Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Test  >  MPPGCL JE Electronics Mock Test Series 2026  >  MPPGCL JE Electronics Mock Test - 10 - Electronics and Communication Engineering (ECE) MCQ

MPPGCL JE Electronics Mock Test - 10 Free Online Test 2026


Full Mock Test & Solutions: MPPGCL JE Electronics Mock Test - 10 (100 Questions)

You can boost your Electronics and Communication Engineering (ECE) 2026 exam preparation with this MPPGCL JE Electronics Mock Test - 10 (available with detailed solutions).. This mock test has been designed with the analysis of important topics, recent trends of the exam, and previous year questions of the last 3-years. All the questions have been designed to mirror the official pattern of Electronics and Communication Engineering (ECE) 2026 exam, helping you build speed, accuracy as per the actual exam.

Mock Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 120 minutes
  • - Total Questions: 100
  • - Analysis: Detailed Solutions & Performance Insights
  • - Sections covered: Electronics, General Awareness and Aptitude

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MPPGCL JE Electronics Mock Test - 10 - Question 1

The following circuit is of a:

Detailed Solution: Question 1

The following circuit is of a: Bubble AND gate

Key Points

Alternative Logic Gates:

An alternative logic gate is an alternate logic gate that produces the same output as the original logic gate and can be used during the unavailability of the original logic gate to serve the same purpose.

  • Alternative logic gates are also called as Alternate Gates.
  • Alternative logic gates are also called as Bubbled Gates since they contain bubbles in them.

MPPGCL JE Electronics Mock Test - 10 - Question 2

Which information is displayed in the fire control room?

Detailed Solution: Question 2

Fire Control Room

  • For all buildings 15 m in height and above and apartment buildings with a height of 30 m and above, there shall be a Control Room on the entrance floor of the building with a communication system (suitable public address system) to all floors and facilities for receiving the message from different floors.
  • Details of all floor plans along with the details of fire fighting equipment and installations shall be displayed in the fire control room.
  • The fire control room shall also have facilities to detect the fire on any floor through indicator boards connection; fire detection and alarm systems on all floors.
  • The fire staff in charge of the fire control room shall be responsible for maintenance of the various services and the fire fighting equipment and installations in coordination with security, electrical and civil staff of the building.

MPPGCL JE Electronics Mock Test - 10 - Question 3

Programmable Logical Controllers are the devices that were invented basically to replace

Detailed Solution: Question 3

A programmable logic controller (PLC) or programmable controller is an industrial digital computer which has been adopted for the control of manufacturing processes, such as assembly lines, or robotic devices, or any activity that requires high-reliability control and ease of programming and process fault diagnosis.

Early (PLC) was designed to replace relay logic systems. These PLCs were programmed in "ladder logic", which strongly resembles a schematic diagram of relay logic.

MPPGCL JE Electronics Mock Test - 10 - Question 4

The following is the circuit diagram of a/an _______ using IC555 timer.

Detailed Solution: Question 4

Concept:

Astable multivibrator:

It is also known as a square wave generator. 

In astable multivibrator, 

T = Ton + Toff 

Ton = 0.69 (R+ Rb)C , and  Toff = 0.69 (RbC) 

T = 0.69 (2R+ Ra)C

Output frequency is given by 

Analysis:

Given circuit diagram is a stable multivibrator or a square wave generator.

MPPGCL JE Electronics Mock Test - 10 - Question 5

In a PNP transistor, due to doping the emitter conductivity is 10 times base conductivity. Electron diffusion length in a more heavily doped emitter is half the hole diffusion length in the base. The base width is one-tenth of the hole diffusion length in the base. The a and b of the transistor are _____ and ______.

Detailed Solution: Question 5

The current gain β of the transistor is given by:

 

DE: Diffusion constant of minority carriers in the emitter

DB: Diffusion constant of minority carries in Base

NE: emitter doping concentration

NB: Base doping concentration

LE: Length of diffusion of minority carriers in emitter

WB: Base width.

