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MPPGCL JE Electronics Mock Test - 8 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test - MPPGCL JE Electronics Mock Test - 8

MPPGCL JE Electronics Mock Test - 8 for Electronics and Communication Engineering (ECE) 2025 is part of Electronics and Communication Engineering (ECE) preparation. The MPPGCL JE Electronics Mock Test - 8 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The MPPGCL JE Electronics Mock Test - 8 MCQs are made for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MPPGCL JE Electronics Mock Test - 8 below.
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MPPGCL JE Electronics Mock Test - 8 - Question 1

Which of the following options about tunnel diodes is INCORRECT?

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 1

Tunnel Diode:

  • Symbol:

  • It is a highly doped PN Junction diode, used for low voltage high-frequency switching applications.
  • It works on the tunneling principle.
  • It has a high charge carrier velocity.
  • When compared to a normal p-n junction diode, the tunnel diode has a narrow depletion width.
  • In normal forward-biased operation, it exhibits the “Negative resistance region” as shown:

  • The negative differential resistance in their operation, allows them to be used as oscillators, amplifiers, and switching circuits.
  • Their low capacitance allows them to function at microwave frequencies.
  • Microwave frequencies range between 109 Hz (1 GHz) to 1000 GHz
  • Tunnel diodes are not good rectifiers, as they have relatively high leakage current when reverse biased.
MPPGCL JE Electronics Mock Test - 8 - Question 2

A MOSFET carries a drain current of 1 mA with VDS = 0.5 V in saturation. If channel length modulation parameters λ = 0.1 V-1. Then the drain current if VDS is 1V is ______ 

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 2

Concept:

Drain current in saturation region is given by

Calculation:

= 1.047 mA

MPPGCL JE Electronics Mock Test - 8 - Question 3

The term control system means:

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 3

Control system:

A Control system is defined as a device that manages, commands direct, or regulates the behavior of other devices or systems to achieve the desired result.

It is also defined as A system with a provision for controlling the response.

There are two types of control systems.

Open-loop control system:

The open-loop control system can be described by a block diagram as shown in the figure below.

These are the systems in which the control action is independent of output.

Ex: Traffic signals, washing machine and bread toaster, systems having no sensor, etc.

Advantages:

  • Simple construction and design
  • Cost is less
  • Maintenance is easy
  • No problem of instability
  • Convenient to measure when the output is difficult to measure

Disadvantages:

  • These are less accurate, and their accuracy depends on the calibration.
  • Inaccurate results are obtained with parameter variations within the system.
  • Recalibration of the controller is required from time to time to maintain accuracy.

Closed-loop control system:

The open-loop control system can be described by a block diagram as shown in the figure below.

These are the systems in which the control action depends on the output. These systems have a tendency to oscillate.

Ex: Temperature controllers, speed control of the motor, systems having sensors, etc.

Advantages:

  • More accurate
  • Non-linear distortions are less
  • Output is less sensitive to parameter changes within the system
  • Bandwidth increases

Disadvantages:

  • Design is complicated
  • More expensive
  • May become unstable, if there are malfunctions in the feedback.
MPPGCL JE Electronics Mock Test - 8 - Question 4

At cut-off frequency, the phase velocity ‘f’ of a waveguide is:

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 4

Concept:

Phase velocity of a Rectangular waveguide is defined as:

Also,  where θ is the angle with which the wave enters the waveguide as shown:

Calculation:

Given, f = fc

So, 

Implication:

vp = ∞,

means that cos θ = 0, i.e. θ = 90°.

This situation is shown as below:

This implies that at f = fc, the wave will oscillate between the walls as shown.

MPPGCL JE Electronics Mock Test - 8 - Question 5

Find the source voltage VGS for the given NMOS FET with parameters . Assume that the MOSFET is not operating in the cutoff mode, and also assume an ideal MOSFET with gate current equal to zero.

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 5

Concept:

For a MOSFET in saturation, the current Equation is given as:

For a MOSFET in active mode, the current equation is given as:

Calculation:

Let us assume that the MOSFET is in saturation:

i.e. VDS ≥ VGS – Vth

Given, Vth = 1 V

 (Voltage Division Rule)

So, ID(sat) = 125 μ (5 – Vs - 1)2

⇒ ID(sat) = 12.5 μ (4 - Vs)2

Also, Vs = iD(sat) Rs

So, ID(sat) = 12.5 μ (4 – iD(sat).50K)2

⇒ iD(sat) = 12.5μ (16 + i2D(sat).2500 × 106 – 400 × 103 iD(sat))

⇒ iD(sat) = 12.5 μ (16 + i2D(sat) 2500 × 106 – 4 × 105 iD(sat))

Solving the above quadratic equation,

⇒ iD(sat) = 0.043 mA or 0.1489 mA

For iD = 0.1489 mA

VD = 10 – 0.1489 m × 75 K

= -1.167 V

Since, the MOSFET is not operating in the cut off mode, VD = -1.167 V is not possible.

