Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Test  >  MPPGCL JE Electronics Mock Test Series 2026  >  MPPGCL JE Electronics Mock Test - 7 - Electronics and Communication Engineering (ECE) MCQ

MPPGCL JE Electronics Mock Test - 7 Free Online Test 2026


Full Mock Test & Solutions: MPPGCL JE Electronics Mock Test - 7 (100 Questions)

You can boost your Electronics and Communication Engineering (ECE) 2026 exam preparation with this MPPGCL JE Electronics Mock Test - 7 (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

Sign up on EduRev for free and get access to these mock tests, get your All India Rank, and identify your weak areas to improve your marks & rank in the actual exam.

MPPGCL JE Electronics Mock Test - 7 - Question 1

Multiplexing scheme which uses carrier phase shifting and synchronous detection to permit two DSB signals to occupy the same frequency band is called

Detailed Solution: Question 1

Digital to Analog Modulation technique is as shown.

Important Points

1) As shown, QAM is a mixture of both ASK and PSK.

2) QAM uses two carrier signals with the same frequency but with which are in quadrature. Quadrature here means out phase by 90°.

3) Since the two signals have the same frequency, they are detected using synchronous detection.

4) Hence, amplitude and the phase of the carrier frequency both vary with the message signal.

A constellation diagram of a QAM signal with 2 different amplitude levels and 8 different phases is shown:

MPPGCL JE Electronics Mock Test - 7 - Question 2

The frequency deviation in phase modulated carrier is proportional to

Detailed Solution: Question 2

The frequency deviation in a phase-modulated carrier is proportional to the amplitude and frequency of the modulating signal both.

Derivation:

A general expression for a phase-modulated wave is:

xPM (t) = A cos [2πfct + kpm(t)]

The instantaneous angle is given as:

ϕi(t) = 2πfct + kp m(t)

The instantaneous frequency (in Hz) will be obtained as:

Frequency deviation is given by:

For a single tone modulation:

m(t) = Am sin ωmt

The frequency deviation becomes:

Δf ∝ Afm

MPPGCL JE Electronics Mock Test - 7 - Question 3

The modulation used in video transmission of commercial TV transmission

Detailed Solution: Question 3

Concept:

Vestigial Sideband Modulation is a type of amplitude modulation where a part of the signal called vestige is modulated, along with one sideband. The spectrum requirement is between DSB-SC and SSB (less than DSB-SC and more than SSB)

Vestigial Sideband Modulation (VSB) is used for video modulation in TV transmission due to the following reasons:

1) A workable compromise between the spectrum-conserving characteristics of single-sideband modulation and the demodulation simplicity of double-sideband amplitude modulation is found in the vestigial-sideband modulation.

2) Video signal exhibits a large bandwidth and significant low-frequency content which suggests the use of VSB.  

The spectrum of a vestigial sideband is as shown:

Important Point

In TV Transmission, FM is used for Audio transmission and AM for Video transmission.

MPPGCL JE Electronics Mock Test - 7 - Question 4

For the given circuit, which output gives logic function A.B.C.D.?

Detailed Solution: Question 4

Concept:

We know that:

  • AND gate output is '0' (If any input is '0').
  • AND gate output is '1' (If all inputs are '1').
  • OR gate output is '1' (If any input is '1').
  • OR gate output is '0' (If all inputs are '0').

Analysis:

In the given logic circuit:

Inputs: A, B, C, D

Output: x, y, z

Let us assume , name of AND gates are A1,  A2,  A3

Solution:

Output of A1 = 

Output of A=

Output of A3 =

Output x =Output of A1 +Output of A2 

  

Output y =Output of A2 +Output of A3 

  

Output z =Output of A3

 = A.B.C.D

MPPGCL JE Electronics Mock Test - 7 - Question 5

A signal m(t) = 2 cos (2π × 103t) phase modulates (PM) a 1 MHz carrier to produce peak frequency deviation of 4 kHz. What is the value of phase deviation constant kp

Detailed Solution: Question 5

Concept:

The general expression for a frequency modulated wave is given by:

SPM(t) = Ac [cosωct + kpm(t)]

The instantaneous phase is given by:

Phase deviation is defined as:

Phase deviation = [Max [kp m(t)]]

Instantaneous frequency is given by:

From the above, the frequency deviation will be:

Frequency deviation maximum

Calculation:

kp = 2 rad/volts

MPPGCL JE Electronics Mock Test - 7 - Question 6

A telephone line of bandwidth of 4 kHz. What data rate is supported for full roll-off?

