MPPGCL JE Electronics Mock Test - 3 Free MCQ Practice Test with Solutions - Electronics and Communication Engineering (ECE) (2024)

In the case of optical fiber transmission link, LED’s and LASER’s are used as ______

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In the case of optical fiber transmission link, LED’s and LASER’s are used as ______

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

Detectors

", "comment": { "@type": "Comment", "text": "

In the case of optical fiber transmission link, LED’s and LASER’s are used as ______

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

Sources

", "comment": { "@type": "Comment", "text": "

In the case of optical fiber transmission link, LED’s and LASER’s are used as ______

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

Repeaters

", "comment": { "@type": "Comment", "text": "

In the case of optical fiber transmission link, LED’s and LASER’s are used as ______

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

Amplifiers

", "comment": { "@type": "Comment", "text": "

In the case of optical fiber transmission link, LED’s and LASER’s are used as ______

" }}],"acceptedAnswer": { "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "b", "comment": { "@type": "Comment", "text": "

In the case of optical fiber transmission link, LED’s and LASER’s are used as ______

" }, "answerExplanation": { "@type": "Comment", "text": "

Being a single wavelength light source with uniform phase:

Laser light travels smoothly with verylittle dispersion.
Makes itidealfor long-distance communication.
For fiber optics with glass fibers, light in the infrared region is widely used which has wavelengths longer than visible light, typically of 850, 1300, and 1550 nm.
Transmitters (Lasers or LEDs) emit in infrared regions and are used as a light source for fiber transmission.
The laser acts as the best source in optical communication becauseit is a single wavelength light source.
Ordinary light contains many different wavelengths of light, differences emerge in the speed of the transmission, reducing the number of signals that can be transmitted in any set time.
Receivers (photodetectors) at these particular wavelengths are also easy to construct.

Notes:

We use infrared because the attenuation of the fiber is much less at those wavelengths.
The attenuation of glass optical fiber is caused by two factors, absorption, and scattering.
Absorption occurs in several specific wavelengths called water bands due to the absorption by minute amounts of water vapor in the glass.

." }}},{"@type": "Question","typicalAgeRange": "4-35","educationalLevel": "intermediate","eduQuestionType": "Multiple choice","learningResourceType": "Practice problem","name": "Recollect the concept of MPPGCL JE Electronics Mock Test - 3 to solve the ques", "comment": { "@type": "Comment", "text": "‘Bluetooth' technology is" },"encodingFormat": "text/html","text": "‘Bluetooth' technology is","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "Satellite communication", "comment": { "@type": "Comment", "text": "‘Bluetooth' technology is" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "Wired communication between devices", "comment": { "@type": "Comment", "text": "‘Bluetooth' technology is" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "Wireless communication between devices", "comment": { "@type": "Comment", "text": "‘Bluetooth' technology is" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "Communication between landline to mobile phone", "comment": { "@type": "Comment", "text": "‘Bluetooth' technology is" }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "‘Bluetooth' technology is" }, "answerExplanation": { "@type": "Comment", "text": "

The correct answer isWireless communication between devices.

Key Points

Bluetoothtechnology

Bluetooth is a Wireless Personal Area Network (WPAN) technology and is used for exchanging data over smaller distances.
Bluetooth technology allows devices to communicate with each other without cables or wires. Bluetooth relies on short-range radio frequency, and any device that incorporates the technology can communicate as long as it is within the required distance.
Bluetooth is used for short-range communications between device peers, as well as a device to the peripheral. The number and variety of peripherals that communicate via Bluetooth are immense—wireless headsets for hands-free cell phone use, keyboards, mice, videogame controllers, audio speakers, you name it.

The condition for the low-loss transmission line is

" },"encodingFormat": "text/html","text": "

The condition for the low-loss transmission line is

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

R >>ωC and G >>ωL

", "comment": { "@type": "Comment", "text": "

The condition for the low-loss transmission line is

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

R <<ωL and G <<ωC

", "comment": { "@type": "Comment", "text": "

The condition for the low-loss transmission line is

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

The condition for the low-loss transmission line is

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

RG >> LC

", "comment": { "@type": "Comment", "text": "

The condition for the low-loss transmission line is

" }}],"acceptedAnswer": { "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "b", "comment": { "@type": "Comment", "text": "

The condition for the low-loss transmission line is

" }, "answerExplanation": { "@type": "Comment", "text": "

Concept:

Propagation constant for the transmission line is given by:

For lossless line α = 0

If R ≪ ωL and a ≪ ωC, the propagation constant becomes:

---(1)

Comparing equation (1) with α + jβ:

α = 0. Hence the line is lossless.

Important points:

A transmission line is also lossless when:

R = 0 and G = 0.

Here also α = 0 and the line is lossless

∴ The conditions for lossless line includes:

1) R = 0; G = 0, or

2) ωL ≫ R & ωC ≫ G

If [A] Matrix is Incidence matrix then which one of the following is true?

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If [A] Matrix is Incidence matrix then which one of the following is true?

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

|A| = 0 (For closed loop)

", "comment": { "@type": "Comment", "text": "

If [A] Matrix is Incidence matrix then which one of the following is true?

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

Adj [A] / |A| = 0 (For closed loop)

", "comment": { "@type": "Comment", "text": "

If [A] Matrix is Incidence matrix then which one of the following is true?

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

[A] = 1 (For closed loop)

", "comment": { "@type": "Comment", "text": "

If [A] Matrix is Incidence matrix then which one of the following is true?

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

|A| = 1 (For closed loop)

", "comment": { "@type": "Comment", "text": "

If [A] Matrix is Incidence matrix then which one of the following is true?

" }}],"acceptedAnswer": { "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "a", "comment": { "@type": "Comment", "text": "

If [A] Matrix is Incidence matrix then which one of the following is true?

" }, "answerExplanation": { "@type": "Comment", "text": "

Incidence matrix:

It is the matrix that gives a relation between the branches and nodes.
The rows of the incidence matrix [A] represent the number of nodes and the column of the matrix represents the number of branches in the given graph.
If there are‘n’ number of rows in a given incidence matrix[A], that means in a graph there are ‘n’ number of nodes.
Similarly, if there are‘b’ numbers of columns in that given incidence matrix[A], that means in that graph there are ‘b’ number of branches.
We can construct the incidence matrix for the directed graph. We can draw a graph with the help of the incidence matrix.
The algebraic sum of elements of all the columns is zero.
The rank of the incidence matrix is (n–1).
The determinant of the incidence matrix of a closed loop is zero.

Hence, an Only statement I is correct.

Example:

Consider the below-closed graph:

The incidence matrix is written as:

We observe that the algebraic sum of elements of all the columns is zero.

The determinant of the given closed path is calculated as:

∴ The determinant of the incidence matrix of a closed loop is zero.

If for a BJT the value ofα is 0.98, then the value ofβ will be _______.

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If for a BJT the value ofα is 0.98, then the value ofβ will be _______.

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

If for a BJT the value ofα is 0.98, then the value ofβ will be _______.

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

If for a BJT the value ofα is 0.98, then the value ofβ will be _______.

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

101

", "comment": { "@type": "Comment", "text": "

If for a BJT the value ofα is 0.98, then the value ofβ will be _______.

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

120

", "comment": { "@type": "Comment", "text": "

If for a BJT the value ofα is 0.98, then the value ofβ will be _______.

" }}],"acceptedAnswer": { "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "a", "comment": { "@type": "Comment", "text": "

If for a BJT the value ofα is 0.98, then the value ofβ will be _______.

" }, "answerExplanation": { "@type": "Comment", "text": "

Concept:

Where,

β = common-emitter current gain

α = Common base current gain

Calculation:

Common base current gain = α = 0.98

Note:

Where I_C= Collector current

I_E= Emitter current

I_B= Base current

Important Points

The DC current gain of a common collector is therefore given by the ratio of emitter current to the base current, i.e.

I_E= Emitter Current

I_B= Base Current

Also, I_E= I_B+ I_C

DC current gain will be:

The DC current gain for a common emitter configuration is defined as:

Equation (1) now becomes:

Important Differences between different transistor configuration is as shown:

Overdamping will make instruments:

" },"encodingFormat": "text/html","text": "

Overdamping will make instruments:

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

robust

", "comment": { "@type": "Comment", "text": "

Overdamping will make instruments:

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

accurate

", "comment": { "@type": "Comment", "text": "

Overdamping will make instruments:

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

fast

", "comment": { "@type": "Comment", "text": "

Overdamping will make instruments:

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

slow

", "comment": { "@type": "Comment", "text": "

Overdamping will make instruments:

" }}],"acceptedAnswer": { "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "d", "comment": { "@type": "Comment", "text": "

Overdamping will make instruments:

" }, "answerExplanation": { "@type": "Comment", "text": "

Concept:

Classification of instruments on the basis of damping:

1.) Overdamping:

The value of the damping ratio is greater than 1.
In case of overdamping, the instrument will become slow and lethargic and rise very slowly from its zero position to a steady state position.

2.) Undamped:

The value of the damping ratio is equalto 0.
In this case, the instrumentnever settles down and always oscillates about its mean position.

3.) Under-damped:

The value of the damping ratio lies between 0 to 1.
In this case, the system initially oscillates about its mean position but settles down to a steady state.

4.) Critical-damping:

The value of the damping ratio is equalto 1.
This damping returns the system to equilibrium as fast as possible without any oscillations.

." }}},{"@type": "Question","typicalAgeRange": "4-35","educationalLevel": "intermediate","eduQuestionType": "Multiple choice","learningResourceType": "Practice problem","name": "Recollect the concept of MPPGCL JE Electronics Mock Test - 3 to solve the ques", "comment": { "@type": "Comment", "text": "The ALU uses ______ to store intermediate result." },"encodingFormat": "text/html","text": "The ALU uses ______ to store intermediate result.","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "Cache", "comment": { "@type": "Comment", "text": "The ALU uses ______ to store intermediate result." }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "Registers", "comment": { "@type": "Comment", "text": "The ALU uses ______ to store intermediate result." }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "Accumulators", "comment": { "@type": "Comment", "text": "The ALU uses ______ to store intermediate result." }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "Stack", "comment": { "@type": "Comment", "text": "The ALU uses ______ to store intermediate result." }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "The ALU uses ______ to store intermediate result." }, "answerExplanation": { "@type": "Comment", "text": "

Concept:

ALU (Arithmetic logic unit) is a circuit to perform arithmetic and logic operations.

Explanation:

ALU uses accumulator (a register) for storing intermediate results of arithmetic and logic operations.
Control unit provides the data to the ALU and directs the ALU to perform specific operations.
ALU stores the result in accumulator which is a type of register. Accumulator is a kind of temporary storage device which stores the intermediate results and if some previous results are already stored on the accumulator then it automatically overwrites the new results and previous results are removed.

." }}},{"@type": "Question","typicalAgeRange": "4-35","educationalLevel": "intermediate","eduQuestionType": "Multiple choice","learningResourceType": "Practice problem","name": "Recollect the concept of MPPGCL JE Electronics Mock Test - 3 to solve the ques", "comment": { "@type": "Comment", "text": "Radar principle is used in" },"encodingFormat": "text/html","text": "Radar principle is used in","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "detection of aircraft", "comment": { "@type": "Comment", "text": "Radar principle is used in" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "telephony", "comment": { "@type": "Comment", "text": "Radar principle is used in" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "electron microscope", "comment": { "@type": "Comment", "text": "Radar principle is used in" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "all of the above", "comment": { "@type": "Comment", "text": "Radar principle is used in" }}],"acceptedAnswer": { "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "a", "comment": { "@type": "Comment", "text": "Radar principle is used in" }, "answerExplanation": { "@type": "Comment", "text": "

The radar principle is used in thedetection of aircraft.

Radar:

Radar is an acronym for Radio Detection and Ranging.
The term 'radio'refers to the use of electromagnetic waves with wavelengths in the so-calledradiowave portion of the spectrum, which covers a wide range from 104km to 1 cm.
It is basically an electromagnetic system used to detect the location and distance of an object from the point where the RADAR is placed.
Itworks by radiating energy into space and monitoring the echo or reflected signal from the objects.
It works on radio frequency in the range of about 3 kHz to 300 GHz.
But most of the RADAR operates between 220 MHz to 35 GHz.

Radar systems have been used in military applications for

Ground surveillance
Missile control, fire control,
Air traffic control (ATC),
Moving target indication (MTI),
Weapons location
Vehicle search

The phenomenon of pulse spreading in optical fibre is known as:

" },"encodingFormat": "text/html","text": "

The phenomenon of pulse spreading in optical fibre is known as:

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

Scattering

", "comment": { "@type": "Comment", "text": "

The phenomenon of pulse spreading in optical fibre is known as:

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

Dispersion

", "comment": { "@type": "Comment", "text": "

The phenomenon of pulse spreading in optical fibre is known as:

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

Attenuation

", "comment": { "@type": "Comment", "text": "

The phenomenon of pulse spreading in optical fibre is known as:

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

Inter symbol Interference

", "comment": { "@type": "Comment", "text": "

The phenomenon of pulse spreading in optical fibre is known as:

" }}],"acceptedAnswer": { "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "b", "comment": { "@type": "Comment", "text": "

The phenomenon of pulse spreading in optical fibre is known as:

" }, "answerExplanation": { "@type": "Comment", "text": "

Scattering:

Scattering is the loss of signal caused by the diffusion of a light beam, where the diffusion itself is caused by microscopic variations in the transmission medium
Scattering typically happens when a light signal hits an impurity in the fibre

Dispersion:

Dispersion is defined as pulse spreading in an optical fibre
As a pulse of light propagates through a fibre, elements such as numerical aperture, core diameter, refractive index profile, wavelength, and laser linewidth cause the pulse to broaden

Attenuation:

The attenuation of an optical fibre measures the amount of light lost between input and output
Total attenuation is the sum of all losses
For a given fibre, these losses are wavelength-dependent

Inter-symbol Interference:

This is the effect of dispersion on optical fibre
Intersymbol interference occurs when the pulse spreading caused by dispersion causes the output pulses of a system to overlap

." }}},{"@type": "Question","typicalAgeRange": "4-35","educationalLevel": "intermediate","eduQuestionType": "Multiple choice","learningResourceType": "Practice problem","name": "Recollect the concept of MPPGCL JE Electronics Mock Test - 3 to solve the ques", "comment": { "@type": "Comment", "text": "A _________stores such instructions that are required to start a computer." },"encodingFormat": "text/html","text": "A _________stores such instructions that are required to start a computer.","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "ROM", "comment": { "@type": "Comment", "text": "A _________stores such instructions that are required to start a computer." }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "RAM", "comment": { "@type": "Comment", "text": "A _________stores such instructions that are required to start a computer." }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "Mother Board", "comment": { "@type": "Comment", "text": "A _________stores such instructions that are required to start a computer." }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "PROM", "comment": { "@type": "Comment", "text": "A _________stores such instructions that are required to start a computer." }}],"acceptedAnswer": { "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "a", "comment": { "@type": "Comment", "text": "A _________stores such instructions that are required to start a computer." }, "answerExplanation": { "@type": "Comment", "text": "

The correct answer is option 1 i.e ROM

The startup instructions are stored on ROM in a computer. It isa part of BIOS (Basic Input Output System)

The amplifier in which each input will be multiplied by a different factor and then summed together is called as:

" },"encodingFormat": "text/html","text": "

The amplifier in which each input will be multiplied by a different factor and then summed together is called as:

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

non-weighted amplifier

", "comment": { "@type": "Comment", "text": "

The amplifier in which each input will be multiplied by a different factor and then summed together is called as:

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

scaling amplifier

", "comment": { "@type": "Comment", "text": "

The amplifier in which each input will be multiplied by a different factor and then summed together is called as:

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

averaging amplifier

", "comment": { "@type": "Comment", "text": "

The amplifier in which each input will be multiplied by a different factor and then summed together is called as:

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

multiple-way amplifier

", "comment": { "@type": "Comment", "text": "

The amplifier in which each input will be multiplied by a different factor and then summed together is called as:

" }}],"acceptedAnswer": { "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "b", "comment": { "@type": "Comment", "text": "

The amplifier in which each input will be multiplied by a different factor and then summed together is called as:

" }, "answerExplanation": { "@type": "Comment", "text": "

One type of summing amplifier is the SCALING AMPLIFIER.