So, iD(sat) = 0.043 mA, for which,

VD = 10 – 0.043 m × 75 K

= 6.775 V

Vs = 0.043 m × 50 K

= 2.15 V

VDS = 6.775 – 2.15 = 4.625

VGS = 5 – 2.15 = 2.85

Since, VDS > VGS – Vt, (our assumption is correct)

So, Vs = 2.15 V

VGS = Vs = 5 – 2.15 = 2.85 V

MPPGCL JE Electronics Mock Test - 8 - Question 6

The bandwidth of AM broadcasting channel is equal to _____

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 6

The frequency spectrum of an Amplitude Modulated wave is given by:

Upper sideband = fc + fm

Lower sideband = fc - fm

Bandwidth in AM = Upper sideband - Lower sideband

Bandwidth = (fc + fm) - (fc - fm) = 2fm

∴ In AM the bandwidth is equal to twice the maximum frequency component.

If sideband's highest frequency is fm, Bandwidth is 2fm

MPPGCL JE Electronics Mock Test - 8 - Question 7
In position control systems, the device used for providing rate – feedback voltage is called
Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 7
In position control systems, the device is used for providing rate feedback voltage is called tachogenerator.
MPPGCL JE Electronics Mock Test - 8 - Question 8

Determine the range of values of Vi that will maintain the Zener diode of figure in the 'ON' state -

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 8

Concept:

A Zener diode is a silicon semiconductor device that permits current to flow in either a forward or reverse direction. The diode consists of a special, heavily doped p-n junction, designed to conduct in the reverse direction when a certain specified voltage is reached.

Voltage Vz: The Zener voltage refers to the reverse breakdown voltage.

Current Iz (max.): Maximum current at the rated Zener voltage Vz.

Current Iz (min.): Minimum current required for the diode to break down.

Calculation:

Open circuit the zener diode to calculate the voltage appearing across the diode.

Vx = Vi × 1.2 K/( 1.2 K + 0.22 K)

To operate in zener breakdown region:  

Vi(min) × 1.2 K/( 1.2 K + 0.22 K) > 20 V

Vi(min) > [20( 1.2 K + 0.22 K)]/1.2 K 

Vi(min) > 23.66 V

Current through the zener diode will be maximum if maximum possible input voltage is applied.

⇒ IR = IZmax + IL

⇒ IR = 60mA + (VL/RL) [∵ VL = VZ = 20 V]

⇒ I= 60mA + (20/1.2 K)

⇒ I= 76.66 mA

(Vi(max) - VZ)/ 0.22K = IR

⇒ (Vi(max) - 20)/ 0.22K = 76.66

⇒ Vi(max) = 36.87 V

MPPGCL JE Electronics Mock Test - 8 - Question 9

Which is the sequence for PCM?

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 9

PCM is a technique by which an analog signal gets converted into digital form to have signal transmission through a digital network.

The major steps involved in PCM is sampling, quantizing, and encoding.

Low Pass Filter:ccc

  • Here, the message signal which is the continuous-time form is allowed to pass through a low pass filter (LPF).
  • This LPF whose cutoff frequency is fm eliminates the high-frequency components of the signal and passes only the frequency components that lie below fm.

Sampler:

  • The output of the LPF is then fed to a sampler where the analog input signal is sampled at regular intervals.
  • This sampling frequency is so selected that it must follow the sampling theorem that is expressed as:

          

  • The output of the sampler is a signal that is discrete-time continuous amplitude signal denoted as nTs which is nothing but a PAM signal.

Quantizer:

  • A quantizer is a unit that rounds off each sample to the nearest discrete level.
  • The Sampler provides a continuous range signal and hence still an analog one.
  • The quantizer performs the approximation of each sample thus assigning it a particular discrete level.

Encoder:

  • An encoder performs the conversion of the quantized signal into binary codes.
  • This unit generates a digitally encoded signal which is a sequence of binary pulses that acts as the modulated output.
  • At it is a binary encoder thus generates a binary code sequence. That is transmitted through the transmission path.