 

 ratio of conductivity from emitter to base = 10 (Given)

LEn (minority carries in emitter)  (hole diffusion length in base)

 

 

Taking the case of Si:

β = 2.1 × 10 × 5 = 105

Taking case of Si:

β = 2.6 × 10 × 5 = 130

None of the given option match, but if we check the option by satisfying the relation  

Only option 1 is satisfying

Hence option 1 is correct.

MPPGCL JE Electronics Mock Test - 10 - Question 6

For an isotropic radiator, electric field intensity at a distance R is measured as 3 V/m. What will be the electric field intensity at a distance 3 R?

Detailed Solution: Question 6

Concept:

The electric field for an isotropic antenna varies with the distance as given:

1/r term is the most dominant. 

Calculation:

Given,

E = 3 V/m for r = R

For r2 = 3R,

Electric field E2 will be 

MPPGCL JE Electronics Mock Test - 10 - Question 7

The Eddy current loss is proportional to the

Detailed Solution: Question 7

Concept:

  • When an alternating magnetic field is applied to a magnetic material, an emf is induced in the material itself according to Faraday’s law of Electromagnetic induction.
  • Since the magnetic material is a conducting material, these EMF’s circulates current within the body of the material. These circulating currents are called Eddy currents. They are produced when the conductor experiences a changing magnetic field.

Eddy current loss in the transformer is given by:

Pe = Ke Bm2. t2. f2. V Watts

Where;

K - coefficient of eddy current. Its value depends upon the nature of magnetic material

Bm - Maximum value of flux density in Wb/m2

t - Thickness of lamination in meters

f - Frequency of reversal of the magnetic field in Hz

V - Volume of magnetic material in m3

From the above formula we conclude that the Eddy current loss is proportional to the square of the frequency.

MPPGCL JE Electronics Mock Test - 10 - Question 8

Moving iron instruments are scaled at ______.

Detailed Solution: Question 8

Moving Iron Instrument

  • MI instruments are used for the measurement of the RMS value of current and voltage.
  • These instruments have a non-uniform scale.
  • It works on the principle of attraction of a single piece of soft iron into the magnetic field.
  • In the MI instrument, when the current is passed through the solenoid, a magnetic field is set up inside the solenoid. Then the iron piece gets magnetized which in turn deflects the pointer. 
  • The deflecting torque produced is directly proportional to the square of the current.
  • The controlling torque is provided by the spring control.
  • The damping torque is provided by air friction damping.

MPPGCL JE Electronics Mock Test - 10 - Question 9

IRS series satellite are used for -

Detailed Solution: Question 9

The correct answer is Remote sensing.

  • IRS series satellite is used for Remote sensing.

Key Points

  • The Indian Remote Sensing (IRS) system is the largest constellation of remote sensing satellites for civilian use in operation today in the world, with 11 operational satellites.
  • All these are placed in polar Sun-synchronous orbit and provide data in a variety of spatial, spectral, and temporal resolutions.
  • Indian Remote Sensing Programme completed its 25 years of successful operations on March 17, 2013.
  • IRS data applications:
    • Drought monitoring and assessment based on vegetation condition.
    • Flood risk zone mapping and flood damage assessment.
    • Hydro-geomorphological maps for locating underground water resources for drilling well.
    • Irrigation command area status monitoring.
    • Snow-melt run-off estimates for planning water use in downstream projects.
    • Land use and land cover mapping.
    • Urban planning.
    • Forest survey.
    • Wetland mapping.
    • Environmental impact analysis.

Additional Information

  • ​Following are the remote sensing satellites planned by ISRO to be launched:
    • RESOURCESAT-3
      • A follow on to Resourcesat-2, it will carry a more advanced LISS-III-WS (Wide Swath) Sensor having similar swath and revisit capability as Advanced Wide Field Sensor (AWiFS).
      • The satellite would also carry Atmospheric Correction Sensor (ACS) for quantitative interpretation and geophysical parameter retrieval.
      • It slated to be launched in 2021.
    • OCEANSAT-3
      • Oceansat-3 would carry Thermal IR Sensor, 12 channel Ocean Color Monitor, Scatterometer, and Passive Microwave Radiometer.
      • IR Sensor and Ocean Color Monitor would be used in the analysis for operational Potential Fishing Zones.
      • Satellite is mainly for Ocean biology and sea state applications.
      • It is slated to the launched aboard PSLV in January 2020.
    • GISATs
      • GISAT-1 is planned for launch in May 2020 and GISAT-2 is planned for launch in August 2020. 
      • They are expected to provide images from geostationary orbit during disasters. 