Detailed Solution: Question 6

Concept:

Calculation:

Given:

B.W = 4000 Hz

α = 1

∴ B.W 

fb = 4 kbps

MPPGCL JE Electronics Mock Test - 7 - Question 7

In FSK, the carrier frequency is switched between ______ extremes.

Detailed Solution: Question 7

ASK, PSK and FSK are signaling schemes used to transmit binary sequences through free space. In these schemes, bit-by-bit transmission through free space occurs.

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’ → S1 (A) = Acos 2π fHt

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

Important Point

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’ → S(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 - 7 - Question 8

Which of the following method is used to improve the switching speed of a PN-diode?

(i) Length of p-region is shortened

(ii) Length of n-region is shortened

(iii) Bismuth dopants are used

(iv) Gold dopants are used

Detailed Solution: Question 8

Concept:

  • By shortening the length of the p-region the reverse recovery time (tn) is reduced considerably.
  • The stored charges can be reduced by adding gold dopants, thereby reducing storage time (ts).
  • Hence, gold dopants are also called lifetime killer.

Generally, T = Tn = ts + tt (t→ storage time, t→ transition time)

Also, frequency 

Hence, on reducing the storage time and reverse recovery time, the frequency is increased, thereby increasing the switching speed of the diode.

MPPGCL JE Electronics Mock Test - 7 - Question 9

A piece of gold has a resistance of 0.8 ohm. Find the resistance of a wire of the same size and resistivity 25 times that of gold.

Detailed Solution: Question 9

Concept:

We know,

where,

R = Resistance

ρ = Resistivity of conductor

L = Length of conductor

A = Cross-sectional area of the conductor.

So, Resistance ∝ Resistivity

Calculation:

Given, Resistance of gold = 0.8 Ω

And Resistivity of measuring wire = 25 × gold resistivity

∴ Resistance of wire = 25 × gold resistance

= 25 × 0.8 Ω

= 20 Ω

MPPGCL JE Electronics Mock Test - 7 - Question 10

Pulse code modulation employing 4-bit code is used to transmit a data signal having frequency components from dc to 2 KHz. The minimum bandwidth of the carrier channel should be

Detailed Solution: Question 10

Given that 

Nyquist Rate 

Bit rate 

Bandwidth is ½ times of bit rate

Hence bandwidth is ½ (16) = 8 kHz

MPPGCL JE Electronics Mock Test - 7 - Question 11

Consider the discrete time signal x(n) = {1, 1, 1, 1, 0.5, 0.5} ⋅ y(n) = conv (δ(n – 1), x(n)) is: 

Detailed Solution: Question 11

Concept:

Shifting property of convolution

Convolution of the signal x(t) with the impulse gives the same signal as a result after the convolution.

x(t) ∗ δ(t) = x(t)

x(t) ∗ δ(t – t0) = x(t – t0)

Substitute ‘t – t0’ in the place of ‘t’

calculation:

Given:

x(n) = {1, 1, 1, 1, 0.5, 0.5}

y(n) = conv (δ(n – 1), x(n)) 

Using property of impilse function:

y(n) = x(n - 1)

Important Point

Properties of δ(t):

1. 

2. 

3. x(t) δ(t – t0) = x(t0) δ(t – t0)

4. 

5. 

6. 

Properties of δ(n):

1. x[n] δ[n] = x[0] δ[n]

2. 