Comparator is not used in pulse code modulation.

MPPGCL JE Electronics Mock Test - 8 - Question 10

Two point charges exert on each other a force F, when they are placed r distance apart in air. When they are placed R distance apart in medium of dielectric constant K, they exert the same force. The distance R equals

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 10

The correct answer is 

Key Points

  • Given two point charges exert a force F when placed r apart in constant K, the force between them is the same. We have to find R.
  • let q1 and q2 be two point charges.
  • 1/4πE0q1*q2/r2 = 1/4πE0q1*q2/K*R2 (where E0= permittivity of air and K = dielectric constant)
  • 1/r2 = 1/K*R2
  • R= 
MPPGCL JE Electronics Mock Test - 8 - Question 11

Which of the following statements are true?

(a) Advanced Mobile Phone System (AMPS) is a second-generation cellular phone system.

(b) IS - 95 is a second generation cellular phone system based on CDMA and DSSS.

(c) The Third generation cellular phone system will provide universal personnel communication.

Code:
Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 11

Statement a:

Advanced Mobile Phone System (AMPS) is a second-generation cellular phone system.

This is statement incorrect

Reason:

  • An advanced mobile phone system is an analog cellular system. It uses FDMA to separate channels in a link. It operates in ISM 800 MHz band.
  • It uses two separate analog channels one for forwarding communication and one for reverse communication.
  • It is a first-generation cellular phone system. So, the given statement is incorrect.

Statement b:

IS - 95 is a second-generation cellular phone system based on CDMA and DSSS.

This statement is correct

Reason:

  • IS – 95 is a second-generation cellular phone system.
  • It is based on CDMA and DSSS. It uses two bands for duplex communication. IS 95 has two different transmission techniques: one for use in the forward direction and another in a reverse direction.
  • In this, each voice channel is digitized. It uses FDMA to separate channels in the link.

Statement c:

The Third-generation cellular phone system will provide universal personnel communication.

This statement is correct

Reason:

The third generation of cellular telephony refers to a combination of technologies that provide a variety of services. Using a small portable device, a person should be able to talk to anyone else in the world with a voice quality similar to that of an existing fixed telephone network. It provides universal personal communication.
MPPGCL JE Electronics Mock Test - 8 - Question 12
Which of the following devices forwards data packets to all connected ports?
Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 12

Hubs:

Hubs are used in networks that use twisted-pair cabling to connect devices.

Hubs can also be joined together to create larger networks.

Hubs are simple devices that direct data packets to all devices connected to the hub, regardless of whether the data package is destined for the device. This makes them inefficient devices and can create a performance bottleneck on busy networks.
MPPGCL JE Electronics Mock Test - 8 - Question 13
Program counter for any counter:
Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 13

Program counter for any counter Stores the address of the next instruction to be executed.

Additional Information

Program counter:

  • It is a 16-bit register as 8085 has 16 address lines. The program counter is updated by the processor and points to the next instruction after the processor has fetched the complete instruction.
  • A program counter is used to keep track of the execution of the program. It contains the memory address of the next instruction to be fetched.
  • The program counter contains the address of the next instruction to be fetched from the main memory when the previous instruction has been successfully completed.
  • The program counter also functions to count the member instruction.
  • As each instruction gets fetched, the program counter increases its stored value by 1.
MPPGCL JE Electronics Mock Test - 8 - Question 14

Mobile telephone is

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 14

Content: 

Mobile telephony is the most common mode of communication these days. A mobile phone is like a handy computer equipped with the internet. It has greatly changed our lives. The operation frequency range of telephone is Ultra High-Frequency range. The frequency of outgoing signals and incoming signals is not the same. This prevents intermixing of signals and makes simultaneous communication possible. It works both as a transmitter as well as a receiver which can both send and receive signals in the radio frequency range. 

The most common applications of mobile telephony are-

  • Receiving and sending signals (calling facility)
  • Taking photos and videos
  • Sending text messages
  • Playing games
  • receiving e-mails
  • Listening to music and so on.

Explanation:

A mobile phone is low powered transceiver i.e., a combination of both transmitter and receiver. It can wirelessly send and receive signals.

The correct answer is option (3)

MPPGCL JE Electronics Mock Test - 8 - Question 15

Which of the following are the advantages of FM over AM?