This circuit multiplies each input by a factor (the factor is determined by circuit design) and then adds these values together.

The factor that is used to multiply each input is determined by the ratio of the feedback resistor to the input resistor.

For example, we can design a circuit that would produce the following output from three inputs (E₁, E₂, E₃)

= [(k₁× E₁) + (k₂× E₂) + (k₃× E₃)]

Using input resistors R₁ for input number one (E₁), R₂ for input number two (E₂), R₃ for input number three (E₃), and R₄ for the feedback resistor,

We can calculate the values for the resistors.

Booths algorithm is used for

" },"encodingFormat": "text/html","text": "

Booths algorithm is used for

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

floating point division

", "comment": { "@type": "Comment", "text": "

Booths algorithm is used for

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

restoring division

", "comment": { "@type": "Comment", "text": "

Booths algorithm is used for

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

integer multiplication

", "comment": { "@type": "Comment", "text": "

Booths algorithm is used for

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

square root

", "comment": { "@type": "Comment", "text": "

Booths algorithm is used for

" }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "

Booths algorithm is used for

" }, "answerExplanation": { "@type": "Comment", "text": "

Booth’s algorithm is a multiplication algorithm that multiplies two signed binary numbers in 2’s complement notation.
Booth used desk calculators that were faster at shifting than adding and created the algorithm to increase their speed. Booth’s algorithm is of interest in the study of computer architecture.

Here’s the implementation of the flowchart.

The deflecting torque of a moving iron instrument is

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The deflecting torque of a moving iron instrument is

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

The deflecting torque of a moving iron instrument is

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

The deflecting torque of a moving iron instrument is

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

The deflecting torque of a moving iron instrument is

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

The deflecting torque of a moving iron instrument is

" }}],"acceptedAnswer": { "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "b", "comment": { "@type": "Comment", "text": "

The deflecting torque of a moving iron instrument is

" }, "answerExplanation": { "@type": "Comment", "text": "

Moving Iron Instruments are the most common type of ammeter and voltmeter used at power frequencies in laboratories.

These instruments are very accurate, cheap, and rugged as compared to other AC instruments.

Working Principle of Moving Iron Instruments:

In Moving Iron Instruments, a plate or van of soft iron or of high permeability steel forms the moving element of the system.

The iron van is so situated that it can move in the magnetic field produced by a stationary coil.

The below figure shows a simple moving iron instrument.

The stationary coil is excited by the current or voltage under measurement.

When the coil is excited, it becomes an electromagnet, and the iron van moves in direction of offering a low reluctance path.

Thus the force of attraction is always produced in a direction to increase the inductance of the coil.

Mind that as the van follows the low reluctance path, the net flux in the air gap will increase which means increased flux linkage of the coil, and hence inductance of coil will increase.

It shall also be noticed that the inductance of the coil is variable and depends on the position of the iron van.

Torque Equation of Moving Iron Instruments:

Deflecting torque in Moving iron Instruments is given as

Td =(1/2)I²(dL/dƟ)

In moving iron instruments, the controlling torque is provided by spring. Controlling torque due to spring is given as

Tc = KƟ

Where K = Spring constant

Ɵ = Deflection in the needle

In equilibrium state, deflecting and controlling torque shall be equal as below.

Deflecting Torque = Controlling Torque

⇒ Td = Tc

⇒ (1/2)I²(dL/dƟ) = KƟ

⇒Ɵ = (1/2)(I²/K)(dL/dƟ)

What is the reason to have secondary memory?

" },"encodingFormat": "text/html","text": "

What is the reason to have secondary memory?

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

Limited capacity of main memory

", "comment": { "@type": "Comment", "text": "

What is the reason to have secondary memory?

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

Main memory is expensive

", "comment": { "@type": "Comment", "text": "

What is the reason to have secondary memory?

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

Main memory possesses highly volatility

", "comment": { "@type": "Comment", "text": "

What is the reason to have secondary memory?

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

All of the above

", "comment": { "@type": "Comment", "text": "

What is the reason to have secondary memory?

" }}],"acceptedAnswer": { "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "d", "comment": { "@type": "Comment", "text": "

What is the reason to have secondary memory?

" }, "answerExplanation": { "@type": "Comment", "text": "

We require secondary memory for our computer systems as:

Limited capacity of main memory - Although we can add main memory to a system but it is extremely difficult for general users. And we also want to store all our programs and data inside the main memory. But main memory having limited capacity,can accommodate only a limited amount of data and small number of programs.
Main memory is expensive -Main memory is faster than secondary memory and hence, it is highly expensive. Therefore, we can have multiple secondary storages in the cost of a single main memory.
Main memory possesses highly volatility - Main memory is volatile storage whereas secondary memory is non-volatile storage, that is, main memory will lose its content as soon as its power is turned off whereas secondary memory will retain the content even when the power is lost.

All these reasons makes us to use secondary memory along with main memory.

." }}},{"@type": "Question","typicalAgeRange": "4-35","educationalLevel": "intermediate","eduQuestionType": "Multiple choice","learningResourceType": "Practice problem","name": "Recollect the concept of MPPGCL JE Electronics Mock Test - 3 to solve the ques", "comment": { "@type": "Comment", "text": "The stage used to eliminate the noise effects in a FM detector is called –" },"encodingFormat": "text/html","text": "The stage used to eliminate the noise effects in a FM detector is called –","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "Limiter", "comment": { "@type": "Comment", "text": "The stage used to eliminate the noise effects in a FM detector is called –" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "Discriminator", "comment": { "@type": "Comment", "text": "The stage used to eliminate the noise effects in a FM detector is called –" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "AFC", "comment": { "@type": "Comment", "text": "The stage used to eliminate the noise effects in a FM detector is called –" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "Noise eliminator", "comment": { "@type": "Comment", "text": "The stage used to eliminate the noise effects in a FM detector is called –" }}],"acceptedAnswer": { "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "a", "comment": { "@type": "Comment", "text": "The stage used to eliminate the noise effects in a FM detector is called –" }, "answerExplanation": { "@type": "Comment", "text": "

At the transmitter the amplitude of the modulated signal is constant, noise and interfering stations produce amplitude changes on the modulated signal
The purpose of the limiter is to provide a constant level of the signal to the FM demodulator, thus reducing the effect of signal level changes in the output

In which modulation technique does the phase of the carrier signal is changed by varying the sine and cosine inputs at a particular time.

" },"encodingFormat": "text/html","text": "

In which modulation technique does the phase of the carrier signal is changed by varying the sine and cosine inputs at a particular time.

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

Frequency modulation

", "comment": { "@type": "Comment", "text": "

In which modulation technique does the phase of the carrier signal is changed by varying the sine and cosine inputs at a particular time.

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

Phase shift key modulation

", "comment": { "@type": "Comment", "text": "

In which modulation technique does the phase of the carrier signal is changed by varying the sine and cosine inputs at a particular time.

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

Analog modulation

", "comment": { "@type": "Comment", "text": "

In which modulation technique does the phase of the carrier signal is changed by varying the sine and cosine inputs at a particular time.

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

Pulse code modulation

", "comment": { "@type": "Comment", "text": "

In which modulation technique does the phase of the carrier signal is changed by varying the sine and cosine inputs at a particular time.

" }}],"acceptedAnswer": { "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "b", "comment": { "@type": "Comment", "text": "

In which modulation technique does the phase of the carrier signal is changed by varying the sine and cosine inputs at a particular time.

" }, "answerExplanation": { "@type": "Comment", "text": "

PSK(Phase Shift Keying):

InPSK (phase shift keying),binary 1 is represented with a carrier signal and binary 0is represented with 180° phase shift of a carrier, i.e. the phase of the carrier signal is changed by varying the sine and cosine inputs at a particular time.

For binary ‘1’ → S₁(A) = Acos 2π f_Ct

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

The Constellation Diagram Representation is as shown:

Important Point

ASK, PSK and FSKare 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):

InFSK (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 f_Ht

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

ASK(Amplitude Shift Keying):

InASK (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 f_ct

For binary ‘0’ → S₂(t) = 0

The Constellation Diagram Representation is as shown:

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

Find the voltage across 30 Ω resistor in the given circuit.

" },"encodingFormat": "text/html","text": "

Find the voltage across 30 Ω resistor in the given circuit.

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

105 V

", "comment": { "@type": "Comment", "text": "

Find the voltage across 30 Ω resistor in the given circuit.

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

45 V

", "comment": { "@type": "Comment", "text": "

Find the voltage across 30 Ω resistor in the given circuit.

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

75 V

", "comment": { "@type": "Comment", "text": "

Find the voltage across 30 Ω resistor in the given circuit.

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

30 V

", "comment": { "@type": "Comment", "text": "

Find the voltage across 30 Ω resistor in the given circuit.

" }}],"acceptedAnswer": { "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "b", "comment": { "@type": "Comment", "text": "

Find the voltage across 30 Ω resistor in the given circuit.

" }, "answerExplanation": { "@type": "Comment", "text": "

The given circuit is redrawn as:

Applying KVL around the loop, we get:

100 - 8I - 40 - 30I - 2I = 0

60 = 40 I

I = 1.5 A

Now, the voltage across the 30Ω resistor will be:

V_30Ω = 1.5× 30

V_30Ω = 45 V

." }}},{"@type": "Question","typicalAgeRange": "4-35","educationalLevel": "intermediate","eduQuestionType": "Multiple choice","learningResourceType": "Practice problem","name": "Recollect the concept of MPPGCL JE Electronics Mock Test - 3 to solve the ques", "comment": { "@type": "Comment", "text": "Find out even and odd part of signal x(t) = (2 + sin t)²" },"encodingFormat": "text/html","text": "Find out even and odd part of signal x(t) = (2 + sin t)²","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "4 + sin²t, 4 sin t", "comment": { "@type": "Comment", "text": "Find out even and odd part of signal x(t) = (2 + sin t)²" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "2 + sin²t, 2 sin t", "comment": { "@type": "Comment", "text": "Find out even and odd part of signal x(t) = (2 + sin t)²" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "4, 4 sin t + sin² t", "comment": { "@type": "Comment", "text": "Find out even and odd part of signal x(t) = (2 + sin t)²" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "2, sin t", "comment": { "@type": "Comment", "text": "Find out even and odd part of signal x(t) = (2 + sin t)²" }}],"acceptedAnswer": { "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "a", "comment": { "@type": "Comment", "text": "Find out even and odd part of signal x(t) = (2 + sin t)²" }, "answerExplanation": { "@type": "Comment", "text": "

Concept:

Any given signal x(t) can be written as the sum of its even part and odd part, i.e.

x(t) = xe(t) + xo(t)

xe(t) = Even part of g(t) calculated as:

xo(t) = Odd part of x(t) calculated as:

Calculation:

Given:

x(t) = (2 + sin t)²

x(t) = 4 + sin²t + 4 sin t

x(-t) =4 + sin2t-4 sin t

Even part of g(t) calculated as:

= 4 +sin2t

Odd part of x(t) calculated as:

= 4 sin t

The Transfer characteristic of the different types of MOSFETs is shown below, where I_D is drain current and V_GS is the Gate-Source voltage, the correct combination of MOSFET w.r.t to transfer characteristics is:

Types of MOSFET

(P) P – Channel Enhancement MOSFET

(Q) P – Channel Depletion MOSFET

(R) N – Channel Enhancement MOSFET

(S) N – Channel Depletion MOSFET

" },"encodingFormat": "text/html","text": "

Types of MOSFET

(P) P – Channel Enhancement MOSFET

(Q) P – Channel Depletion MOSFET

(R) N – Channel Enhancement MOSFET

(S) N – Channel Depletion MOSFET

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

P – 4, Q – 2, R – 1, S – 3

", "comment": { "@type": "Comment", "text": "

Types of MOSFET

(P) P – Channel Enhancement MOSFET

(Q) P – Channel Depletion MOSFET

(R) N – Channel Enhancement MOSFET

(S) N – Channel Depletion MOSFET

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

P – 3, Q – 2, R – 4, S – 1

", "comment": { "@type": "Comment", "text": "

Types of MOSFET

(P) P – Channel Enhancement MOSFET

(Q) P – Channel Depletion MOSFET

(R) N – Channel Enhancement MOSFET

(S) N – Channel Depletion MOSFET

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

P – 1, Q – 3, R – 2, S – 4

", "comment": { "@type": "Comment", "text": "

Types of MOSFET

(P) P – Channel Enhancement MOSFET

(Q) P – Channel Depletion MOSFET

(R) N – Channel Enhancement MOSFET

(S) N – Channel Depletion MOSFET

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

P – 2, Q – 4, R – 3, S - 1

", "comment": { "@type": "Comment", "text": "

Types of MOSFET

(P) P – Channel Enhancement MOSFET

(Q) P – Channel Depletion MOSFET

(R) N – Channel Enhancement MOSFET

(S) N – Channel Depletion MOSFET

" }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "

Types of MOSFET

(P) P – Channel Enhancement MOSFET

(Q) P – Channel Depletion MOSFET

(R) N – Channel Enhancement MOSFET

(S) N – Channel Depletion MOSFET

" }, "answerExplanation": { "@type": "Comment", "text": "

TheEnhancement mode MOSFET's are equivalent to a "Normally Open" switch that requires gate-source voltage to switch ON the device.
If the positive voltage (+V_GS) is applied to an n-channel gate terminal, then only the channel will conduct and the drain current starts to flow through the channel.
If thebias voltage is zero or negative(- V_GS)then the transistorswitches OFFand the channel stays in the non-conductive state resulting in the drain current to be zero.

Transfer characteristics of an n-channel depletion MOSFET is as shown:

We observe that for V_GS= 0, drain current is flowing.

Transfer characteristics of p-channel depletion MOSFET is as shown:

Transfer characteristics of n-channelEnhancement MOSFET is as shown:

We observe that for V_GS= 0, there is no drain current flowing as there is no conduction region at V_GS= 0, in enhancement mode MOSFET.

