitude of voltage gain (V0/Vin) will be ___ Solution: First we draw the AC equivalent of circuit  Voltage gain (Av) = -gmRc QUESTION: 25

Calculate the hysteresis width (in Volts) for an inverting Schmitt trigger with feedback fraction 0.5. Assume the supply voltage to be 1

Solution: For a inverting Schmitt trigger.

Given: β = 0.5, Vcc = 10V

Hysteresis Voltage VH = VUT - VLT

VUT = +β x Vcc

VLT = -β x Vcc

VH = 2 x β x Vcc

VH = 2 x 0.5 x 10

VH = 10V

QUESTION: 26

For a given sample-and-hold circuit, if the value of the hold capacitor is increased, then

Solution: Capacitar drap rate = dv/dt

For a capacitar, Drap rate decreases as capacitar value is increased

For a capacitar,

Q = cv = i × t

⇒ t ∝ c

∴ Acquisition time increases as capacitar value increased

QUESTION: 27

The amplitude of a random signal is uniformly distributed between -5 V and 5 V.

If the signal to quantization noise ratio required in uniformly quantizing the signal is 43.5 dB, the step size (in V) of the quantization is approximately

Solution: Which is close to 0.0667 Volts

QUESTION: 28

The Potential (scalar) distribution in free space is given as V = 10y4 + 20x3. if ε0 is permittivity of free space then the charge density ρ at the point (2,0) is

Solution: Here, we have to use Passion’s Equation,  At paint (2,0)

ρ = -240ε0

QUESTION: 29

For a telephone line, β = 0.02rad/km. If frequency is 1 kHz. Calculate time taken to travel 3.14 km by the wave in

Solution: QUESTION: 30

Consider an air-filled rectangular waveguide with inside dimensions 5 cm x 2cm. If the wave impedance is 222.24 Ω for TM11 mode of propagation. Then the operating frequency (in GHz)is given by

Solution: Given a = 5cm, b 2cm, ηTM11 = 222.24Ω  *Answer can only contain numeric values
QUESTION: 31

A 1MHz uniform plane wave propagates through fresh water for which σ = 0, μr = 1,εr = Calculate wavelength in meters.

Solution: As σ = 0, α will also be 0. but μ = μ0μr and ε = ε0εrAlso Ω = 2π x 106 λ = 2π/ β = 2π/0.185

λ = 33.963m

QUESTION: 32

Which of the following is correct for 1110100 ÷ 1010?

Solution: By simple divide method, QUESTION: 33

Which of the following coding scheme is most suitable for coding the successive positions in 8-position digital shaft encoder?

Solution: In digital shaft encoder only one bit change in successive codes, so Gray Code is preferred for digital shaft encoder.

Hence option B is correct as it signify only 1-bit change between two successive codes.

000→001 →011→010→ 110→111→101→111

QUESTION: 34

12 signal each band-limited to 5 KHz are to be transmitted over a single channel by freq. division multiplexing. If AM-SSNB modulated guard band of 1 KHz is used, then the bandwidth of the multiplexed signal will be

Solution: The total signal Bankwith = 5 × 12 = 60 KHz

There would be 11 guard band between 12 signal,

So, guard band width

Total band width = 60 + 11 = 71 KHz

QUESTION: 35

ln INTEL 8085, suppose the peripheral mapped l/O has address length of M and memory mapped l/O address length of N. Then M+N =_______.

Solution: Peripheral mapped l/O has address length

M = 8

Memory mapped l/O has address length

N = 16

M + N = 8 + 16 = 24

QUESTION: 36

Find the bandwidth (in rad/s) of a parallel GLC resonant circuit whose transfer function is Solution: The standard representation of transfer function of a parallel GLC resonant circuit is  On comparing the given T.F with the standard T.F,

Bandwidth = G/C = 2 rad/s.

QUESTION: 37

A 70 kHz bandwidth is to accommodate 7 AM broadcast signals simultaneously. The maximum modulation frequency must each station be limited to _____ kHz. Assume that a guard band of 5 kHz is also present.

Solution: QUESTION: 38

A carrier frequency modulated with a sinusoidal signal of frequency 4 kHz resulting in a maximum frequency deviation of 7kHz. The bandwidth of the modulated signal is

Solution: Using Carson’s rule

BW = 2(β + 1)fm QUESTION: 39

Consider two real valued signals, x(t) band-limited to [−500 Hz, 500Hz] and y(t) bandlimited to [−1kHz, 1kHz]. For z (t) = x(t). y(t), the Nyquist sampling frequency (in kHz) is __________

Solution:

x(t) is band limited to [−500Hz, 500Hz] y(t )is band limited to [−1000Hz, 1000Hz]

z(t) = x (t).y(t)

Multiplication in time domain results convolution in frequency domain.