1. Better noise immunity is provided

2. Lower bandwidth is required

3. Transmitted power is more useful

4. Less modulating power is required

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 15

Advantages of FM over AM are:

  • Improved signal to noise ratio (about 25dB) w.r.t. to manmade interference.
  • Smaller geographical interference between neighboring stations.
  • Less radiated power.
  • Well defined service areas for given transmitter power.
  • The transmitted power can be used only for significant sidebands. This helps in less power distributed to the carrier and the majority of the power to the sidebands that are actually transmitting the power. This results in an efficient use of power. (Statement 3 is correct)

Disadvantages of FM:

  • Much more Bandwidth (as much as 20 times as much). The theoretical bandwidth requirement is infinite.
  • More complicated receivers and transmitters required.

Other important differences are elaborated in the table below:

MPPGCL JE Electronics Mock Test - 8 - Question 16

Frequency Shift Keying is used mostly in

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 16

FSK (Frequency Shift Keying):

It is used in the voice frequency telegraph system and for wireless telegraphy in the high-frequency bands. 

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) = A (cos 2π fHt)

For binary ‘0’ → S2 (t) = A (cos 2π fLt)

 The constellation diagram is as shown:

∴ Option 2 is the most appropriate. 

Notes:

Radio transmission: The two most common types of modulation used in radio are amplitude modulation (AM) and frequency modulation (FM).

Telephony: The type of modulation used in digital telephony is Pulse code modulation (PCM). 

Television: For television broadcasting, Vestigial sideband modulation is used for video transmission and Frequency modulation is used for audio transmission.

MPPGCL JE Electronics Mock Test - 8 - Question 17

MAC destination address field in an Ethernet Frame specified in IEEE 802.3 consist of ____ bit and ____ byte?

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 17

Layer 2 Ethernet Frame Format:

MAC address consists of 6 bytes

1 byte = 8 bits

∴ 6 bytes = 6 × 8 bits = 48 bits

MAC address has 48 bit or 6 bytes

MPPGCL JE Electronics Mock Test - 8 - Question 18
For a linear block code, it is known that the code can detect 2 error and can correct only 1 error. The minimum hamming distance for the code is ________.
Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 18

Concept:

Let, s = No of error that can be detected

t = No of error that can be corrected

dmin = Minimum hamming distance for the code

Then,

To detect up to 's' errors

1) dmin ≥ s + 1

To correct up to 't' errors

2) dmin ≥ 2t + 1

it can detect up to 's' errors and correct up to 't' errors if and only if:

3) dmin ≥ t + s + 1

Calculations:

Given:

Error detected = 2 = s

Error corrected = 1 = t

From(1),

dmin ≥ 3

From(2),

dmin ≥ 3

We need to find for both correction and detection and hence, 

Using formula (3)

dmin ≥ t + s + 1

dmin ≥ 2 + 1 + 1

dmin = 4
MPPGCL JE Electronics Mock Test - 8 - Question 19
Instruction pipelining improves CPU performance due to
Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 19

Concept:

  • The main goal of pipelining is to balance the length of each pipelined stage.
  • If the stages are perfectly balanced, then the time for instruction on the pipelined machine is reduced.
  • Pipelining does not decrease the time for individual instruction execution. Instead, it increases instruction throughput.
  • This is achieved by using the processor hardware efficiently.
MPPGCL JE Electronics Mock Test - 8 - Question 20
Which type of bridge is not suitable for measuring the quality factor of coil? 
Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 20

The correct answer is option 2.

Maxwell’s inductance bridge is used for the measurement of the self-inductance of the circuit.

The quality factor is measured for a coil, which is the combination of inductance and capacitance.

Hence, it is measured by the Maxwell Inductance Capacitance bridge.

Measurement of resistance, inductance, and capacitance

MPPGCL JE Electronics Mock Test - 8 - Question 21
In which of the following values are period and frequency, respectively, specified in a CRO measurement?
Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 21

Cathode Ray Oscilloscope (CRO)

The cathode ray oscilloscope is an electronic test instrument.

It is used to obtain waveforms when different input signals are given to the CRO.

Thus CRO measures the time period and the frequency of the obtained waveform.

Time period

The number of seconds taken by a signal to complete its one cycle is known as the time period.

Therefore time period is nothing but the seconds/cycle of a waveform.

Frequency

The number of cycles completed in one second is known as the frequency.

Therefore frequency is nothing but the cycle/seconds of a waveform.

The relationship between time and frequency is:

where, T = Time period

f = Frequency

MPPGCL JE Electronics Mock Test - 8 - Question 22

A logic family shows the following values:

VOH = 5 V, VOL = 1 V, VIH = 3 V, and VIL = 2 V.