Transfer characteristics of p-channelEnhancement MOSFET is as shown:

If 1 is added to the denominator of a fraction it becomes1/2. If 1 is added to the numerator it becomes 1. The product of numerator and denominator of the fraction is

" },"encodingFormat": "text/html","text": "

If 1 is added to the denominator of a fraction it becomes1/2. If 1 is added to the numerator it becomes 1. The product of numerator and denominator of the fraction is

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

If 1 is added to the denominator of a fraction it becomes1/2. If 1 is added to the numerator it becomes 1. The product of numerator and denominator of the fraction is

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

If 1 is added to the denominator of a fraction it becomes1/2. If 1 is added to the numerator it becomes 1. The product of numerator and denominator of the fraction is

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

If 1 is added to the denominator of a fraction it becomes1/2. If 1 is added to the numerator it becomes 1. The product of numerator and denominator of the fraction is

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

If 1 is added to the denominator of a fraction it becomes1/2. If 1 is added to the numerator it becomes 1. The product of numerator and denominator of the fraction is

" }}],"acceptedAnswer": { "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "d", "comment": { "@type": "Comment", "text": "

If 1 is added to the denominator of a fraction it becomes1/2. If 1 is added to the numerator it becomes 1. The product of numerator and denominator of the fraction is

" }, "answerExplanation": { "@type": "Comment", "text": "

Let the numerator = x and denominator = y

y + 3 = 2y ⇒ y = 3

x +1= 3 ⇒ x = 2

∴ xy = 2 × 3 = 6

Directions: Study the following question carefully and choose the right answer.

In a certain code ‘TERMINAL’ is written as ‘SDQLJOBM’. How is CREDIBLE written in that code?

" },"encodingFormat": "text/html","text": "

Directions: Study the following question carefully and choose the right answer.

In a certain code ‘TERMINAL’ is written as ‘SDQLJOBM’. How is CREDIBLE written in that code?

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

BQDCJCMF

", "comment": { "@type": "Comment", "text": "

Directions: Study the following question carefully and choose the right answer.

In a certain code ‘TERMINAL’ is written as ‘SDQLJOBM’. How is CREDIBLE written in that code?

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

BQDCHAKD

", "comment": { "@type": "Comment", "text": "

Directions: Study the following question carefully and choose the right answer.

In a certain code ‘TERMINAL’ is written as ‘SDQLJOBM’. How is CREDIBLE written in that code?

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

DSFEJCMF

", "comment": { "@type": "Comment", "text": "

Directions: Study the following question carefully and choose the right answer.

In a certain code ‘TERMINAL’ is written as ‘SDQLJOBM’. How is CREDIBLE written in that code?

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

DSFEHAKD

", "comment": { "@type": "Comment", "text": "

Directions: Study the following question carefully and choose the right answer.

In a certain code ‘TERMINAL’ is written as ‘SDQLJOBM’. How is CREDIBLE written in that code?

" }}],"acceptedAnswer": { "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "a", "comment": { "@type": "Comment", "text": "

Directions: Study the following question carefully and choose the right answer.

In a certain code ‘TERMINAL’ is written as ‘SDQLJOBM’. How is CREDIBLE written in that code?

" }, "answerExplanation": { "@type": "Comment", "text": "

Hence, option A is correct.

Sunny can complete a work in 15 days. If Bobby is 60% as efficient as Sunny, in how many days can Bobby complete the work?

" },"encodingFormat": "text/html","text": "

Sunny can complete a work in 15 days. If Bobby is 60% as efficient as Sunny, in how many days can Bobby complete the work?

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

Sunny can complete a work in 15 days. If Bobby is 60% as efficient as Sunny, in how many days can Bobby complete the work?

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

Sunny can complete a work in 15 days. If Bobby is 60% as efficient as Sunny, in how many days can Bobby complete the work?

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

Sunny can complete a work in 15 days. If Bobby is 60% as efficient as Sunny, in how many days can Bobby complete the work?

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

Sunny can complete a work in 15 days. If Bobby is 60% as efficient as Sunny, in how many days can Bobby complete the work?

" }}],"acceptedAnswer": { "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "b", "comment": { "@type": "Comment", "text": "

Sunny can complete a work in 15 days. If Bobby is 60% as efficient as Sunny, in how many days can Bobby complete the work?

" }, "answerExplanation": { "@type": "Comment", "text": "

Direction: If a Paper (Transparent Sheet) is folded in a manner and a design or pattern is drawn. When unfolded this paper appears as given below in the answer figure. Choose the correct answer figure given below.

A round punched paper is given as shown in the question figure. Figure out from the four alternatives as to how it will appear when folded.
Question Figure

Answer figure

" },"encodingFormat": "text/html","text": "

A round punched paper is given as shown in the question figure. Figure out from the four alternatives as to how it will appear when folded.
Question Figure

Answer figure

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

A round punched paper is given as shown in the question figure. Figure out from the four alternatives as to how it will appear when folded.
Question Figure

Answer figure

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

A round punched paper is given as shown in the question figure. Figure out from the four alternatives as to how it will appear when folded.
Question Figure

Answer figure

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

A round punched paper is given as shown in the question figure. Figure out from the four alternatives as to how it will appear when folded.
Question Figure

Answer figure

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

A round punched paper is given as shown in the question figure. Figure out from the four alternatives as to how it will appear when folded.
Question Figure

Answer figure

" }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "

A round punched paper is given as shown in the question figure. Figure out from the four alternatives as to how it will appear when folded.
Question Figure

Answer figure

" }, "answerExplanation": { "@type": "Comment", "text": "

The shadow of a building is 10 m long when the point of rise of the sun is 60°. Discover the building's stature.

" },"encodingFormat": "text/html","text": "

The shadow of a building is 10 m long when the point of rise of the sun is 60°. Discover the building's stature.

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

16.32m

", "comment": { "@type": "Comment", "text": "

The shadow of a building is 10 m long when the point of rise of the sun is 60°. Discover the building's stature.

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

17.32 m

", "comment": { "@type": "Comment", "text": "

The shadow of a building is 10 m long when the point of rise of the sun is 60°. Discover the building's stature.

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

18.32 m

", "comment": { "@type": "Comment", "text": "

The shadow of a building is 10 m long when the point of rise of the sun is 60°. Discover the building's stature.

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

19.32m

", "comment": { "@type": "Comment", "text": "

The shadow of a building is 10 m long when the point of rise of the sun is 60°. Discover the building's stature.

" }}],"acceptedAnswer": { "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "b", "comment": { "@type": "Comment", "text": "

The shadow of a building is 10 m long when the point of rise of the sun is 60°. Discover the building's stature.

" }, "answerExplanation": { "@type": "Comment", "text": "

Let AB be the building and AC be its shadow.
Then, AC = 20m and ∠ACB = 60°. Let AB = x m.
Presently AB/AC = tan 60° = √3 => x/10 = √3
=> x = 10√3m = (10 * 1.732) m = 17.32m.
∴ Height of the building is 17.32m.

Direction: Study the following information carefully and answer the given questions besides.

Eight students Kirti, Ayushi, Manu, Sona, Aarti, Bobby, Ankita and Divya are sitting around the square table. Four students sit on each side while rest on each corner. Students at the side faces away from the centre and students at the corner faces toward the centre.
Divya sits third to the left of Ayushi. Sona sits immediate left of Ankita, who sits on the side of the table. Kirti sits opposite to Manu and both faces away from the centre. Neither Kirti nor Manu are beside Ayushi. Bobby is not an immediate right of Ayushi. Bobby is not adjacent to Manu.
Q. Who sits opposite to Bobby?

" },"encodingFormat": "text/html","text": "

Direction: Study the following information carefully and answer the given questions besides.

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

Aarti

", "comment": { "@type": "Comment", "text": "

Direction: Study the following information carefully and answer the given questions besides.

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

Manu

", "comment": { "@type": "Comment", "text": "

Direction: Study the following information carefully and answer the given questions besides.

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

Kirti

", "comment": { "@type": "Comment", "text": "

Direction: Study the following information carefully and answer the given questions besides.

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

Sona

", "comment": { "@type": "Comment", "text": "

Direction: Study the following information carefully and answer the given questions besides.

" }}],"acceptedAnswer": { "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "d", "comment": { "@type": "Comment", "text": "

Direction: Study the following information carefully and answer the given questions besides.

" }, "answerExplanation": { "@type": "Comment", "text": "

From the common explanation, we get Sona sits opposite Bobby.

Hence, Option E is correct.

Final Arrangement:

Common Explanation:

References:

Eight students Kirti, Ayushi, Manu, Sona, Aarti, Bobby, Ankita and Divya are sitting around the square table.

Four students sit on each side while rest on each corner.

Students at the side face away from the centre and students at the corner faces towards the centre.

Inferences:

So, we need to arrange as follows to accommodate the information as stated in the question,

References:

Divya sits third to the left of Ayushi.

Kirti sits opposite to Manu and both faces away from the centre.

Inferences:

From the above references, we get the following two different cases:

Case 1:

Case 2:

References:

Neither Kirti nor Manu are beside Ayushi.

Sona sits immediate left of Ankita, who sits on the side of the table.

Bobby is not an immediate right of Ayushi.

Bobby is not adjacent to Manu.

Inferences:

So, from this case 2 will be eliminated as neither Kirti nor Manu are beside Ayushi.

Now, from case 1 we get the following final arrangements:

The HCF of two numbers is 15 and their LCM is 300. If one of the number is 60, the other is :

" },"encodingFormat": "text/html","text": "

The HCF of two numbers is 15 and their LCM is 300. If one of the number is 60, the other is :

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

The HCF of two numbers is 15 and their LCM is 300. If one of the number is 60, the other is :

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

The HCF of two numbers is 15 and their LCM is 300. If one of the number is 60, the other is :

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

100

", "comment": { "@type": "Comment", "text": "

The HCF of two numbers is 15 and their LCM is 300. If one of the number is 60, the other is :

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

The HCF of two numbers is 15 and their LCM is 300. If one of the number is 60, the other is :

" }}],"acceptedAnswer": { "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "a", "comment": { "@type": "Comment", "text": "

The HCF of two numbers is 15 and their LCM is 300. If one of the number is 60, the other is :

" }, "answerExplanation": { "@type": "Comment", "text": "

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
First number × Second number = HCF × LCM

Directions: Read the given information carefully and answer the questions given beside:

A certain number of people are sitting in a row watching F1 racing. All of them are facing towards the south direction. The distance between any two adjacent persons is the same.
The number of people sitting between A and C is the same as the number of people sitting between X and C. At most 25 people are sitting in a row. X is an immediate neighbour of D, who sits on one of the ends. Two people are sitting between A and Z. The number of people sitting between D and C is the same as the number of people sitting between C and Y. C sits fourth to the right of Z. The number of people sitting to the right of A is the same as the number of people sitting to the left of Z.

Q.Who sits two places away from Y?

" },"encodingFormat": "text/html","text": "

Directions: Read the given information carefully and answer the questions given beside:

Q.Who sits two places away from Y?

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

Directions: Read the given information carefully and answer the questions given beside:

Q.Who sits two places away from Y?

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

Directions: Read the given information carefully and answer the questions given beside:

Q.Who sits two places away from Y?

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

Directions: Read the given information carefully and answer the questions given beside:

Q.Who sits two places away from Y?

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

either B or C option

", "comment": { "@type": "Comment", "text": "

Directions: Read the given information carefully and answer the questions given beside:

Q.Who sits two places away from Y?

" }}],"acceptedAnswer": { "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "d", "comment": { "@type": "Comment", "text": "

Directions: Read the given information carefully and answer the questions given beside:

Q.Who sits two places away from Y?

" }, "answerExplanation": { "@type": "Comment", "text": "

Following the common explanation, we get “C and Z sits two places away from Y”.
Hence, option D is correct.
Final Arrangement:

Common Explanation:
References:
1. Two people are sitting between A and Z.
2. C sits fourth to the right of Z.
3. The number of people sitting to the right of A is the same as the number of people sitting to the left of Z.
4. The number of people sitting between A and C is the same as the number of people sitting between X and C.
5. X is an immediate neighbour of D, who sits on one of the ends.
6. The number of people sitting between D and C is the same as the number of people sitting between C and Y.
7. At most 25 people are sitting in a row.
Inferences:
From reference 1, we get two possible cases.
From reference 7, case2 was eliminated.
Hence, Case 1 is the final arrangement.

Which revolution in the world inspired the Indians to set up a socialist economy?

" },"encodingFormat": "text/html","text": "

Which revolution in the world inspired the Indians to set up a socialist economy?

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

French Revolution

", "comment": { "@type": "Comment", "text": "

Which revolution in the world inspired the Indians to set up a socialist economy?

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

Turkish Revolution

", "comment": { "@type": "Comment", "text": "

Which revolution in the world inspired the Indians to set up a socialist economy?

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

Russian Revolution

", "comment": { "@type": "Comment", "text": "

Which revolution in the world inspired the Indians to set up a socialist economy?

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

American War of Independence

", "comment": { "@type": "Comment", "text": "

Which revolution in the world inspired the Indians to set up a socialist economy?

" }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "

Which revolution in the world inspired the Indians to set up a socialist economy?

" }, "answerExplanation": { "@type": "Comment", "text": "

During that time Lenin introduced the idea of a socialist, cooperative economy and emphasized collectivisation .Hence, these values inspired Indians to set up a socialist economy.

When did Zimbabwe attain independence and from whom?

" },"encodingFormat": "text/html","text": "

When did Zimbabwe attain independence and from whom?

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

1970, from Black minority rule

", "comment": { "@type": "Comment", "text": "

When did Zimbabwe attain independence and from whom?

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

1880, from White minority rule

", "comment": { "@type": "Comment", "text": "

When did Zimbabwe attain independence and from whom?

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

1980, from Americans

", "comment": { "@type": "Comment", "text": "

When did Zimbabwe attain independence and from whom?

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

1980, from White minority rule

", "comment": { "@type": "Comment", "text": "

When did Zimbabwe attain independence and from whom?

" }}],"acceptedAnswer": { "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "d", "comment": { "@type": "Comment", "text": "

When did Zimbabwe attain independence and from whom?

" }, "answerExplanation": { "@type": "Comment", "text": "

The country gained official independence as Zimbabwe on18 April 1980. The government held independence celebrations in Rufaro stadium in Salisbury, the capital.

Which of the following does not include election procedure?

" },"encodingFormat": "text/html","text": "

Which of the following does not include election procedure?

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

Voting

", "comment": { "@type": "Comment", "text": "

Which of the following does not include election procedure?

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

Nomination of Candidate

", "comment": { "@type": "Comment", "text": "

Which of the following does not include election procedure?

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

Booth capturing

", "comment": { "@type": "Comment", "text": "

Which of the following does not include election procedure?

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

Canvassing

", "comment": { "@type": "Comment", "text": "

Which of the following does not include election procedure?

" }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "

Which of the following does not include election procedure?