3. x[n] δ[n – n0] = x[n0] δ[n – n0]

4. 

5. δ[an] = δ[n]

MPPGCL JE Electronics Mock Test - 7 - Question 12

The collector to emitter cut off current (ICEO) of a transistor is related to collector to base cut off current (ICBO) as:

(α is CB current gain of the transistor)

Detailed Solution: Question 12

Relation between ICEO and ICBO:

IE = IB + IC

IC = IC (majority) + IC (minority)

= αI + ICBO

= α (IB + IC) + ICBO

⇒ IC (1 – α) = αIB + ICBO

For IB = 0, IC = ICEO

⇒ ICEO (1 – α) = ICBO

MPPGCL JE Electronics Mock Test - 7 - Question 13

The Off-push button is connected as ___________ of the motor starter that controls the wiring.

Detailed Solution: Question 13

The correct answer is option 2): Series in the control circuit

Concept:

  • The push button switch is usually used to turn on and off the control circuit.
  • It is used in electrical automatic control circuits to manually send control signals to control contactors, relays, electromagnetic starters, etc. 
  • It is connected in series with the motor control circuit

Additional Information

  • The threaded neck inserts through a hole cut into a metal or plastic panel, with a matching nut to hold it in place.
  • Thus, the button faces the human operator(s) while the switch contacts reside on the other side of the panel.
  • When pressed, the downward motion of the actuator breaks the electrical bridge between the two NC contacts, forming a new bridge between the NO contacts.
  • The common industrial push button shown below

MPPGCL JE Electronics Mock Test - 7 - Question 14

What type of operation is generally used in GMAW?

Detailed Solution: Question 14

Gas metal arc welding (GMAW) or Metal inert gas arc welding (MIG):

  • In this process the arc is formed between a continuous, automatically fed, metallic consumable electrode and welding job in an atmosphere of inert gas, and hence this is called metal inert gas arc welding (MIG) process.
  • The type of operation is generally used in GMAW is Semi automatic.

  • MIG welding power sources control voltage; this is done by either voltage stepped switches, wind handles, or electronically.

  • The amperage that the power source produces is controlled by the cross sectional area of the wire electrode and the wire speed, i.e. the higher the wire speed for each wire size, the higher the amperage the power source will produce.

  • Argon is used in MIG welding as shielding gas.
  • Argon is denser than air and settles over the joint to protect the molten pool from atmospheric gas contamination.
  • In addition, argon is easy to ionize, so it handles a long arc at low voltages well.
  • When the inert gas is replaced by carbon dioxide then it is called CO2 arc welding or metal active gas (MAG) arc welding.
  • The common name for this process is gas metal arc welding (GMAW). 

MPPGCL JE Electronics Mock Test - 7 - Question 15

Select correct answer -

A 400W carrier is modulated to a depth of 75 percent. Calculate the total power in the modulated wave.

Detailed Solution: Question 15

Concept:

The total transmitted power for an AM system is given by:

Pc = Carrier Power

μ = Modulation Index

Calculation:

With PC = 400 W and μ = 0.75, the total power in the modulated wave will be:

Pt = 512.5 W

NoteThe transmitted power is independent of the modulation index in the case of FM and PM.

MPPGCL JE Electronics Mock Test - 7 - Question 16

Match the following:

Detailed Solution: Question 16

Concept:

For M-array PSK or M-array QAM the bit-rate(Rb) is given as:

Bit-rate (Rb) is directly proportional to log2M.

Analysis:

FSK is used for slow speed Modem so bit rate of FSK is too low

for 4 - DPSK modem: M = 4

for 8 - DPSK modem: M = 8

for 16 QAM modem: M = 16

The correct sequence of Bit-rate is given as:

FSK < 4-DPSK < 8-DPSK < 16-QAM

MPPGCL JE Electronics Mock Test - 7 - Question 17

Two indentical FETs each characterized by the parameters gm and rd are connected in parallel. The composite FET is then characterized by the parameters. 