MPPGCL JE Electronics Mock Test - 10 - Question 10

A voltmeter has resistance of 2000 Ω, when it is connected across a DC circuit its power consumption is 2 mW. Suppose this voltmeter is replaced by a voltmeter of 4000 Ω resistance, the power consumption will be:

Detailed Solution: Question 10

The correct answer is option "4"
Concept:

The  power consumption is given by 

P =V/ R watts as the voltage across it will be constant.

where 

V is voltage in Volts

R is the resistance

Calculation:

R1 = 2000 Ω

P1 =  2 mW

P = V2/R1

V2 = PR1 = 2 × 10-3 × 2000 = 4 V

V = 2 V

for R2 = 4000 Ω

P = V2/R2

= 22 / 4000 = 1 mW

MPPGCL JE Electronics Mock Test - 10 - Question 11

Which of the following parameters are affected due to short-channel MOSFET geometry

I. Mobility of carriers

II. Threshold voltage

III. Drain current

Detailed Solution: Question 11

Short-channel effect in MOSFET:

  • If the channel length of MOSFET becomes comparatively equal to the depletion layer widths of source and drain junction then it is called that MOSFET suffers from short-channel effect.
  • Due to this effect, the following parameters are affected
  • Mobility of charge carrier in the channel
  • The threshold voltage changed due to a shorter length of the channel.
  • Drain current also changed as threshold voltage changed

So, based on the above points all statements are true

MPPGCL JE Electronics Mock Test - 10 - Question 12

Find I in the network shown.

Detailed Solution: Question 12

Concept:

According to Kirchhoff’s current law (KCL), the algebraic sum of the electric currents meeting at a common point is zero. I.e. the sum of currents entering a node is equal to the sum of currents leaving the node. It is based on the conservation of charge.

Mathematically we can express this as:

Where in represents the nth current

Calculation:

Now by applying KCL

I + 0.4 I = 14 A

1.4 I = 14 A

⇒ I = 10 A

Additional Information

Kirchhoff’s voltage law (KVL), the algebraic sum of all the voltages around any closed path is zero. It is based on the conservation of energy.

Mathematically we can express this as:

Where Vn represents the nth Voltage

MPPGCL JE Electronics Mock Test - 10 - Question 13

Find Z parameter Z11, Z22 in given network -

Detailed Solution: Question 13

Concept:

Z parameter:

We will get the following set of two equations by considering the variables V1 & Vas dependent and I1 & I2 as an independent. The coefficients of independent variables, I1 and I2 are called Z parameters.

V1 = Z11I+Z12I2

V2 = Z21I1+ Z22I2

Calculation:

Apply KVL on loop 1:

⇒ V1 = 5I1 + 20(I1 + I2)

⇒ V1 = 25I1 + 20I2 ------------ (1)

Apply KVL on loop 2:

⇒  V2 = 10I2 + 20(I1 + I2)

⇒ V= 20I1 + 30I2 -------------(2)

Compare equation 1 and 2 with Z parameter equations:

 Z11 = 25Ω, Z22 = 30Ω

Alternate Method 
For a T equivalent network as shown below 

Z11 = (Z1 + Z3) = (5+20) = 25Ω 

Z22 = (Z2 + Z3) = (10 + 20) = 30Ω 

Z12 = Z21 = Z3 = 20Ω 

MPPGCL JE Electronics Mock Test - 10 - Question 14

If radiation resistance is 200 Ω and operating with 6 A of current at 4 GHz frequency, then what is the power radiated by an antenna?

Detailed Solution: Question 14

Radiated power(P) is a measure of how much power is radiated by an antenna when the antenna is connected to a transmitter.