" }, "answerExplanation": { "@type": "Comment", "text": "

Booth capturing is a type of electoral fraud in which party loyalists "capture" a polling booth and vote in place of legitimate voters to ensure that their candidate wins.

Which of the following is not an agent of erosion and deposition

" },"encodingFormat": "text/html","text": "

Which of the following is not an agent of erosion and deposition

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

Running water

", "comment": { "@type": "Comment", "text": "

Which of the following is not an agent of erosion and deposition

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

Glaciers

", "comment": { "@type": "Comment", "text": "

Which of the following is not an agent of erosion and deposition

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

Earthquake

", "comment": { "@type": "Comment", "text": "

Which of the following is not an agent of erosion and deposition

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

Wind

", "comment": { "@type": "Comment", "text": "

Which of the following is not an agent of erosion and deposition

" }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "

Which of the following is not an agent of erosion and deposition

" }, "answerExplanation": { "@type": "Comment", "text": "

There are four main agents of erosion. Moving water, wind, gravity, and ice wear away or break up rocks, sediments, and soil from the land's surface. When these materials are deposited or dropped in new places, it is called deposition. Erosion and deposition work together.

." }}},{"@type": "Question","typicalAgeRange": "4-35","educationalLevel": "intermediate","eduQuestionType": "Multiple choice","learningResourceType": "Practice problem","name": "Recollect the concept of MPPGCL JE Electronics Mock Test - 3 to solve the ques", "comment": { "@type": "Comment", "text": "State animal of Haryana is:" },"encodingFormat": "text/html","text": "State animal of Haryana is:","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "Black buck", "comment": { "@type": "Comment", "text": "State animal of Haryana is:" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "Asiatic Lion", "comment": { "@type": "Comment", "text": "State animal of Haryana is:" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "Chinkara", "comment": { "@type": "Comment", "text": "State animal of Haryana is:" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "Tiger", "comment": { "@type": "Comment", "text": "State animal of Haryana is:" }}],"acceptedAnswer": { "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "a", "comment": { "@type": "Comment", "text": "State animal of Haryana is:" }, "answerExplanation": { "@type": "Comment", "text": "

The correct answer is theBlackbuck.

Important Points

The Blackbuck is the state animal of Haryana. In India hunting of Blackbuck is ban under theWildlife Protection Act of 1972, due to excessive hunting andhabitat degradation of Blackbucks during the 20th century.
The Blackbuck or an Indian antelope are found in India, Nepal, and Pakistan. They have a lifespan of approximately 10 to 15 years.

Additional Information

Black Francolin of thepheasant family is the State bird of Haryana.
There are 2 National parks located in Haryana viz,Sultanpur National Park located in Gurugram andKalesar National Park located in Yamunanagar.
Chinkara alsoknown as Indian Gazelleis thestate animal of Rajasthan.

While working with 75% of its efficiency a pipe can fill a tank in 80 minutes. In how many minutes can the pipe fill the tank while working with 100% efficiency?

" },"encodingFormat": "text/html","text": "

While working with 75% of its efficiency a pipe can fill a tank in 80 minutes. In how many minutes can the pipe fill the tank while working with 100% efficiency?

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

While working with 75% of its efficiency a pipe can fill a tank in 80 minutes. In how many minutes can the pipe fill the tank while working with 100% efficiency?

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

While working with 75% of its efficiency a pipe can fill a tank in 80 minutes. In how many minutes can the pipe fill the tank while working with 100% efficiency?

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

While working with 75% of its efficiency a pipe can fill a tank in 80 minutes. In how many minutes can the pipe fill the tank while working with 100% efficiency?

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

While working with 75% of its efficiency a pipe can fill a tank in 80 minutes. In how many minutes can the pipe fill the tank while working with 100% efficiency?

" }}],"acceptedAnswer": { "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "b", "comment": { "@type": "Comment", "text": "

While working with 75% of its efficiency a pipe can fill a tank in 80 minutes. In how many minutes can the pipe fill the tank while working with 100% efficiency?

" }, "answerExplanation": { "@type": "Comment", "text": "

Hence, Option B is correct.

Seven persons B, C, D, E, F, G and H are born on seven different months of a year starting from January, but not necessarily in the same order. Who was born in July?

Statement I: C was born just before D. Only two persons are born after H. Three persons are born between H and B. F was born two months after B.
Statement II: E was born in April. D was born after C. Two persons are born between E and B. B was not born in July. C was born one of the months after May.

" },"encodingFormat": "text/html","text": "

Seven persons B, C, D, E, F, G and H are born on seven different months of a year starting from January, but not necessarily in the same order. Who was born in July?

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

If the data in statement I alone is sufficient to answer the question.

", "comment": { "@type": "Comment", "text": "

Seven persons B, C, D, E, F, G and H are born on seven different months of a year starting from January, but not necessarily in the same order. Who was born in July?

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

If the data in statement II alone is sufficient to answer the question.

", "comment": { "@type": "Comment", "text": "

Seven persons B, C, D, E, F, G and H are born on seven different months of a year starting from January, but not necessarily in the same order. Who was born in July?

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

If the data in either statement I or statement II is sufficient to answer the question.

", "comment": { "@type": "Comment", "text": "

Seven persons B, C, D, E, F, G and H are born on seven different months of a year starting from January, but not necessarily in the same order. Who was born in July?

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

If the data in both statement I and statement II is together necessary to answer the question.

", "comment": { "@type": "Comment", "text": "

Seven persons B, C, D, E, F, G and H are born on seven different months of a year starting from January, but not necessarily in the same order. Who was born in July?

" }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "

Seven persons B, C, D, E, F, G and H are born on seven different months of a year starting from January, but not necessarily in the same order. Who was born in July?

" }, "answerExplanation": { "@type": "Comment", "text": "

Checking Statement I:

Using the above references, we get the following arrangement:

Clearly, D was born in July.

Hence data in statement I is sufficient to answer the question.

Checking Statement II:

Using the above references, we get the following arrangement:

Clearly, D was born in July.

Hence data in statement II is sufficient to answer the question.

Here, the data in either Statement I or II alone is sufficient to answer the question.

Hence, Option C is correct.

Sambhu beats Kalu by 30 metres in10 seconds. How much time was taken by Sambhu tocomplete a race 1200 meters.

" },"encodingFormat": "text/html","text": "

Sambhu beats Kalu by 30 metres in10 seconds. How much time was taken by Sambhu tocomplete a race 1200 meters.

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

6 min 30 s

", "comment": { "@type": "Comment", "text": "

Sambhu beats Kalu by 30 metres in10 seconds. How much time was taken by Sambhu tocomplete a race 1200 meters.

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

3 min 15 s

", "comment": { "@type": "Comment", "text": "

Sambhu beats Kalu by 30 metres in10 seconds. How much time was taken by Sambhu tocomplete a race 1200 meters.

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

12 min 10 s

", "comment": { "@type": "Comment", "text": "

Sambhu beats Kalu by 30 metres in10 seconds. How much time was taken by Sambhu tocomplete a race 1200 meters.

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

2 min 5 s

", "comment": { "@type": "Comment", "text": "

Sambhu beats Kalu by 30 metres in10 seconds. How much time was taken by Sambhu tocomplete a race 1200 meters.

" }}],"acceptedAnswer": { "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "a", "comment": { "@type": "Comment", "text": "

Sambhu beats Kalu by 30 metres in10 seconds. How much time was taken by Sambhu tocomplete a race 1200 meters.

" }, "answerExplanation": { "@type": "Comment", "text": "

Kalu’s speed = 3 m/s.
For 1200 m, Kalu would take 400 seconds and Sambhu would take 10 seconds less. Hence, 390
seconds.

A piece of paper is folded and cut as shown below in the question figures. From the given answer figures, indicate how it will appear when opened.
Question Figure

Answer figure

" },"encodingFormat": "text/html","text": "

A piece of paper is folded and cut as shown below in the question figures. From the given answer figures, indicate how it will appear when opened.
Question Figure

Answer figure

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

A piece of paper is folded and cut as shown below in the question figures. From the given answer figures, indicate how it will appear when opened.
Question Figure

Answer figure

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

A piece of paper is folded and cut as shown below in the question figures. From the given answer figures, indicate how it will appear when opened.
Question Figure

Answer figure

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

A piece of paper is folded and cut as shown below in the question figures. From the given answer figures, indicate how it will appear when opened.
Question Figure

Answer figure

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

A piece of paper is folded and cut as shown below in the question figures. From the given answer figures, indicate how it will appear when opened.
Question Figure

Answer figure

" }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "

A piece of paper is folded and cut as shown below in the question figures. From the given answer figures, indicate how it will appear when opened.
Question Figure

Answer figure

" }, "answerExplanation": { "@type": "Comment", "text": "

The average monthly salary of the workers in a workshop is Rs. 8,500. If the average monthly salary of 7 technicians is Rs. 10,000 and the average monthly salary of the rest is Rs. 7,800, then the total number of workers in the workshop is

" },"encodingFormat": "text/html","text": "

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

" }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "

" }, "answerExplanation": { "@type": "Comment", "text": "

Directions to Solve

In each of the following questions find out the alternative which will replace the question mark.

Question -

68 : 130 :: ? : 350

" },"encodingFormat": "text/html","text": "

Directions to Solve

In each of the following questions find out the alternative which will replace the question mark.

Question -

68 : 130 :: ? : 350

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

220

", "comment": { "@type": "Comment", "text": "

Directions to Solve

In each of the following questions find out the alternative which will replace the question mark.

Question -

68 : 130 :: ? : 350

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

224

", "comment": { "@type": "Comment", "text": "

Directions to Solve

In each of the following questions find out the alternative which will replace the question mark.

Question -

68 : 130 :: ? : 350

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

222

", "comment": { "@type": "Comment", "text": "

Directions to Solve

In each of the following questions find out the alternative which will replace the question mark.

Question -

68 : 130 :: ? : 350

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

226

", "comment": { "@type": "Comment", "text": "

Directions to Solve

In each of the following questions find out the alternative which will replace the question mark.

Question -

68 : 130 :: ? : 350

" }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "

Directions to Solve

In each of the following questions find out the alternative which will replace the question mark.

Question -

68 : 130 :: ? : 350

" }, "answerExplanation": { "@type": "Comment", "text": "

As, 68 = (4)³ + 4

130 = (5)³ + 5

and 350 = (7)³ + 7

Therefore, ? = (6)³ + 6 = 222

In a school having roll strength 299, the ratio of boys and girls is 8 : 5. If 25 more girls get admitted into the school, the ratio of boys and girls becomes

" },"encodingFormat": "text/html","text": "

In a school having roll strength 299, the ratio of boys and girls is 8 : 5. If 25 more girls get admitted into the school, the ratio of boys and girls becomes

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

46: 35

", "comment": { "@type": "Comment", "text": "

In a school having roll strength 299, the ratio of boys and girls is 8 : 5. If 25 more girls get admitted into the school, the ratio of boys and girls becomes

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

43: 35

", "comment": { "@type": "Comment", "text": "

In a school having roll strength 299, the ratio of boys and girls is 8 : 5. If 25 more girls get admitted into the school, the ratio of boys and girls becomes

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

56: 33

", "comment": { "@type": "Comment", "text": "

In a school having roll strength 299, the ratio of boys and girls is 8 : 5. If 25 more girls get admitted into the school, the ratio of boys and girls becomes

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

53: 33

", "comment": { "@type": "Comment", "text": "

In a school having roll strength 299, the ratio of boys and girls is 8 : 5. If 25 more girls get admitted into the school, the ratio of boys and girls becomes

" }}],"acceptedAnswer": { "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "a", "comment": { "@type": "Comment", "text": "

In a school having roll strength 299, the ratio of boys and girls is 8 : 5. If 25 more girls get admitted into the school, the ratio of boys and girls becomes

" }, "answerExplanation": { "@type": "Comment", "text": "

'Dogs' is related to 'Bark' in the same way as 'Goats' is related to:

" },"encodingFormat": "text/html","text": "

'Dogs' is related to 'Bark' in the same way as 'Goats' is related to:

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

Bleat

", "comment": { "@type": "Comment", "text": "

'Dogs' is related to 'Bark' in the same way as 'Goats' is related to:

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

Crow

", "comment": { "@type": "Comment", "text": "

'Dogs' is related to 'Bark' in the same way as 'Goats' is related to:

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

Grunt

", "comment": { "@type": "Comment", "text": "

'Dogs' is related to 'Bark' in the same way as 'Goats' is related to:

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

Howl

", "comment": { "@type": "Comment", "text": "

'Dogs' is related to 'Bark' in the same way as 'Goats' is related to:

" }}],"acceptedAnswer": { "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "a", "comment": { "@type": "Comment", "text": "

'Dogs' is related to 'Bark' in the same way as 'Goats' is related to:

" }, "answerExplanation": { "@type": "Comment", "text": "

As the cry of 'Dog' is called 'Bark' in the same way the cry of 'Goat' is called 'Bleat'.

The product of two numbers is 1280 and their H.C.F. is 8. The L.C.M. of the number will be ?

" },"encodingFormat": "text/html","text": "

The product of two numbers is 1280 and their H.C.F. is 8. The L.C.M. of the number will be ?

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

160

", "comment": { "@type": "Comment", "text": "

The product of two numbers is 1280 and their H.C.F. is 8. The L.C.M. of the number will be ?

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

150

", "comment": { "@type": "Comment", "text": "

The product of two numbers is 1280 and their H.C.F. is 8. The L.C.M. of the number will be ?

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

120

", "comment": { "@type": "Comment", "text": "

The product of two numbers is 1280 and their H.C.F. is 8. The L.C.M. of the number will be ?

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

140

", "comment": { "@type": "Comment", "text": "

The product of two numbers is 1280 and their H.C.F. is 8. The L.C.M. of the number will be ?

" }}],"acceptedAnswer": { "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "a", "comment": { "@type": "Comment", "text": "

The product of two numbers is 1280 and their H.C.F. is 8. The L.C.M. of the number will be ?

" }, "answerExplanation": { "@type": "Comment", "text": "

Product of two numbers = HCF × LCM
1280 = 8 × LCM
160 = LCM

Select the related letters from the given alternatives.

ACFOMR : FCARMO :: MNTDEF : ?

" },"encodingFormat": "text/html","text": "

Select the related letters from the given alternatives.

ACFOMR : FCARMO :: MNTDEF : ?

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

TNMFDE

", "comment": { "@type": "Comment", "text": "

Select the related letters from the given alternatives.

ACFOMR : FCARMO :: MNTDEF : ?

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

NMTFED

", "comment": { "@type": "Comment", "text": "

Select the related letters from the given alternatives.

ACFOMR : FCARMO :: MNTDEF : ?

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

TNMFED

", "comment": { "@type": "Comment", "text": "

Select the related letters from the given alternatives.

ACFOMR : FCARMO :: MNTDEF : ?

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

TMNFED

", "comment": { "@type": "Comment", "text": "

Select the related letters from the given alternatives.

ACFOMR : FCARMO :: MNTDEF : ?

" }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "

Select the related letters from the given alternatives.

ACFOMR : FCARMO :: MNTDEF : ?