The range of convolution in frequency domain is [−1500Hz, 1500Hz]

So maximum frequency present in z(t) is 1500Hz Nyquist rate is 3000Hz or 3 kHz

*Answer can only contain numeric values
QUESTION: 40

The laplace transform of a signal x(t) is  then the value of ∅ in degrees is

Solution: Taking inverse laplace We know that Using above equation So, on comparing θ = -30º

QUESTION: 41

A sinusoidal oscillator is built using an amplifier of real gain A, infinite input impedance and the feedback(β) network as shown in the figure.

If Z = jX then Z0 is equal to Solution: For zero phase shift

Z0 + 2 = 0 (As Z0 + 2Z is imaginary)

Z0 = – 2Z = – 2jX

QUESTION: 42

Consider the three transistor current source in figure below. Assume all transistor are matched with finite gain and early voltage VA = ∞. The expression for I0 is Solution:

The circuit is as follows  QUESTION: 43

Two 4-ray signal constellations are shown. It is given that ϕ1 and ϕ2 constitute an orthonormal basis for the two constellations. Assume that the four symbols in both the constellations are equiprobable. Let N0/2 denote the power spectral density of white Gaussian noise. The ratio of the average energy of Constellation 1 to the average energy of Constellation 2 is:

Solution:

Average energy of constellation 1 is Average energy of constellation 2 is QUESTION: 44

In an analogue cellular system, find the number of channels when total spectrum allocation (Bt) is 12.5 MHz per simplex band with guard band (Bguard) as 10 kHz having channel bandwidth (Bc) as 30 kHz.

Solution:

Number of channels in a FDMA system be calculated as N = Bt-Bguard/Bc

where

N is number of channels

Bt is total spectrum allocation

Bguard is guard band

Bc is channel bandwidth

Calculating channels: N = Bt-Bguard/Bc

N= 12.5 x 106 2(103) / 30 x 103

= 416 channels

QUESTION: 45

In synchronous TDM, there are four inputs and data rate of each input connection is 3 kbps. If 1 bit at a time is multiplexed, what is the duration of each frame?

Solution:

Given, N = 4n = 1 bit fs = 3kbps

bit raterb = n.N.fs = 1 x 4 x 3kbps = 12 kbps *Answer can only contain numeric values
QUESTION: 46

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

Solution: = 120 m3/C

Therefore, μ = σRH

=5 × 10-4 × 360

=5 × 10-4 × 120

= 600 cm2/V-sec

QUESTION: 47

Silicon is doped with boron to a concentration of 4 × 1017 atoms/cm3. Assuming the intrinsic carrier concentration of silicon to be 1.5×1010/cm3 and the value of kT/q to be 25 mV at 300 K

Compared to undoped silicon, the Fermi level of doped silicon?

Solution:

E2 – E1 = kTln (NA/ni)

NA = 4 × 1017

ni = 1.5 × 1010

E2 – E1 = 25 × 10-3 e ln 4 × 1017/1.5×1010 = 0.427eV

Hence Fermi level goes down by 0.427 eV as silicon is doped with boron.

*Answer can only contain numeric values
QUESTION: 48

The transmission line shown in figure is 40 m long and has Vg = 15ÃOo Vrms, Zo = 30 + j 60W and VL = 5 Ã–48o Vrms. If the line is matched to the load, then the magnitude of sending – end current Iin is

Solution: QUESTION: 49

Consider the following synchronous counter made up of JK, D and T flip-flops. Find the modulus value of the counter.

Solution: Consider characteristics equation of J-K flip-flop Consider characteristic equation of D- Flip-Flop : QN+1 = D

QN+1 = Q2….(ii)

Consider characteristics equation of T-Flip-flop: QN+1 = T⊕QN

Qx+1 = Q1⊕Q0

Using equations (i), (ii) and (iii)  The number of used states = 5

∴ Modulus value of the counter = 5.

QUESTION: 50

Consider the following k-map, P, Q, R are don’t cares. What should be the values of P, Q, R in order to get the minimized SOP expression? Solution: SOP = yw + xy = y(x + w) so total is 4 variable

So. P, Q, R are all zero’s.

and if we put the p=0 and Q = 1, R = 1 then we get wy+xy here also 4 variable so and is D correct

QUESTION: 51

An antenna consist of 4 identical Hertzian dipoles located along z-axis uniformly with λ/8 space apart.