Radiated resistance(R) is the component of antenna resistance that accounts for the power radiated into space and is equal in ohms.

Power radiated by an antenna is calculated by using formula,

P= I2× R

P = (6 × 6) × 200

P = 36 × 200

P = 7200 W

MPPGCL JE Electronics Mock Test - 10 - Question 15

A single Decimal-to-BCD encoder has _______ outputs.

Detailed Solution: Question 15

  • An encoder is a device that converts numbers, characters, or symbols into a coded format.
  • Decimal-to-BCD encoder is the type of encoder that usually consists of ten input lines and 4 output lines.
  • Each input line corresponds to each decimal digit and 4 outputs correspond to the BCD code which is represented in 4 bits.
  • This encoder accepts the decoded decimal data as an input and encodes it to the BCD output which is available on the output lines.
  • The block diagram of a Decimal-to-BCD encoder is as shown:

Important Point

Different types of other encoders include:

  • 4-to-2 Binary Encoder (4 inputs and 2 coded outputs)
  • Octal to Binary Encoder (Eight inputs and 3 coded outputs)
  • Keyboard Encoder

MPPGCL JE Electronics Mock Test - 10 - Question 16

What type of computer component is a printer:

Detailed Solution: Question 16

The correct answer is "Output Device".

Key Points 

  • Printer-
    • It is a computer peripheral machine (output device) that prints graphics or text, usually on paper.
    • There are different types of printers available. They are- 3D printers, inkjet printers, laser printers, and thermal printers.

Additional Information

  • Output Device-
    • An output device is any piece of computer hardware that converts information into a human-perceptible form or, a physical machine-readable form for use.
    • The computer monitor is the most common output device.
    • For example- Printer, Plotter, Projector, etc.
  • Input Device-
    • In computer terminology, Input Device is a piece of equipment used to feed data and signals to an information processing system, such as a computer.
    •  For example- keyboards, mouse, scanners, cameras, joysticks, microphones, etc.
  • Data Generator-
    • It generates data according to some pattern instead of reading data from a file, databases, or any other data source. 

MPPGCL JE Electronics Mock Test - 10 - Question 17

Assertion A): In satellite communication, uplink and downlink frequencies are kept different.

Reason R): The uplink frequency is chosen to be of lower value than the down link frequency.

Choose the correct answer:

Detailed Solution: Question 17

In satellite communication, uplink and downlink frequencies are kept different.

Uplink frequency > Downlink frequency

Hence, option C is correct.

Explanation:

  • The transmission of a signal from the first earth station to satellite through a channel is called uplink​.
  • The transmission of a signal from the satellite to the second earth station through a channel is called the downlink.
  • Uplink frequency is the frequency at which, the first earth station is communicating with the satellite.
  • The satellite transponder converts this signal into another frequency and sends it down to the second earth station.
  • This frequency is called Downlink frequency. In a similar way, the second earth station can also communicate with the first one

In the atmosphere, the attenuation is directly proportional to frequency.

Since the satellite cannot afford to have high powers downlink is sent at lower frequencies, while the station at earth can afford to transmit at high frequencies.

MPPGCL JE Electronics Mock Test - 10 - Question 18

Number of bits per symbol in a 16 QAM is

Detailed Solution: Question 18

Concept:

Symbol rate

  • The symbol rate is measured in baud (Bd) or symbols per second.
  • In the case of a line code, the symbol rate is the pulse rate in pulses per second.
  • Each symbol can represent or convey one or several bits of data.
  • The symbol rate is related to the gross bitrate expressed in bits per second.

A symbol may consist of one or more bits as determined by the modulation format.

In a BPSK system, each symbol represents 1 bit; in a QPSK system, each symbol represents 2 bits. The Baud Rate (BR) is given by:

Calculation:

Given modulation technique is 16 QAM

The number of QAM states is 2N, as determined by the number of binary bits per symbol.