" }, "answerExplanation": { "@type": "Comment", "text": "

The logic follows here is:-

1st and 3rd letters are interchanged.

4th and 6th letters are interchanged.

And 2nd letter is placed as it is.

Similarly,

Hence, the correct answer is "TNMFED".

In a school 30% of the students play football and 50% of students play cricket. If 40% of the students play neither football nor cricket, what percentage of total students play both the games?

" },"encodingFormat": "text/html","text": "

In a school 30% of the students play football and 50% of students play cricket. If 40% of the students play neither football nor cricket, what percentage of total students play both the games?

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

In a school 30% of the students play football and 50% of students play cricket. If 40% of the students play neither football nor cricket, what percentage of total students play both the games?

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

In a school 30% of the students play football and 50% of students play cricket. If 40% of the students play neither football nor cricket, what percentage of total students play both the games?

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

In a school 30% of the students play football and 50% of students play cricket. If 40% of the students play neither football nor cricket, what percentage of total students play both the games?

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

", "comment": { "@type": "Comment", "text": "

In a school 30% of the students play football and 50% of students play cricket. If 40% of the students play neither football nor cricket, what percentage of total students play both the games?

" }}],"acceptedAnswer": { "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "d", "comment": { "@type": "Comment", "text": "

In a school 30% of the students play football and 50% of students play cricket. If 40% of the students play neither football nor cricket, what percentage of total students play both the games?

" }, "answerExplanation": { "@type": "Comment", "text": "

The number of students who play both the games = (30 + 50 + 40)% – 100% = 20%
Hence, Option D is correct.

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.

A, B, C, D and E are sitting around a circle and facing away from the centre. Who sits second to the right of B?
Statement I:A sits to the immediate left of C. D sits second to the right of A.
Statement II:D sits immediately between E and B. B sits to the immediate left of A.

" },"encodingFormat": "text/html","text": "

","suggestedAnswer": [{ "@type": "Answer", "position": 0, "encodingFormat": "text/html", "text": "

The data in Statement I alone is sufficient to answer the question while the data in Statement II is not sufficient to answer the question.

", "comment": { "@type": "Comment", "text": "

" }},{ "@type": "Answer", "position": 1, "encodingFormat": "text/html", "text": "

The data in both the Statements I and II is not sufficient to answer the question.

", "comment": { "@type": "Comment", "text": "

" }},{ "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "

The data in Statement II alone is sufficient to answer the question while the data in Statement I is not sufficient to answer the question.

", "comment": { "@type": "Comment", "text": "

" }},{ "@type": "Answer", "position": 3, "encodingFormat": "text/html", "text": "

The data in both the Statements I and II together is necessary to answer the question.

", "comment": { "@type": "Comment", "text": "

" }}],"acceptedAnswer": { "@type": "Answer", "position": 2, "encodingFormat": "text/html", "text": "c", "comment": { "@type": "Comment", "text": "

" }, "answerExplanation": { "@type": "Comment", "text": "

From Statement 1:
A sits to the immediate left of C.
D sits second to the right of A.

From the statement, we cannot determine who sits second to the right of B.
From Statement 2:
D sits immediately between E and B.
B sits to the immediate left of A.

From the statement, C sits second to the right of B.
Hence option C is correct.

." }}}] }]

Electronics and Communication Engineering (ECE) Exam>Electronics and Communication Engineering (ECE) Tests>MPPGCL JE Electronics Mock Test - 3 - Electronics and Communication Engineering (ECE) MCQ

Test Description

100 Questions MCQ Test - MPPGCL JE Electronics Mock Test - 3

MPPGCL JE Electronics Mock Test - 3 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The MPPGCL JE Electronics Mock Test - 3 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The MPPGCL JE Electronics Mock Test - 3 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for MPPGCL JE Electronics Mock Test - 3 below.

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MPPGCL JE Electronics Mock Test - 3 - Question 1

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A series RL circuit has a resistance of 5 kΩ and inductive reactance of 5 kΩ. What is the magnitude and phase angle of the effective impedance?

A. 7.07 kΩ, 45°
B. 5 kΩ, 45°
C. 7.07 kΩ, 90°
D. 5 kΩ, 90°

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 1

Concept:

For a series RLC circuit, the net/effective impedance is given by:

X_L = Inductive reactance

X_C = Capacitive reactance

The magnitude of the effective/net impedance is given by:

Also, the phase angle of the effective impedance is given by:

Calculation:

For a series R_L circuit, X_C = 0 Ω

Given, R = 5 kΩ and X_L = 5 kΩ

Themagnitude of the net/effectiveimpedance will be:

Putting on the respective values, we get:

Z = 5√2 = 7.07 kΩ

The phase angle of the effective impedance will be:

ϕ = 45°

MPPGCL JE Electronics Mock Test - 3 - Question 2

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Limitation(s) of control is (are)

A.
external factors
B.
fixing of responsibility
C.
variation and its causes
D.
All of the above

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 2

Controllingcan be defined as that function of management that helps to seek planned results from the subordinates, managers, and at all levels of an organization. Thecontrollingfunction helps in measuring the progress towards the organizational goals & brings any deviations, & indicates corrective action.

Key Points

Limitations of Controlling:

1.Difficulty in setting quantitative standards:

Control system loses its effectiveness when the standard of performance cannot be defined in quantitative terms and it is very difficult to set the quantitative standard for human behavior, efficiency level, job satisfaction, employee’s morale, etc.
In such cases, judgment depends upon the discretion of the manager.

2.No control on external factors:

An enterprise cannot control external factors due to its variation and causessuch as government policy, technological changes, change in fashion, change in competitor’s policy, etc.

3.Resistance from employees:

Employees often resist control and as result effectiveness of control reduces.
Employees feel control reduces or curtails their freedom.
Employees may resist and go against the use of cameras, to observe them minutely.

4.Costly affair:

Control is an expensive process it involves a lot of time and effort as sufficient attention has to be paid to observe the performance of the employees.
To install an expensive control system organizations have to spend a large amount.
Management must compare the benefits of controlling systems with the cost involved in installing them.
The benefits must be more than the cost involved then only controlling will be effective otherwise it will lead to inefficiency.

Therefore, option 4 is the correct answer.

MPPGCL JE Electronics Mock Test - 3 - Question 3

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Which of the following instrument is used in an airplane to measure its height above the ground?

A.
MTI Radar
B.
Doppler Radar
C.
CW Radar
D.
Radar Altimeter

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

The radar altimetermeasures absolute altitude, i.e. the altitude that the aircraft is above the ground (Above ground level)
The radar altimeter system sends out a radio wave, and that radio wave bounces off the ground and is received.
AGL is computed by taking the speed of the radio wave and the amount of time for it to return.
Unlike radar, the signal is not pulsed, but is Frequency modulated and is transmitted at 4300 MHz at 350 milliwatts.
Radar Altimeters operate between - 20 and 2500 feet.

MPPGCL JE Electronics Mock Test - 3 - Question 4

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What is the frame rate of a TDM when the transmission rate is 32 kbps and each slot has 16 bits?

A. 1000 fps
B. 2000 fps
C. 3000 fps
D. 4000 fps

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 4

Concept:

Bitrate in the TDM system is given as:

R_b = bits per frame × frame rate

Units:

Calculation:

Given in a TDM system bitrate R_b = 32 kbps

Each slot has 16 bits i.e., each frame has 16 bits

The frame rate will be:

Frame rate = 2000 fps

MPPGCL JE Electronics Mock Test - 3 - Question 5

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In the given circuit, viewed from AB, the circuit can be reduced to an equivalent circuit as:

A.
1 volt source in series with 10 Ω resistor
B.
7 volts source in series with 2.4Ω resistor
C.
5 volts source in series with 10Ω resistor
D.
15 volts source in series with 2.4 Ω resistor

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 5

Thevenin's Theorem:

Thevenin’s Theorem states that “Any linear circuit containing several voltages and resistances can be replaced by just one single voltage in series with a single resistance connected across the load".

Thevenin's equivalent circuit:

Calculation of R_th:

6Ω and 4Ω are in parallel

R_th= 2.4Ω

Calculation of Vth:

5V_th= 35

V_th= 7 V

MPPGCL JE Electronics Mock Test - 3 - Question 6

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In class AB amplifiers, the current flows through the active device for

A.
Full duration of input cycle
B.
half duration of input cycle
C.
more than half but less than full cycle duration
D.
less than half of the duration of input cycle

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 6

Class AB Power Amplifier:

Class AB amplifiers are a combination of class A and class B amplifiers.
This class of amplifiers is designed to reduce the less efficient problem of class A amplifiers and distortion of the signal at the crossover region in class B amplifiers.
It maintains high-frequency response like in class A amplifiers and good efficiency as in class B amplifiers.
A combination of diodes and resistors are used to provide little bias voltage which reduces the distortion of waveform near the crossover region.

Important Points

In class AB amplifiers, the current flows through the active device formore than half but less than full-cycle duration

MPPGCL JE Electronics Mock Test - 3 - Question 7

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The average value of the waveform is:

A.
8 V
B.
20 V
C.
10 V
D.
4 V

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 7

Concept:

The average value of a periodic waveform is given as

It is also calculated by calculating the area of the given waveform.

Where A is the area

Calculation:

The given figure consists of rectangular pulses.

Area of rectangle = L× B

Where L = Length

B = Breadth

MPPGCL JE Electronics Mock Test - 3 - Question 8

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The figure of merit (FOM) of a logic family is given by the product of

A. Gain and Bandwidth
B. Propagation delay and power dissipation
C. Fan out and propagation delay time
D. Noise margin and power dissipation

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

FOM of a logic Family is defined as the product of the propagation delay (t_pd) and the power dissipation (P_Davg), i.e.

FOM = t_pd × P_D(avg)

For the best IC operation, FOM should be as small as possible.

Units: ns × mW

= pJ (pico Joule)

Propagation delay (t_pd):

t_PHL = delay time in going form High to low logic

t_PLH = Delay time in going from low to High logic

Power dissipation (P_D):

P_D(avg) = I_cc × V_cc

V_CC = power supply

I_CC = avg collector current calculated as the average of the High and low current, i.e.

MPPGCL JE Electronics Mock Test - 3 - Question 9

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The solid angle subtended by the sun as viewed from the earth is Ω = 4 × 10^-5steradian. A microwave antenna designed to be used for studying the microwave radiation from the sun has a very narrow beam whose equivalent solid angle is approximately equal to that subtended by the sun. What is the approximate directivity, D?

A.
10⁵
B.
π × 10⁵
C.
π × 10⁶
D.
10⁶

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 9

Calculation:

The given situation can be visualized as shown:-

The approximate directivity is given by:

MPPGCL JE Electronics Mock Test - 3 - Question 10

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Which of the following is the expression for upper side band SSB signal?

is Hilbert transform of .

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 10

Concept:

SSB (Single Side Band) modulation isa technique that allows only one sideband to pass saving bandwidth and power both.

Practically it is very difficult to pass only a single sideband.

Analysis:

Ex: If we consider a single-tone modulating signal as x(t) = cos ω_mt, the SSB modulated wave withonly the upper sidebandwill be represented as:

cos ((ω_c+ ω_m)t) = cos ω_c.cos ω_mt – sin ω_c.sin ω_mt

andwith lower sideband onlywill be represented as:

cos (ω_c- ω_m)t = cos ω_ct .cos ω_mt + sin ω_ct. sin ω_mt

where sin ω_mt is the Hilbert transform of the original message signal x(t) = cos ω_mt

Generalizing this for any signal m(t), we can say that the SSB modulated signal with only upper sideband will be:

m(t) cos ω_ct – m̂(t) sin ω_ct, (option 1 is the correct answer)

Similarly, we can say that the SSB modulated signal with only lower sideband will be:

m(t) cos ω_ct + m̂(t) sin ω_ct,

where m̂(t) is the Hilbert transform of m(t).

SSB/SC is generated using a balanced modulator circuitas shown:

MPPGCL JE Electronics Mock Test - 3 - Question 11

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Three resistances of 1ohm, 2 ohms and 3ohms are connected in delta. These resistances are to be replaced by star connection as shown in the figure below, maintaining the same terminal conditions.

The value of highest resistance in star will be:

A.
1/4Ω
B.
1/3Ω
C.
1/2Ω
D.
1Ω

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 11

Delta to star conversion:

Calculation:

Given,

R_a= 3Ω, R_b= 2Ω, R_c= 1Ω

R₁= (2 × 1) / (1 + 2 + 3)

= 2/6

= 1/3Ω

R₂= (3 × 1) / (1 + 2 +3)

= 3/6

= 1/2Ω

R₃= (3 × 2) / (1 +2 +3)

= 6/6

= 1Ω

Hence,The value of highest resistance in star will be 1Ω

Points to remember:

Star-delta conversion

MPPGCL JE Electronics Mock Test - 3 - Question 12

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What is the octal equivalent of (2F.C4)₁₆?

A.
(44.37)₈
B.
(57.61)₈
C.
(27.19)₈
D.
(63.42)₈

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 12

Given number =2F.C4

Step 1: Converting HD to Decimal:

Now we have Decimal Number (D) will be,

D =

D = 32 + 15 + 0.75 +0.015625

D =47.765625

(2F.C4)₁₆= (47.765625)_D

Step 2: Converting Decimal to Octal:

Taking 47:

(47)D = 57

Taking 0.765625:

(0.765625)D = 61

He nce,(2F.C4)₁₆ = (47.765625)_D = (57.61)₈

Alternate MethodStep 1: Convert HD to Binary of 4 bit:

Step 2: Rearrange 4 bits into 3 bits group:

Step 3: Converting 3 bits of Binary into Octal:

Hence,

(2F.C4)₁₆= (57.61)₈

MPPGCL JE Electronics Mock Test - 3 - Question 13

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ROC of x(n) = -αⁿ u(-n-1) is

A. |z| < |α|
B. |z| = |α|
C. |z| = 1
D. |z| > |α|

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 13

Concept:

Z- transform of x(n) is given as:

Region of convergence (ROC) is defined as the region where the Z-transform exists.

Application:

Given:

x(n) =-αnu(-n-1)

For the Z-transform to exist,

MPPGCL JE Electronics Mock Test - 3 - Question 14

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If the differential voltage gain and the common mode voltage gain of a differential Amplifier are48 dB and 2 dB respectively, then its common mode rejection ratio is

A. 23 dB
B. 25 dB
C. 46 dB
D. 50 dB

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 14

Concept:

The Common Mode rejection ratio is given by:

Ad= differential voltage gain

Ac= common-mode voltage gain

In dB the voltage gain is expressed as:

Gain (dB) = 20 log₁₀(A)

Calculation:

Given:

Differential voltage gain = 48 dB

Common mode voltage gain = 2 dB

CMRR = 20 log(A_d) - 20 log(A_c)

= 48 - 2 = 46 dB

MPPGCL JE Electronics Mock Test - 3 - Question 15

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A loss-less transmission line having characteristic impedance of 100 ohm is connected to a load of 200 ohm. The voltage reflection coefficient is:

A.
1
B.
3
C.
1/3
D.
2/3

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 15

Concept:

The reflection coefficient gives the percentage of incident voltage reflected from the load.