The group pattern function is_________?

Solution:  Hence option B is correct.

QUESTION: 52

The maximum phase lag provided by the given lag compensator is? Solution:

Given network can be expressed as QUESTION: 53

Find the particular integral of

(D2 – 4D +4) y = e2x sinx cosx

Solution: QUESTION: 54

For a transmission line at frequency of 300 MHz. We have, first maxima occurs at 1/8 m from the load. ZL = jRL , Where RL = k.R0 and R0 = characteristic impedance of line.

Then k = _________? (positive value of k)

Solution: Hence, we see that R is purely imaginary As R is purely imaginary

R|real = 0

∴ k2 − 1 = 0

∴ k = ±1

∴ k = 1............positive value

QUESTION: 55

A signal m(t) is shown This signal is applied to both phase modulator (with kp as phase constant) and frequency modulator (with Kf as frequency constant).

The ratio of kf/kp for same maximum phase deviation is _______?

Solution:

In phase modulation maximum phase deviation = kp∣m(t)Imax

Δϕ = 4.kp…(1)

In frequency modulation

Maximum phase deviation = 2πkf∫m(t)dt∣max Given

The maximum phase deviation for both techniques is same

i.e., from (1) and (2) Hence option B is correct.

QUESTION: 56

Calculate the drift velocity of free electrons in a conductor of cross-section area of 10-4 m2 and current of 400 A is flowing through it, assuming free electron density of conductor to be 9 × 1028 m-3.

Solution:  Hence option A is correct

QUESTION: 57

The inverse z-transform of the signal x(z) = log(1 + z–1) |z| > 1 is?

Solution:

Given,

x(z) = log⁡(1 + z − 1)

Differentiating w.r.t z we get QUESTION: 58

The input voltage Vi/P is 5 + 4sinωt. Assume diode is ideal. Then the average value of output voltage is? Solution:

It is a negative clamper circuit and capacitor blocked dc voltage. Hence only Vi/P = 4 sinωt

Hence waveforms are: Hence avg VO/P = -4 volt

Hence option D is correct.

QUESTION: 59

A die is rolled ‘n’ times until 4 appears then the probability that die must rolled more than 5 times before getting a “4”,

Solution:

probability of Getting ‘4’ on rolling a die is = ⅙

So, Probability of Not getting ‘4’ is = ⅚

To getting success (get-4), die must be rolled more than 5 times

So,

S

P(x > 5) = 1 − P(x ≤ 5)

Getting success in st trial So  *Answer can only contain numeric values
QUESTION: 60

For a p-n-p BJT with emitter doping concentration of 1017 / cm3, Base doping of 5 x 1015 / cm3 and collector doping of 3 x 1014 /cm3. Then under no external voltage VCE is ___ volt?

Solution: (0.10, 0.25)   *Answer can only contain numeric values
QUESTION: 61

The value of the Integral is (accurate upto 3 decimal place)

Solution: The integral is of the form  Putting the value of s = 3 *Answer can only contain numeric values
QUESTION: 62

The open-loop transfer function unity feedback system is given by The breakaway point of the root locus plot is given by

Solution: But s = −1.577 doesn't lie on the root locus

Hence s = −0.423 is the breakaway point.

*Answer can only contain numeric values
QUESTION: 63

An antenna a power input of 50 π watt and an efficiency of 90%.

The radiation intensity has been found to have a 6-maximum value of 140W per solid angle.

The directivity of antenna is ____________.

Solution: *Answer can only contain numeric values
QUESTION: 64

Power across 20 Ω resistor is 10 watt. Then the power factor of the coil for the given connection is _____ Solution: We know  QUESTION: 65

For given code executed in 8085 μp After the PSW value is?

MVI A, 63H

ORA A

ANI 36 H

DAA

Solution:

A = 63

ORA A: A = 63

Flags: P = 1

Z = 0

S = 0

AC = 0

C = 0

ANI: 63 AND 36

= 22 (0010 0010)

Flags: P = 1

Z = 0

S = 0

AC = 1

C = 0

DAA:

as AC = 1 add 6 (0110) to lower nibble

As C = 0 do not add 6 to upper nibble.

∴ ACC ⇒ 0010 0010

+ 0000 0110

0010 1000

= 28 H

Flags: cy = 0

P = 1

AC = 0

Z = 0

S = 0 Use Code STAYHOME200 and get INR 200 additional OFF Use Coupon Code