Detailed Solution: Question 17

The composite FET (Two FET connected in parallel can be drawn as)

∵ Both FET are identical

Overall gm = 2gm

Now, 

∵ both are parallel

MPPGCL JE Electronics Mock Test - 7 - Question 18

Which of the following is used to prevent, detect, and remove malicious programs?

Detailed Solution: Question 18

Concept:

  • Antivirus software, also known as anti-malware, is a computer program used to prevent, detect, and remove malware.
  • Examples of common antivirus programs include MacAfee, Norton Antivirus, Kaspersky Anti-Virus, and Zone Alarm Antivirus.

Important Points

  • Cipher: It is a method of writing in which a secret pattern of a particular set of letters or symbols is used to represent other letters, symbols, etc.
  • Adware: It is also called advertising-supported software by its developers, is software that generates revenue for its developer by automatically generating online advertisements on users’ screen or installation process
  • Firewall: It is a network security system that monitors and controls the incoming and outgoing network traffic based on predefined security rules

MPPGCL JE Electronics Mock Test - 7 - Question 19

A stack-based CPU organization uses_______ address instructions

1) 2

2) 0

3) 1

4) 3

Detailed Solution: Question 19

The correct answer is 0.

Stack-based CPU organization uses zero address instruction.

Key Points

  • The computers which use Stack-based CPU Organization are based on a data structure called a stack. 
  • It makes use of the Last In First Out (LIFO) access technique, which is the most common in most CPUs.
  • The address of the highest member of the stack, known as the Stack pointer, is stored in a register (SP).
  • Push and Pop are the two most common operations done on the stack's operators. These two operations are only carried out from one end.
  • This instruction contains the opcode only with no address field. It pops the two top data from the stack, subtracting the data, and pushing the result into the stack at the top. 
  • PDP-11, Intel’s 8085, and HP 3000 are some examples of stack-organized computers. 

The advantages of Stack-based CPU organization – 

  1. Efficient computation of complex arithmetic expressions. 
  2. Execution of instructions is fast because operand data are stored in consecutive memory locations. 
  3. The length of instruction is short as they do not have an address field. 

The disadvantages of Stack-based CPU organization –

  • The size of the program increases. 

MPPGCL JE Electronics Mock Test - 7 - Question 20

LAN Topology contains

Detailed Solution: Question 20

LAN Topology contains Bus, star, and rings.

Key Points

A star topology is a topology for a Local Area Network (LAN) in which all nodes are individually connected to a central connection point, like a hub or a switch.

In networking a bus is the central cable i.e. the main wire that connects all devices on a local area network (LAN).

A ring topology is a topology for a Local Area Network (LAN) in which every device has exactly two ​neighbors for communication purposes.

MPPGCL JE Electronics Mock Test - 7 - Question 21

Which of the following pairs of friends are good, both in English and Science?

Detailed Solution: Question 21

Four friends W, X, Y and Z are students of Class 10th.

i) W and X are good in Hindi but poor in English.

ii) W and Y are good in Science but poor in Mathematics.

iii) Y and Z are good in English but poor in Social Studies.

iv) Z and X are good in Mathematics as well as in Science.

Thus we can see Y and Z pairs of friends are good, both in English and Science

Hence, the correct answer is "Y and Z".

MPPGCL JE Electronics Mock Test - 7 - Question 22

DirectionsIn questions given below out of four alternatives, choose the one which can be substituted for the given word/sentence.

Q. State in which the few govern the many.

Detailed Solution: Question 22

One word substitution is Oligarchy.

Monarchy: A form of government with a monarch at the head.

Oligarchy: A small group of people having control of a country or organization.

Plutocracy: A state or society governed by the wealthy.

Autocracy: A system of government by one person with absolute power.

MPPGCL JE Electronics Mock Test - 7 - Question 23

“Michelle, I’m very upset about what you said,” Rosie stated firmly. “You didn’t come right out and accuse me of eating all your candy. On the other hand, your remarks about my love of chocolate were made for a reason. You are insinuating that I’m the candy thief!” What does “insinuate” mean?