The noise margins, NMhigh and NMlow, respectively, will be:

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 22

Concept:

  • In a digital circuit, the Noise Margin is the amount by which the signal exceeds the threshold for a proper ‘0’ or ‘1’.
  • For Ex: a Digital circuit might be designed to swing between 0 and 1.2 Volts, with anything below 0.2 V considered as a ‘0’ and anything above 1 Volt is considered a ‘1’. Then the noise margin for a ‘0’ would be the amount that a signal is below 0.2 Volts, and a noise margin for 1 would be the amount by which a signal exceeds 1 Volt.
  • In this case noise margins are measured as an absolute voltage, not as a ratio.
  • This is schematically explained with the help of the following diagram:

NMlow = VIL – VOL

NMhigh = VOH - VIH

Calculation:

Given voltage levels are VIL = 2 V  VOL = 1 V, VOH = 5 V, VIH = 3 V

NMlow = VIL – VOL = 2 V – 1 V

NMlow = 1 V

NMhigh = VOH - VIH = 5 V – 3 V

NMhigh = 2 V

So option(1) is the correct answer.

MPPGCL JE Electronics Mock Test - 8 - Question 23

Direction: In this question, you need to replace the underlined part of the sentence with the most suitable idiom / expression given as option.

He said that he was a huge fan of the president, although I suspect it was a joke.

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 23

Tongue in cheek: a humorous or sarcastic statement expressed in a serious manner; in an ironic, flippant, or insincere way.
Above someone’s station: higher than suitable for position or rank.
Agitate the gravel: to get angry.
Be thin on the ground: exist in small numbers or amounts.
Hence, the correct answer is option A. 

MPPGCL JE Electronics Mock Test - 8 - Question 24

Direction: Choose the option closest in meaning to the OPPOSITE of the word given in capitals.

OSSIFIED

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 24

To ossify is to harden or become inflexible, hence adaptable is an antonym.

MPPGCL JE Electronics Mock Test - 8 - Question 25

Find out the Synonym of the following word:

WARRIOR

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 25
  • Meaning of Warrior: A brave or experienced soldier or fighter.
  • Meaning of Sailor: a person who goes sailing as a sport or recreation
  • Meaning of Pirate: a person who attacks and robs ships at sea
  • Meaning of Spy: a person employed by a government or other organization to secretly obtain information on an enemy or competitor
MPPGCL JE Electronics Mock Test - 8 - Question 26

Give synonym of the word REPEAL

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 26

Repeal: To revoke or annul a law, regulation, or decision, typically through official or legislative means.

  1. Sanction:

    • As a verb: to give official permission or approval for an action.
    • As a noun: a penalty or measure taken against someone who has violated a rule or law.
  2. Perpetuate: To make something continue indefinitely; to sustain or prolong.

  3. Pass:

    • As a verb: to approve or adopt a proposal or legislation.
    • As a noun: a successful attempt to achieve something; also, a document or ticket granting authorization or access.
  4. Cancel: To revoke or annul something previously scheduled, arranged, or agreed upon; to declare null and void.

Correct option D

MPPGCL JE Electronics Mock Test - 8 - Question 27

What is the minimum age required to contest an election to Lok Sabha?

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 27

Article 84 (b) of Constitution of India provides that the minimum age for becoming a candidate for Lok Sabha election shall be 25 years.

MPPGCL JE Electronics Mock Test - 8 - Question 28

In the following question number of triangle are

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 28

 

The main triangle shown is in the given figure and this the total no. of triangle is 15. remaing triangle we can find out in the drawing the triangle in the image.

MPPGCL JE Electronics Mock Test - 8 - Question 29

They felt safer

P: to watch the mountain
Q: of more than five miles
R: as they settled down
S: from a distance

The Proper sequence should be:

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 29

The correct order will be:
They felt safer → as they settled down  (R)→ to watch the mountain (P)→ from a distance (S)→ of more than five miles.(Q)

MPPGCL JE Electronics Mock Test - 8 - Question 30

Directions: In the following question, a sentence is given with a blank. You have to fill the blank with one of the words given as options in order to make the sentence contextually and grammatically correct.

Human monkey pox was first identified in 1970 and named after the disease caused by the virus was _______________ in captive monkeys more than a decade before.

Detailed Solution for MPPGCL JE Electronics Mock Test - 8 - Question 30

The segment is in passive voice, thus only a verb in third form is eligible for the blank. Among all the given options, only ‘discovered’ fits the blank as the virus already existed in the monkeys but was found in 1970.

Hence, option D is correct.

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