Mathematically, this is defined as:

Γ = Reflection coefficient

Z_L=Load Impedance

Z_o= Characteristic Impedance.

Application:

Given Z₀ = 100Ωand Z_L = 200Ω

Voltage reflection coefficient will be:

MPPGCL JE Electronics Mock Test - 3 - Question 16

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The relationship between f_T, f_βand h_fein a high frequency model is given by:

A.
f_T= f_β/ h_fe
B.
f_β= f_T+h_fe
C.
f_β= f_Th_fe
D.
f_T= f_βh_fe

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 16

The correct answer is option 4): (f_T= f_βh_fe)

Concept:

In BJT, at the high–frequency end, there are two factors that define the – 3dB cutoff point:

The network capacitance ( parasitic and introduced) and the frequency dependence of hfe (β )

The frequency at which |hfe|dB = 0 dB is indicated by f_T.

f_T= f_βh_fe

The hybridπ model for high-frequency analysis is given as

MPPGCL JE Electronics Mock Test - 3 - Question 17

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In Ethernet Frame Format number of bits needed in Frame check sequence Field?

A.
8
B.
16
C.
32
D.
64

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 17

A frame check sequence (FCS) refers to an error-detecting code added to a frame in a communications protocol

Layer 2 Frame Format:

Ethernet Frame Format consists of Cyclic Redundancy Check (CRC) in Frame check sequence Field.

CRC size → 4 bytes

1 byte = 8 bit

∴ 4 byte = 4 × 8 bits = 32 bits

MPPGCL JE Electronics Mock Test - 3 - Question 18

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Hay’s bridge is used for measuring:

A. admittance
B. capacitance
C. inductance of high Q factor
D. resistance

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 18

Hay’s bridge is used for measurement of inductance of coils with high quality factor.

MPPGCL JE Electronics Mock Test - 3 - Question 19

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Which type does an X-Y recorder belong to?

A. Digital recorder
B. Oscillographic recorder
C. Magnetictape recorder
D. Graphic recorder

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 19

X-Y recorder:

X-Y recorders is type of graphic recorders.
An XY recorder plots the instantaneous relation between two variables.
The X-Y recordersrecords the variation in one dependent variable with respect to an independent variable.
In X-Y recorders, an EMFis plotted as a function of another EMF.
This is done by having one self-balancing potentiometer control the position of the rolls while another self- balancing potentiometer controls the position of the recording pen.

MPPGCL JE Electronics Mock Test - 3 - Question 20

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For an n-channel MOSFET, if conduction parameter (k_n) is 0.249 mA/V², gate to source voltage V_QS is 2V_TN where V_TN = 0.75V. The current will be

A. 0.160 mA
B. 0.150 mA
C. 0.140 mA
D. 0.170 mA

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 20

Concept:

For an n-channel MOSFET, the current in triode and saturation region is given by:

ID= Kn[2VDS(VGS- Vth) - VDS2)] when VDS< VGS- Vth

when VDS> VGS- Vth

VDS= Drain to Source Voltage

VGS= Gate to source voltage

Vth= Threshold Voltage

Kn= Conduction parameter

Analysis:

Given:Kn= 0.249 mA / V2

VGS= 2VTN

VTN= 0.75 V

∵ VGS– VTN= 0.75 V

And VDSis usually more than that

∴ VDS> VGS- VTN

And the MOSFET is in saturation.

I_D = 0.249 (2V_TN - V_TN)²

= 0.249 (V_TN)²

= 0.249 (0.75)²

= 0.14 mA

MPPGCL JE Electronics Mock Test - 3 - Question 21

Save

The algebric sum of the voltages around any closed path is equal to

A.
1
B.
Infinite
C.
D.
Indefinite

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 21

Concept:

Kirchhoff’s Voltage Law (KVL):

It states that the sum of the voltages or electrical potential differences in a closed network is zero.

Mathematically we can express this as:

Where V_nrepresents the n^thVoltage

M is the total number of voltage elements.

Example:

according to KVL

apply KVL to the above circuit

- V + V₁+V₂= 0

V - V₁-V₂= 0

Note:KCL is based on the law ofconservation of charge.

MPPGCL JE Electronics Mock Test - 3 - Question 22

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In control systems, excessive bandwidth is NOT employed because:

A. noise is proportional to bandwidth
B. it leads to low relative stability
C. it leads to slower response
D. noise is proportional to the square of the bandwidth

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 22

Noise power is given as:

Pn= KTB

Where,

K = Boltz man's constant

T = temperature

B = Bandwidth over which noise is measured

When the bandwidth is increased, then system becomes fast as it reduces rise time, increases stability as well as reduces oscillations in the response but according to the above equation, the noise also increases.

MPPGCL JE Electronics Mock Test - 3 - Question 23

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Two incandescent light bulbs of 40 W and 60 W rating are connected in series across the mains, then

A.
The bulbs together consume 100 W
B.
The bulbs together consume 50 W
C.
The 60 W bulb glows brighter
D.
The 40 W bulb glows brighter

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 23

Concept:

We know that

P = I²R

Where

P = Power

I = Current

R = Resistance

In this, the currentthrough thebulbs is fixed.

The resistance of the bulb is inversely proportional to the wattage of the bulb.

Calculation:

Given

R₆₀ < R₄₀

The resistance of 40 W is more than 60 W bulbs.

Hence R₆₀Consumes less power and is less bright

So, the 40 bulb glows brighter.

MPPGCL JE Electronics Mock Test - 3 - Question 24

Save

The output of the 4 to 1 Mux is

A.
x + y
B.
C.
x.y
D.
x̅ + y̅

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 24

Concept:

In a 4 × 1 MUX

Truth-Table

MUX contains AND gate followed by OR gate

Calculation:

By re-drawing the circuit diagram

∴ I₀= 1, I₁= 0, I₂= 0, I₃= 0& (x= S₂, y= S₁)

Now output of 4 × 1 MUX is

Y = F = S̅₁S̅₂I₀+ S̅₁S₂I₁+ S₁S̅₂I₂+ S₁S₂I₃

F = x̅ y̅ 1+ x̅ y 0+ x y̅ 0+ x y 0

MPPGCL JE Electronics Mock Test - 3 - Question 25

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If a transmission line is terminated with a resistance equal to its characteristic impedance

A. reflection coefficient will be unity
B. standing wave ratio will be minimum
C. the line loss will be maximum
D. The input impedance will be twice the terminating resistance

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 25

Concept:

The voltage standing wave ratio is defined as the ratio of the maximum voltage (or current) to the minimum voltage (or current).

VSWR is also given by:

Γ = Reflection coefficient, defined as:

Z_L= Load impedance

Z₀= Characteristic Impedance

ForΓ_Lvarying from 0 to 1, VSWRvaries from 1 to∞.

Application:

Given Z_L = Z₀

The reflection coefficient is calculated to be:

VSWR forΓ = 0 equals 1, which is the minimum value (because it varies from 1to∞)

MPPGCL JE Electronics Mock Test - 3 - Question 26

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If the modulation index of an AMwave is changed from 0to 1, the transmitted power

A. increases by 100%
B. increases by 50%
C. unchanged
D. increases by 66.66%

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 26

Concept:

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

Pc= Carrier Power

μ = Modulation Index

Analysis:

Whenμ= 0, the transmitted power will be:

Whenμ = 1, the transmitted power will be:

The % increase in the modulated signal power is given by:

MPPGCL JE Electronics Mock Test - 3 - Question 27

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Which of the following statements are true -

I. Negative resists on exposure of light becomes more soluble.

II. Positive resists on exposure of light becomes more soluble.

III. Chrome glass is used as mask for pattern printing.

A. I & II
B. only II
C. II & III
D. only III

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 27

Key Points

ANegative Photoresistis the one which becomes less soluble in developer when exposed to light.

While thePositive Photoresistbecomes more soluble when exposed to light.

Chrome glass is used as mask for Photolythography

MPPGCL JE Electronics Mock Test - 3 - Question 28

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Find the peak-to-peak ripple voltage of half-wave rectifier and filter circuit which has a 680 μF filter capacitor, an average output voltage of 30 V, and a 220Ω load resistance. The mains frequency is 50 Hz.

A.
3.51 V
B.
4.01 V
C.
3.83 V
D.
2.84 V

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 28

Concept:

The Peak to peak ripple voltage of Half wave rectifies is given by:

Where, I_dc = Average current

f₀ = mains freqeucny

C = capacitor

Calculation:

Given f₀ = 50 Hz,C = 680 μf, V_dc = 30 V, and R_L = 220 Ω, we get:

I_dc = 0.136

MPPGCL JE Electronics Mock Test - 3 - Question 29

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For a 10 bit PCM system the signal to quantization noise ratio is 62 dB. If the number of bits is increased by 2, then the signal to quantization noise ratio will

A.
Increase by 6 dB
B.
Increase by 12 dB
C.
Decrease by 6 dB
D.
Decrease by 12 dB

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 29

Concept:

TheSignal to Noise (SNR)Ratio for a PCM system is given by:

SNR = (1.8 + 6n) dB

Where ‘n’ is the number of bits per sample.

Calculation:

For nnumber of bits, the Signal to Noise Ratio will be:

SNR₁= (1.8 + 6n)dB

With the increase in two-bit( "n+2" bits), the SNR becomes:

SNR₂= (1.8 + 6(n+2)) dB

SNR₂- SNR₁= 12 dB

i.e if the number of bits increased from 10 bits to 12 bits, then the SNR increases by 12 dB.

Important Point

Some salient features of a PCM system are:

Immunity to transmission noiseand interference.
It is possible toregeneratethe coded signal along the transmission path.
TheQuantization Noise depends on the number of quantization levelsandnoton thenumber of samples produced per second.

MPPGCL JE Electronics Mock Test - 3 - Question 30

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Class C amplifier operates

A.
Entire cycle of i/p signal
B.
Half of the cycle of i/p signal
C.
Slightly more than half of the cycle of i/p signal
D.
Less than half of the cycle of i/p signal

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 30

Class C amplifier:

Class C power amplifier is a type of amplifier where the active element (transistor) conduct for less than one-half cycle of the input signal.
Less than one-half cycle means the conduction angle is less than 180° and its typical value is 80° to 120°.
The reduced conduction angle improves the efficiency to a great extends but causes a lot of distortion.
The theoretical maximum efficiency of a Class C amplifier is around 90%.
In a Class C Amplifier efficiency and distortion, both are maximum.

Important Point

MPPGCL JE Electronics Mock Test - 3 - Question 31

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The contents to the register pair DE after the execution of the following Intel 8085 instructions, is

PUSH H; PUSH D; POP H; POP D

A.
[H] → [D] ; [L] → [E]
B.
[A] → [D] ; [B] → [E]
C.
[H] → [D] ; [A] → [E]
D.
[L] → [D] ; [H] → [E]

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 31

PUSH RP (Pre-decrement):

This instruction copies the contents of RP to stack.

For every PUSH:

Decrement SP and PUSH RP_H
Decrement SP and PUSH RP_L

POP RP (Pre-increment):

Used for retrieving the elements stored in the stack.

For every POP:

Get 1 byte from stack to RP_Land increment.
Get 1 byte from stack to RP_H and increment.

Let stack pointer (SP) = FFFF H and DE = 3025H ; HL = B315H

(a) PUSH H and

(b) PUSH D are as shown:

PUSH D: contents of DE register is copied to (FFFC and FFFB) H

PUSH H: contents of HL = B315 register, is copied to (FFFE & FFFD)H

Similarly,

c) POP J and

d) POP D are as shown:

POP H retrieves the contents to HL register

POP D: retrieves the contents to the DE register.

∴ After the execution of instructions, the contents of the DE register pair is DE = B315 H, which is

equal to HL before execution.

∴ [H] → [D]

[L] → [E]

MPPGCL JE Electronics Mock Test - 3 - Question 32

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Following constellation diagram represents:

A.
16 PSK
B.
16 QAM
C.
16 APSK
D.
All of above

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 32

PSK: It stands for phase shift keying.
In PSK, binary 1 is represented with the actual carrier, and 0 is represented with the 180^ophase shift of carrier.
1 is represented as s(t) = Accos(2πf_ct)
0 is represented as s(t) = - Accos(2πf_ct)
As in PSK, one bit is transmitted in a specific time constant.
Similarly, in 16 PSK, log₂(16) = 4 bits are transmitted in a specific time constant which is known as 16-ary PSK.

B.W. of PSK= 2R_b

Whereas for M-ary PSK,

Hence M-ary is preferred due to less bandwidth.

In 16 APSK, Amplitude and Phase both vary in a pattern as shown in the figure.
16 QAMis a signal in which two carriers shifted in phase by 90 degrees are modulated and combined. As a result of their 90° phase difference, they are in quadrature.
Often one signal is called the In-phase or “I” signal, and the other is the quadrature or “Q” signal.

Hence the given figure is 16 QAM.

So the solution is option (2).

MPPGCL JE Electronics Mock Test - 3 - Question 33

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Consider the RLC circuit shown below. At a certain frequency f₀ the circuit behaves like a purely resistive circuit. The equivalent resistance of the parallel combination at that particular frequency f₀ is.

A.
2.0 Ω
B.
1.0 Ω
C.
0.5 Ω
D.
3.0 Ω

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 33

Concept:

Any given circuit will be at the resonant frequency when the net impedance is purely resistive.

Calculation:

Calculating the net-admittance of the given combination because the impedances are in parallel.

Y_net = Y₁ + Y₂

Replacing s with jω, we get;

At resonance, the imaginary part is zero and Y_net will be only real.

Given, R = 1Ω

L = 2μH

C = 1μF.

1 + ω₀² (4μ²) = 2

ω₀² × 4μ² = 1

ω₀ = 1/2μ = 0.5 Mrad/sec

Now, at resonance:

Y_net = ½

R_net = 2Ω

MPPGCL JE Electronics Mock Test - 3 - Question 34

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The third-generation mobile phone are digital and based on

A. AMPS
B. Broadband CDMA
C. CDMA
D. D-AMPS

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 34

There are three generations in cellular telephony.

First Generation: It was designed for voice communication using analog signals. Example: AMPS (Advanced mobile phone system) is one of the leading analog cellular system. It uses FDMA to separate channels in a link. AMPS operates in the ISM 800 MHz band.
Second Generation: To provide higher quality mobile voice communication, these generation of cellular phone are required. These are used for digital voice communication. Example: D- AMPS, CDMA, GSM. The product of evolution of analog AMPS into a digital system is digital AMPS. D-AMPS uses TDMA and FDMA. GSM stands for global system for mobile communication. It provides duplex communication by using two digital bands.
Third generation: It refers to a combination of technologies that provide a variety of services. It can provide both digital data and voice communication. Main goal of third generation cellular telephony is to provide universal personal communication. These are based on broadband CDMA. Example: IMT-DS, IMT-MC etc

MPPGCL JE Electronics Mock Test - 3 - Question 35

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The range of frequency generated by VHF oscillator is -

A.
300 MHz - 3 GHz
B.
Above 3 GHz
C.
30 MHz - 300 MHz
D.
20 kHz - 30 MHz

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 35

Explanation:

Very High Frequency:

It has a range of30MHz to 300MHz and are used intelevision broadcast.