Thus a 16-QAM system (N = 4) is one for which the (microwave) carrier is modulated into any one of 16 different amplitude and phase states

MPPGCL JE Electronics Mock Test - 10 - Question 19

________ is mostly preferred for telegraphy.

Detailed Solution: Question 19

  • FSK is a system of frequency modulation in which the nominal unmodulated carrier frequency corresponds to the mark condition, and space is represented by a downward frequency shift
  • In the FSK generator, the frequency shift may be obtained by applying the varying dc output of the telegraph machine to a varactor diode in a crystal oscillator
  • At the receiving end, the signal is demultiplexed and applied to a standard phase discriminator
  • This is how a telegraph works with FSK modulation

MPPGCL JE Electronics Mock Test - 10 - Question 20

Spread-spectrum signaling technique used in the following mobile communication system

Detailed Solution: Question 20

CDMA (Code Division Multiple Access) uses a digital modulation technique called spread spectrum which spreads voice data over a very wide spectrum using a user or cell specific pseudo-random codes.

MPPGCL JE Electronics Mock Test - 10 - Question 21

Companding is used to

Detailed Solution: Question 21

The smaller the signal strength, the stronger it is hit by the quantization process. Companding is used to protect these small strength signals from quantization noise.

Companding:

  • Companding, also known as Companded PCM, is a non-uniform quantization technique.
  • It is implemented to improve the signal to quantization noise ratio of weak signals.
  • The compression of the signal at the transmitter and expansion at the receiver is combined to be called companding. This is shown in the form of a block diagram:

           

  • Companders like A-law and μ-law Companders are used in PCM to compress signals before input to the ADC in the transmitter stage and expand it after the input to DAC at the receiver.

MPPGCL JE Electronics Mock Test - 10 - Question 22

Which of the following operation is mainly performed by RAM?

Detailed Solution: Question 22

The term RAM refers to read and write memory, that is, you can both write data into RAM and read data from RAM. Most RAM is volatile, which means that it requires a steady flow of electricity to maintain its contents. Read-only memory (ROM) refers to computer memory on which data has been prerecorded. Once data has been written onto a ROM chip, it cannot be removed and can only be read.

MPPGCL JE Electronics Mock Test - 10 - Question 23

The z-transform of  is 

Detailed Solution: Question 23

The z-transform for 1/n does not exist for n=0, but it exists for n > 0 and for n < 0

Let n > 0

X(z) = 

Differentiating both sides w.r.t z we get,

   |z-1| < 1 or |z| > 1

Integrating both sides, we get

 for |z| > 1

Let n < 0

X(z) = 

Differentiating both sides w.r.t z we get,

Let n = -p

     |z| < 1

Integrating both sides, we get

 for |z| < 1

∴ By looking at the options the correct answer is 3 (given for n > 0)

MPPGCL JE Electronics Mock Test - 10 - Question 24

The signals x1(t) and x2(t) are band-limited to 4π rad/sec and 10π rad/sec respectively. The minimum sampling rate required for sampling the signal x1­(2t) + x2(t/2) is:

Detailed Solution: Question 24

Concept:

Sampling Theorem: According to the sampling theorem, for the signal to be recovered back, the minimum sampling rate required for sampling any signal is the Nyquist sampling rate which is twice the maximum band of frequency.

Effect of scaling:

The scaling in the time domain results in an inverse of scaling in the frequency domain, i.e.

For a signal x(t) bandlimited to fm Hz, the signal x(at), will be bandlimited to (a × fm) Hz.

Calculation:

We are given a signal x1(t) is band limited to 4π rad/sec then signal x1(2t) will band limited to:

This is because x(2t) is the compressed version of x(t) and hence in the frequency domain the spectrum will expand.

Similarly, for signal x2(t), band-limited to 10π rad/sec, x2(t/2) will band limited to:

This is because x(t/2) is the expanded version of x(t) and hence in the frequency domain the spectrum will get compressed.

Now a signal x(t) = x1(2t) + x2(t/2) will be bandlimited to 8π rad/sec.

The Nyquist or minimum sampling rate of signal x(t) will twice the maximum rate present in it, i.e.