Detailed Solution: Question 23

Insinuate means suggest or hint (something bad) in an indirect and unpleasant way
Some insinuate that he doesn't love India enough to defend it wholeheartedly
It's degrading for people like Romney to insinuate that they don't contribute

MPPGCL JE Electronics Mock Test - 7 - Question 24

Find out the Synonym of the following word:

FAKE

Detailed Solution: Question 24

  • Meaning of Fake: Not genuine, imitation or counterfeit.
  • Meaning of Original: present or existing from the beginning; first or earliest
  • Meaning of Trustworthy: able to be relied on as honest or truthful
  • Meaning of Loyal: giving or showing firm and constant support or allegiance to a person or institution

MPPGCL JE Electronics Mock Test - 7 - Question 25

Directions: A sentence is given here with a blank and you need to fill the blank choosing the word/words given below. If all the words given can fill the blank appropriately, choose ‘All are correct’ as your answer.

For the first time in its _______________, Oxford Dictionaries allowed the public to vote for ‘the Word of the Year for 2022’.

I. History
II. Course
III. Age

Detailed Solution: Question 25

The idea is to express that it is the first time that Oxford has allowed public to vote. Thus, only ‘history’ is the best fit while ‘course’ and ‘age’ are incorrect.

Hence, the correct answer is option D.

MPPGCL JE Electronics Mock Test - 7 - Question 26

Directions: Each of the following consists of a question and two statements numbered I and II given below it. You have to decide whether the data provided in the statements are sufficient to answer the question.

Which statement is required to derive the conclusion - 'No Player is a Champion'?
Statement I:
All Stars are Goalkeepers.
Statement II: No Champion is a Player.

Detailed Solution: Question 26

Checking Statement I:

Since there is no data given about Player and Champion, therefore, No player is Champion can not be derived.

Clearly, Statement I alone is not sufficient to answer the question.

Checking Statement II:

Statement II is the converse statement of the required conclusion.

Clearly, Statement II alone is sufficient to answer the question.

Hence, Option B is correct.

MPPGCL JE Electronics Mock Test - 7 - Question 27

About how many ministers are there in the Cabinet?

Detailed Solution: Question 27

In the 2024 Council of Ministers, there are 30 Cabinet Ministers, 5 Ministers of State (Independent Charge) and 36 Ministers of State.

MPPGCL JE Electronics Mock Test - 7 - Question 28

In this questions, a number series is given with one term missing. Choose the correct alternative that will continue the same pattern and fill in the black spaces.

Q. 1, 4, 9, 16, 25, (____)

Detailed Solution: Question 28

1st term = 12 = 1
2nd term = 22 = 4
3rd term = 32 = 9
4th term = 42 = 16
5th term = 52 = 25
Thus, Missing number = 6= 36

MPPGCL JE Electronics Mock Test - 7 - Question 29

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

Vanity

Detailed Solution: Question 29

The word Vanity (Noun) means : too much pride in your own appearance, abilities or achievements; arrogance.
The word Humility (Noun) means: the quality of being humble. 

MPPGCL JE Electronics Mock Test - 7 - Question 30

Directions: In each of the following questions, a sentence has been given in Active (or Passive) Voice. Out of the four alternatives suggested, select the one that best expresses the same sentence in Passive/ Active Voice.

Darjeeling grows tea.

Detailed Solution: Question 30

The original sentence "Darjeeling grows tea" is in the active voice.

The correct option that expresses the same sentence in the passive voice is:

D. Tea is grown in Darjeeling.

View more questions
10 tests
Information about MPPGCL JE Electronics Mock Test - 7 Page
In this test you can find the Exam questions for MPPGCL JE Electronics Mock Test - 7 solved & explained in the simplest way possible. Besides giving Questions and answers for MPPGCL JE Electronics Mock Test - 7, EduRev gives you an ample number of Online tests for practice
Download as PDF