Some ultra high frequencies are also used in television broadcasting.

Additional Information

The frequency spectrum for the complete range is:

Hence option (3) is the correct answer.

MPPGCL JE Electronics Mock Test - 3 - Question 36

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Calculate the value of Norton's current and Norton's equivalent resistance for the given circuit across the terminal AB.

A.
250 mA, 20Ω
B.
125 mA, 26.6 Ω
C.
350 mA, 25.6 Ω
D.
200 mA, 10 Ω

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 36

Norton's Theorem :

Any linear bilateral network consisting of multiple resistances and sources can be replaced by a resistor R_Nin parallel of Current source I_N.

Calculation:

R_N:

Disable all independent sources and find equivalent resistance.

R_N= ( 20 || 10 ) + 20 = 26.6 ohm

I_N:

Short circuited current

Let node voltage of 10 ohm is V

Apply nodal analysis

V = 2.5V

I_N= V /20 = 125 mA

MPPGCL JE Electronics Mock Test - 3 - Question 37

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Which of the following methods suitable for the stability analysis of sampled data control systems?

A.
Jury’s stability test
B.
Root locus technique
C.
Bilinear transformation
D.
All of the above

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 37

Sampled data system:

When the signal or information at any or some points in a system is in the form of discrete pulses, then the system is called discrete data system or sampled data system.

The following methods suitable for the stability analysis of sampled data control systems.

Jury’s stability test
Root locus technique
Bilinear transformation

Merits:

Systems are highly accurate, fast and flexible
Use of time sharing concept of digital computer results in economical cost and space.
Digital transducers used in the system have better resolution.
The digital components are less affected by noise, non-linearities and transmission errors of noisy channel.

Demerits:

Conversion of analog signals to discrete-time signals and reconstruction introduce noise and errors in the signal.
Additional filters have to be introduced in the system if the components of the system does not have adequate filtering characteristics.

MPPGCL JE Electronics Mock Test - 3 - Question 38

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In a radar system, the term 'Rate-Race' is used in connection with

A.
Modulator
B.
Pulse characteristics
C.
Receiver Bandwidth
D.
Duplexer

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 38

The hybrid ring is used primarily in high-power radar and communications systems where it acts as a duplexer - allowing the same antenna to be used for transmitting and receiving functions.
The hybrid ring is also called as rat-race circuits.
It consists of an annular line of proper electrical length to sustain standing waves, to which four arms are connected at proper intervals by means of series or parallel junctions.

A duplexer is an electronic device that allows bi-directional (duplex) communication over a single path.
In radar and radio communications systems, it isolates the receiver from the transmitter while permitting them to share a common antenna.
Most radio repeater systems include a duplexer.

MPPGCL JE Electronics Mock Test - 3 - Question 39

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How many 3 × 8 decoders are required to Construct a 4 × 16 decoder?

A.
Two
B.
Three
C.
Four
D.
Five

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 39

Concept:

A decoder is a combinational circuit that converts n lines of input into 2nlines of output.

Decoder expansion

Number of D₂decoder required is given as:

∑ K = K₁+ K₂+ K₃+ ------

Calculation:

Given decoder 1 is 3 × 8 and the second decoder is 4× 16

Number of 3 × 8 decoders = 2+ 0

Number of 3 × 8 decoders = 2

The implementation is shown below.

Important Point

MPPGCL JE Electronics Mock Test - 3 - Question 40

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Simplify the expression

A.
= 1
B.
= 0
C.
= A
D.
= A̅

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 40

Concept:

De Morgan’s law states that:

Application:

Applying De-Morgan's law in the above expression can be written as:

F = 0

Important Point

MPPGCL JE Electronics Mock Test - 3 - Question 41

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Which of the following statements about TCP/IP are incorrect?

A. TCP/IP has high-latency.
B. Data packets sent over TCP/IP are private.
C. TCP is used by Secure Shell (SSH)
D. TCP/IP uses checksum to check error correction.

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 41

The correct answer isData packets sent over TCP/IP are private.

Key Points

Transmission Control Protocol (TCP) is a communications standard that enables application programs and computing devices to exchange messages over a network.
It is designed to send packets across the internet and ensure the successful delivery of data and messages over networks.
TCP guarantees the integrity of the data being communicated over a network. Before it transmits data, TCP establishes a connection between a source and its destination, which it ensures remains live until communication begins.
- It then breaks large amounts of data into smaller packets, while ensuring data integrity is in place throughout the process.
Data packets sent over TCP/IP are not private, which means they can be seen or intercepted. Hence option 2 is correct.
So, it is recommended to avoid using public Wi-Fi networks to send private data and ensure information is encrypted. One way to encrypt data being shared through TCP/IP is through a virtual private network (VPN).
TCP/IP communications provide end-to-end error checking of data transmission for reliable and sequential exchange of data, guarantees the sequential delivery of data packets and so it comes with high latency.

Additional Information

Error control in TCP is mainly done through the use of three simple techniques: Checksum, Acknowledgement andRetransmission.
- Every segment contains a checksum field which is used to find corrupted segments. If the segment is corrupted, then that segment is discarded by the destination TCP and is considered lost.
- In acknowledgement affirm that the data segments have been delivered. Control segments that contain no data but have sequence numbers will be acknowledged as well but ACK segments are not acknowledged.
- In retransmission, when a segment is missing, delayed in delivering to a receiver or corrupted when it is checked by the receiver then that segment is retransmitted again.
- Segments are retransmitted only during two events: when the sender receives three duplicate acknowledgements (ACK) or when a retransmission timer expires.
High-level protocols that need to transmit data like File Transfer Protocol (FTP), Secure Shell (SSH), and Telnet use TCP Protocol.

MPPGCL JE Electronics Mock Test - 3 - Question 42

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A LTI system is described by the difference equation 3y[n] = 2y [n - 2] – 2x[n] + x[n - 1]

The system is

A. Causal but not stable
B. Causal and stable
C. Non - causal and stable
D. None of the above

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 42

Since output of system for any value of n is only depends on values of input at n or previous values the system, it is causal system

3y[n] = 2y [n - 2] – 2x[n] + x [n - 1]

Taking z – transform

3y[z] = 2y (z) z^{- 2} – 2x[z] + x (z) z^{- 1}

It has poles at

Since poles lie in unit circle |z_p| < 1 the system is stable

MPPGCL JE Electronics Mock Test - 3 - Question 43

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What is an operating system?

A. A program
B. A hardware
C. A memory chip
D. RAM

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 43

The Correct Answer isOption (1)i.e A program.

An operating system is a software or a program that manages computer hardware, software resources, and provides common services for computer programs.
An operating system is an interface between the computer user and computer hardware.
The dominant desktop operating system is Microsoft Windows with a market of around 82.74%, followed by macOS by Apple Inc.
Some popular operating system include Linux operating system, Windows operating system, VMS, OS/400, macOS, etc.
Some of the important functions of an operating system are :
- Memory Management
- Processor Management
- Device Management
- File Management
- Security
- Job Accounting
- Co-ordination between other software and users.

MPPGCL JE Electronics Mock Test - 3 - Question 44

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Transport layer can be

A. Connectionless
B. Connection Oriented
C. Bothaandb
D. None of These

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 44

The transport layer can be either connectionless or connection oriented. A connectionless transport layer treats each segment as an independent packet and delivers it to the transport layer at the destination machine. A connection oriented transport layer makes a connection with the transport layer at the destination machine first before delivering the packets. After all the data are transferred, the connection is terminated.

MPPGCL JE Electronics Mock Test - 3 - Question 45

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In the TCP/IP model, encryption and decryption are functions of ____ layer

A.
data link
B.
network
C.
Transport
D.
Application

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 45

TCP/IP model, RFC 1122

Therefore option 4 is correct.

MPPGCL JE Electronics Mock Test - 3 - Question 46

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Which one of the following logic family comprises of BJTs?

A.
TTL
B.
CMOS
C.
NMOS
D.
FET

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 46

Logic families are the logic circuits having identical electrical parameters.

It is a group of compatible ICs with the same logic levels and supply voltages for performing various logic functions.
They are fabricated using a specific circuit configuration which is referred to as a Logic family.
The logic family is designed by considering the basic electronic components such as resistors, diodes, transistors, and MOSFET; or combinations of any of these components.

Different logic families of digital ICs that have been introduced commercially are listed in the table.

TTL:

Important Point

The differences between different logic families are as follows:

MPPGCL JE Electronics Mock Test - 3 - Question 47

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For the circuit shown in the figure, calculate the maximum output voltage V_L -

A.
−5 V
B.
+10 V
C.
+5 V
D.
+3.33 V

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 47

Solution

Concept

During the +ve cycle of input

Diode D₂conducts and D₁ is off
from the voltage division rule, the voltage across the 2kΩ resistance around which output is calculated is
V = 0.5 V_in

During the -ve cycle of input

D₂is off and D₁conducts
from voltage division rule
V = -0.5 V_in

Hence during the full cycle of input,

we will observe the following output voltage waveform

We could observe that it is a full-wave rectified wave

Calculation

The maximum output voltage is the

V_max= |0.5 V_in|

V_max= + 5volt

Hence the correct answer is + 5 volt

The correct option is 3.

MPPGCL JE Electronics Mock Test - 3 - Question 48

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Which one of the following is the advantage of FIR filter over IIR filter ?

A.
FIR filter can have an exact linear phase.
B.
FIR filter is always unstable.
C.
For FIR filter, the design methods are non-linear.
D.
FIR filter cannot be realized efficiently in hardware

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 48

Properties of FIR filter:

Impulse Response isof finite duration
It is a non-recursive system because there is no feedback path present in the FIR filter.
They are always be designed as a linear phase.
Stability is guaranteed is the FIR system.
In FIR, to obtain the same frequency response as IIR large number of additions & multiplication is reviewed, so the speed is very slow.

Properties of IIR filter:

The impulse response is of infinite duration.
IIR system is also known as a recursive system because there is a feedback path from output to input.
IIR system cannot be designed as a linear phase system.
Stability cannot be guaranteed.
In the IIR system, fewer multiplications and additions are reviewed, so processing speed is very fast.

Linear phase:

FIR can be easily designed to have a exact linear phase
No phase distortion is introduced into the signals to be filtered.
All frequencies are shifted in time by the same amount.
Increasing the order of the filter, but keeping all else the same, increases the sharpness of the filter roll-off
The sharpness of the FIR and IIR filters is very different for the same order.
Because of the recursive nature of an IIR filter, where part of the filter output is used as an input, IIR filters have sharper roll-off with the same order FIR filter.

The characteristics of analog and digital filters are summarized below:

MPPGCL JE Electronics Mock Test - 3 - Question 49

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A low earth orbit satellite can provide large signal strength at an earth station because:

A. Path loss is low
B. These orbits are immune to noise
C. Large solar power can be generated at these orbits
D. Lower microwave frequencies in s-band can be used

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 49

A low earth orbit satellite can provide large signal strength at the earth station because path loss is low since low earth orbit (LEO) is the nearest orbit to the earth so path loss is minimum.

Free space path loss

Where R is the distance between stations as we can see.

Free path loss ∝ (distance)²

As (distance ↑) → (Pathloss ↑) (R_x signal strength ↓)

MPPGCL JE Electronics Mock Test - 3 - Question 50

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Power dissipation is negligibly small in:

A.
SCR
B.
BJT
C.
MOSFET
D.
CMOS

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 50

Complementary Metal-oxide-semiconductor (CMOS) uses complementary & symmetrical pair of P-type & n-type MOSFETS.

The two important characteristics of CMOS devices arehigh noise immunity and low power dissipation.
CMOS devices dissipate less power than NMOS devices because the CMOS dissipates power only when switching (“dynamic power), whereas N channel MOSFET dissipates power whenever the transistor is on because there is a current path from Vddto Vss.
In a CMOS, only one MOSFET is switched on at a time. Thus, there is no path from voltage source to ground so that a current can flow. Current flows in a MOSFET only during switching.
Thus, compared to N-channel MOSFET has the advantage of lower drain current from the power supply, thereby causing less power dissipation.

MPPGCL JE Electronics Mock Test - 3 - Question 51

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In one of the pairs of protocols given below, both the protocols can use multiple TCP connections between the same client and the server. Which one is that?

A. HTTP, FTP
B. HTTP, TELNET
C. FTP, SMTP
D. HTTP, SMTP

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 51

HTTP: Multiple TCP connections can be used at client and server end.

FTP: For FTP also multiple connections can be used at the same time.

TELENT: Only one TCP connection is allowed at a time.

SMTP: SMTP also allows only one TCP connection at a time.

MPPGCL JE Electronics Mock Test - 3 - Question 52

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In QAM, both ________ of a carrier frequency vary.

A.
frequency and amplitude
B.
phase and amplitude
C.
phase and frequency
D.
Phase and frequency

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 52

Digital to Analog Modulation technique is shown.

Asshown QAM is the mixture of ASK and PSK.

Hence,amplitude and the phase of thecarrier frequency bothvary with the message signal.

Constellation diagram of a QAM signal with 2 different amplitude levels and 8 different phases is shown.

MPPGCL JE Electronics Mock Test - 3 - Question 53

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The condition for a resistor to have the same value of resistance at medium frequency as with D.C. is -

A.
CR²= L
B.
CR²=2L
C.
CR²= ωL
D.
CR²= 2 ωL

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 53

The resistance has interelectrode capacitances and inductances which lead to offer reactance to the circuit under non-DCconditions

At high and medium frequencies the resistor is represented as shown below

The Equivalent impedance is

Z_eq=

solving we get

Z_eq=+

Comparing with R_eq+L_eq

We have R_eq=

If resistance is needed to be independent of frequency and maintain the same value or resistance value as in DC,

Then

⇒ CR²- 2L must be 0 ⇒ CR²- 2L = 0

⇒CR²=2L and R_eq= R

Therefore the correct option is 2.

MPPGCL JE Electronics Mock Test - 3 - Question 54

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In a semiconductor device, if the Fermi level (E_F) is positioned at the conduction band (E_c). Determine the approximate probability of finding electrons in states at (E_c + kT)

(where k is Boltzmann constant and T is device temperature in Kelvin)

A. 0.18
B. 0.27
C. 0.38
D. 0.52

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 54

Concept:

In a semiconductor, the probability of finding an electron at an energy level E is given by:

E_F = fermi level

Calculation:

Given, E_F = E_C

The probability of finding electrons in the energy level E = E_C + KT will be:

f(E_C + KT) = 0.268

MPPGCL JE Electronics Mock Test - 3 - Question 55

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Consider a test signal m(t) defined by a hyperbolic tangent function m(t) = A tanh(βt), Where A and B are constant. Determine the minimum step sideΔ for delta modulation of this signal which is required to avoid slope overload:

A. Δ≥ Aβ T_s
B. Δ≤ Aβ T_s
C. AΔ≤β T_s
D. βΔ≥ A T_s

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 55

Concept:

To avoid slope overload, we require

Calculations:

Given,

m(t) = A tanh(βt)

To avoid slope overload

Since,

It follows the maximum value of sec h(βt) is 1, which occurs at time t = 0.