Nyquist rate = 2 × 8π rad/sec

Nyquist rate = 16π rad/sec

MPPGCL JE Electronics Mock Test - 10 - Question 25

Which of the following types of standard input signals has the Laplace transformation as  ?

Detailed Solution: Question 25

Concept:

Some important Laplace transforms are:

MPPGCL JE Electronics Mock Test - 10 - Question 26

If a modulating signal with frequency fm amplitude modulates a carrier of frequency fc (where fc ≫ fm), the frequencies present at the output of AM is/are (with carrier).

Detailed Solution: Question 26

The general expression of an amplitude-modulated wave for a single-tone message signal is given as:

sAM(T) = [Ac + Amcosωmt] cosωct

where ωm = Frequency of the message signal

ωc = Carrier signal frequency

The expression can be written as:

sAM(T) = Ac cosωct + Amcosωmtcosωct

We observe that the amplitude modulated signal has both the carrier frequency and the shifted frequency as shown:

MPPGCL JE Electronics Mock Test - 10 - Question 27

In moving coil instruments the scale used is:

Detailed Solution: Question 27

The correct answer is 'option 2'

Solution:

In moving-coil instruments, the deflection is directly proportional to the current. i.e. θ ∝ I.

∴ The scale used in the moving coil instruments is linear in nature.

Additional Information

  • In thermocouple instruments, the deflection is directly proportional to the square of the current. i.e. θ ∝ I2
  • In moving iron instruments, the deflection is directly proportional to the square of the current. i.e. θ ∝ I2
  • In hotwire instruments, the deflection is directly proportional to the square of the current. i.e. θ ∝ I2

MPPGCL JE Electronics Mock Test - 10 - Question 28

If the desired transmission bit rate for a coherent binary FSK system is 4 Kbits / sec, the best possible interval between the carriers is

Detailed Solution: Question 28

Given: 

Rb = 4 kbps

In FSK, the best possible interval between the carriers is given as:

ΔT = 0.5 msec

Important Point

FSK (Frequency Shift Keying):

In FSK (Frequency Shift Keying) binary 1 is represented with a high-frequency carrier signal and binary 0 is represented with a low-frequency carrier, i.e. In FSK, the carrier frequency is switched between 2 extremes.

For binary ‘1’ → S(A) = Acos 2π fHt

For binary ‘0’ → S2 (t) = A cos 2π fLt . The constellation diagram is as shown:

ASK(Amplitude Shift Keying):

In ASK (Amplitude shift keying) binary ‘1’ is represented with the presence of a carrier and binary ‘0’ is represented with the absence of a carrier:

For binary ‘1’ → S1 (t) = Acos 2π fct

For binary ‘0’ → S2 (t) = 0

The Constellation Diagram Representation is as shown:

 

where ‘I’ is the in-phase Component and ‘Q’ is the Quadrature phase.

PSK(Phase Shift Keying):

In PSK (phase shift keying) binary 1 is represented with a carrier signal and binary O is represented with 180° phase shift of a carrier

For binary ‘1’ → S1 (A) = Acos 2π fct

For binary ‘0’ → S2 (t) = A cos (2πfct + 180°) = - A cos 2π fct

The Constellation Diagram Representation is as shown:

MPPGCL JE Electronics Mock Test - 10 - Question 29

Direction: In a family, there are six members P, Q, R, S, T, U, and V are a married couple. P being the male member. S is the only son of R, who is the brother of P. T is the sister of S. Q is the daughter-in-law of U, whose husband has died.

Q. Who is R to Q?

Detailed Solution: Question 29

R is brother of P, who is husband of Q. So, R is brother-in-law of Q.

MPPGCL JE Electronics Mock Test - 10 - Question 30

Find out the Synonym of the following word:
MASSIVE

Detailed Solution: Question 30

  • Lump sum: a single payment made at a particular time, as opposed to a number of smaller payments or installments.
  • Strong: having the power to move heavy weights or perform other physically demanding tasks.
  • Little: small in size, amount, or degree.
  • Huge: extremely large; enormous.
  • Massive: exceptionally large.

So, the synonym of Massive is Huge.

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