Hence |sec h(βt)|_max = 1

∴

Δ≥ Aβ T_s.

Hence option (1) is correct.

MPPGCL JE Electronics Mock Test - 3 - Question 56

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The Zeroth generation ‘0 G’ (0G) of mobile communication used:

A. Analog circuit switched technology with FDMA
B. Radio telephones
C. CDMA
D. WCDMA

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 56

0 Generation:

“0G” refers to pre-cellular mobile telephony technology, such as radio telephones that were used in cars before the advent of cell phones.

1^st Generation:

Analogue circuit-switched technology is used for this system, with frequency division multiple access (FDMA).

2^nd Generation:

2G systems are digital cellular systems
2G digital technology is divided into two standards: time division multiple access (TDMA) and code-division multiple access (CDMA)

2.5 Generation:

The mobile technology using GPRS standard has been termed as 2.5G
Provided enhanced data rates for GSM evolution (EDGE)

3^rd Generation:

The 3G technology added multimedia facilities to 2.5G phones
Examples of 3G system are universal mobile telecommunication systems (UMTS) and international mobile telecommunications at 2,000 MHz (IMT-2000)

MPPGCL JE Electronics Mock Test - 3 - Question 57

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In the case of optical fiber transmission link, LED’s and LASER’s are used as ______

A.
Detectors
B.
Sources
C.
Repeaters
D.
Amplifiers

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 57

Being a single wavelength light source with uniform phase:

Laser light travels smoothly with verylittle dispersion.
Makes itidealfor long-distance communication.
For fiber optics with glass fibers, light in the infrared region is widely used which has wavelengths longer than visible light, typically of 850, 1300, and 1550 nm.
Transmitters (Lasers or LEDs) emit in infrared regions and are used as a light source for fiber transmission.
The laser acts as the best source in optical communication becauseit is a single wavelength light source.
Ordinary light contains many different wavelengths of light, differences emerge in the speed of the transmission, reducing the number of signals that can be transmitted in any set time.
Receivers (photodetectors) at these particular wavelengths are also easy to construct.

Notes:

We use infrared because the attenuation of the fiber is much less at those wavelengths.
The attenuation of glass optical fiber is caused by two factors, absorption, and scattering.
Absorption occurs in several specific wavelengths called water bands due to the absorption by minute amounts of water vapor in the glass.

MPPGCL JE Electronics Mock Test - 3 - Question 58

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‘Bluetooth' technology is

A. Satellite communication
B. Wired communication between devices
C. Wireless communication between devices
D. Communication between landline to mobile phone

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 58

The correct answer isWireless communication between devices.

Key Points

Bluetoothtechnology

Bluetooth is a Wireless Personal Area Network (WPAN) technology and is used for exchanging data over smaller distances.
Bluetooth technology allows devices to communicate with each other without cables or wires. Bluetooth relies on short-range radio frequency, and any device that incorporates the technology can communicate as long as it is within the required distance.
Bluetooth is used for short-range communications between device peers, as well as a device to the peripheral. The number and variety of peripherals that communicate via Bluetooth are immense—wireless headsets for hands-free cell phone use, keyboards, mice, videogame controllers, audio speakers, you name it.

MPPGCL JE Electronics Mock Test - 3 - Question 59

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The condition for the low-loss transmission line is

A.
R >>ωC and G >>ωL
B.
R <<ωL and G <<ωC
C.
D.
RG >> LC

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 59

Concept:

Propagation constant for the transmission line is given by:

For lossless line α = 0

If R ≪ ωL and a ≪ ωC, the propagation constant becomes:

---(1)

Comparing equation (1) with α + jβ:

α = 0. Hence the line is lossless.

Important points:

A transmission line is also lossless when:

R = 0 and G = 0.

Here also α = 0 and the line is lossless

∴ The conditions for lossless line includes:

1) R = 0; G = 0, or

2) ωL ≫ R & ωC ≫ G

MPPGCL JE Electronics Mock Test - 3 - Question 60

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If [A] Matrix is Incidence matrix then which one of the following is true?

A.
|A| = 0 (For closed loop)
B.
Adj [A] / |A| = 0 (For closed loop)
C.
[A] = 1 (For closed loop)
D.
|A| = 1 (For closed loop)

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 60

Incidence matrix:

It is the matrix that gives a relation between the branches and nodes.
The rows of the incidence matrix [A] represent the number of nodes and the column of the matrix represents the number of branches in the given graph.
If there are‘n’ number of rows in a given incidence matrix[A], that means in a graph there are ‘n’ number of nodes.
Similarly, if there are‘b’ numbers of columns in that given incidence matrix[A], that means in that graph there are ‘b’ number of branches.
We can construct the incidence matrix for the directed graph. We can draw a graph with the help of the incidence matrix.
The algebraic sum of elements of all the columns is zero.
The rank of the incidence matrix is (n–1).
The determinant of the incidence matrix of a closed loop is zero.

Hence, an Only statement I is correct.

Example:

Consider the below-closed graph:

The incidence matrix is written as:

We observe that the algebraic sum of elements of all the columns is zero.

The determinant of the given closed path is calculated as:

∴ The determinant of the incidence matrix of a closed loop is zero.

MPPGCL JE Electronics Mock Test - 3 - Question 61

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If for a BJT the value ofα is 0.98, then the value ofβ will be _______.

A.
49
B.
51
C.
101
D.
120

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 61

Concept:

Where,

β = common-emitter current gain

α = Common base current gain

Calculation:

Common base current gain = α = 0.98

Note:

Where I_C= Collector current

I_E= Emitter current

I_B= Base current

Important Points

The DC current gain of a common collector is therefore given by the ratio of emitter current to the base current, i.e.

I_E= Emitter Current

I_B= Base Current

Also, I_E= I_B+ I_C

DC current gain will be:

The DC current gain for a common emitter configuration is defined as:

Equation (1) now becomes:

Important Differences between different transistor configuration is as shown:

MPPGCL JE Electronics Mock Test - 3 - Question 62

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Overdamping will make instruments:

A.
robust
B.
accurate
C.
fast
D.
slow

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 62

Concept:

Classification of instruments on the basis of damping:

1.) Overdamping:

The value of the damping ratio is greater than 1.
In case of overdamping, the instrument will become slow and lethargic and rise very slowly from its zero position to a steady state position.

2.) Undamped:

The value of the damping ratio is equalto 0.
In this case, the instrumentnever settles down and always oscillates about its mean position.

3.) Under-damped:

The value of the damping ratio lies between 0 to 1.
In this case, the system initially oscillates about its mean position but settles down to a steady state.

4.) Critical-damping:

The value of the damping ratio is equalto 1.
This damping returns the system to equilibrium as fast as possible without any oscillations.

MPPGCL JE Electronics Mock Test - 3 - Question 63

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The ALU uses ______ to store intermediate result.

A. Cache
B. Registers
C. Accumulators
D. Stack

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 63

Concept:

ALU (Arithmetic logic unit) is a circuit to perform arithmetic and logic operations.

Explanation:

ALU uses accumulator (a register) for storing intermediate results of arithmetic and logic operations.
Control unit provides the data to the ALU and directs the ALU to perform specific operations.
ALU stores the result in accumulator which is a type of register. Accumulator is a kind of temporary storage device which stores the intermediate results and if some previous results are already stored on the accumulator then it automatically overwrites the new results and previous results are removed.

MPPGCL JE Electronics Mock Test - 3 - Question 64

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Radar principle is used in

A. detection of aircraft
B. telephony
C. electron microscope
D. all of the above

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 64

The radar principle is used in thedetection of aircraft.

Radar:

Radar is an acronym for Radio Detection and Ranging.
The term "radio"refers to the use of electromagnetic waves with wavelengths in the so-calledradiowave portion of the spectrum, which covers a wide range from 104km to 1 cm.
It is basically an electromagnetic system used to detect the location and distance of an object from the point where the RADAR is placed.
Itworks by radiating energy into space and monitoring the echo or reflected signal from the objects.
It works on radio frequency in the range of about 3 kHz to 300 GHz.
But most of the RADAR operates between 220 MHz to 35 GHz.

Radar systems have been used in military applications for

Ground surveillance
Missile control, fire control,
Air traffic control (ATC),
Moving target indication (MTI),
Weapons location
Vehicle search

MPPGCL JE Electronics Mock Test - 3 - Question 65

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The phenomenon of pulse spreading in optical fibre is known as:

A.
Scattering
B.
Dispersion
C.
Attenuation
D.
Inter symbol Interference

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 65

Scattering:

Scattering is the loss of signal caused by the diffusion of a light beam, where the diffusion itself is caused by microscopic variations in the transmission medium
Scattering typically happens when a light signal hits an impurity in the fibre

Dispersion:

Dispersion is defined as pulse spreading in an optical fibre
As a pulse of light propagates through a fibre, elements such as numerical aperture, core diameter, refractive index profile, wavelength, and laser linewidth cause the pulse to broaden

Attenuation:

The attenuation of an optical fibre measures the amount of light lost between input and output
Total attenuation is the sum of all losses
For a given fibre, these losses are wavelength-dependent

Inter-symbol Interference:

This is the effect of dispersion on optical fibre
Intersymbol interference occurs when the pulse spreading caused by dispersion causes the output pulses of a system to overlap

MPPGCL JE Electronics Mock Test - 3 - Question 66

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A _________stores such instructions that are required to start a computer.

A. ROM
B. RAM
C. Mother Board
D. PROM

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 66

The correct answer is option 1 i.e ROM

The startup instructions are stored on ROM in a computer. It isa part of BIOS (Basic Input Output System)

MPPGCL JE Electronics Mock Test - 3 - Question 67

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The amplifier in which each input will be multiplied by a different factor and then summed together is called as:

A.
non-weighted amplifier
B.
scaling amplifier
C.
averaging amplifier
D.
multiple-way amplifier

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 67

One type of summing amplifier is the SCALING AMPLIFIER.

This circuit multiplies each input by a factor (the factor is determined by circuit design) and then adds these values together.

The factor that is used to multiply each input is determined by the ratio of the feedback resistor to the input resistor.

For example, we can design a circuit that would produce the following output from three inputs (E₁, E₂, E₃)

= [(k₁× E₁) + (k₂× E₂) + (k₃× E₃)]

Using input resistors R₁ for input number one (E₁), R₂ for input number two (E₂), R₃ for input number three (E₃), and R₄ for the feedback resistor,

We can calculate the values for the resistors.

MPPGCL JE Electronics Mock Test - 3 - Question 68

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Booths algorithm is used for

A.
floating point division
B.
restoring division
C.
integer multiplication
D.
square root

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 68

Here’s the implementation of the flowchart.

MPPGCL JE Electronics Mock Test - 3 - Question 69

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The deflecting torque of a moving iron instrument is

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 69

Moving Iron Instruments are the most common type of ammeter and voltmeter used at power frequencies in laboratories.

These instruments are very accurate, cheap, and rugged as compared to other AC instruments.

Working Principle of Moving Iron Instruments:

In Moving Iron Instruments, a plate or van of soft iron or of high permeability steel forms the moving element of the system.

The iron van is so situated that it can move in the magnetic field produced by a stationary coil.

The below figure shows a simple moving iron instrument.

The stationary coil is excited by the current or voltage under measurement.

When the coil is excited, it becomes an electromagnet, and the iron van moves in direction of offering a low reluctance path.

Thus the force of attraction is always produced in a direction to increase the inductance of the coil.

Mind that as the van follows the low reluctance path, the net flux in the air gap will increase which means increased flux linkage of the coil, and hence inductance of coil will increase.

It shall also be noticed that the inductance of the coil is variable and depends on the position of the iron van.

Torque Equation of Moving Iron Instruments:

Deflecting torque in Moving iron Instruments is given as

Td =(1/2)I²(dL/dƟ)

In moving iron instruments, the controlling torque is provided by spring. Controlling torque due to spring is given as

Tc = KƟ

Where K = Spring constant

Ɵ = Deflection in the needle

In equilibrium state, deflecting and controlling torque shall be equal as below.

Deflecting Torque = Controlling Torque

⇒ Td = Tc

⇒ (1/2)I²(dL/dƟ) = KƟ

⇒Ɵ = (1/2)(I²/K)(dL/dƟ)

MPPGCL JE Electronics Mock Test - 3 - Question 70

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What is the reason to have secondary memory?

A.
Limited capacity of main memory
B.
Main memory is expensive
C.
Main memory possesses highly volatility
D.
All of the above

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 70

We require secondary memory for our computer systems as:

Limited capacity of main memory - Although we can add main memory to a system but it is extremely difficult for general users. And we also want to store all our programs and data inside the main memory. But main memory having limited capacity,can accommodate only a limited amount of data and small number of programs.
Main memory is expensive -Main memory is faster than secondary memory and hence, it is highly expensive. Therefore, we can have multiple secondary storages in the cost of a single main memory.
Main memory possesses highly volatility - Main memory is volatile storage whereas secondary memory is non-volatile storage, that is, main memory will lose its content as soon as its power is turned off whereas secondary memory will retain the content even when the power is lost.

All these reasons makes us to use secondary memory along with main memory.

MPPGCL JE Electronics Mock Test - 3 - Question 71

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The stage used to eliminate the noise effects in a FM detector is called –

A. Limiter
B. Discriminator
C. AFC
D. Noise eliminator

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 71

At the transmitter the amplitude of the modulated signal is constant, noise and interfering stations produce amplitude changes on the modulated signal
The purpose of the limiter is to provide a constant level of the signal to the FM demodulator, thus reducing the effect of signal level changes in the output

MPPGCL JE Electronics Mock Test - 3 - Question 72

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In which modulation technique does the phase of the carrier signal is changed by varying the sine and cosine inputs at a particular time.

A.
Frequency modulation
B.
Phase shift key modulation
C.
Analog modulation
D.
Pulse code modulation

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 72

PSK(Phase Shift Keying):

For binary ‘1’ → S₁(A) = Acos 2π f_Ct

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

The Constellation Diagram Representation is as shown:

Important Point

ASK, PSK and FSKare 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):

For binary ‘1’ → S₁(A) = Acos 2π f_Ht

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

ASK(Amplitude Shift Keying):

InASK (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 f_ct

For binary ‘0’ → S₂(t) = 0

The Constellation Diagram Representation is as shown:

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

MPPGCL JE Electronics Mock Test - 3 - Question 73

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Find the voltage across 30 Ω resistor in the given circuit.

A.
105 V
B.
45 V
C.
75 V
D.
30 V

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 73

The given circuit is redrawn as:

Applying KVL around the loop, we get:

100 - 8I - 40 - 30I - 2I = 0

60 = 40 I

I = 1.5 A

Now, the voltage across the 30Ω resistor will be:

V_30Ω = 1.5× 30

V_30Ω = 45 V

MPPGCL JE Electronics Mock Test - 3 - Question 74

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Find out even and odd part of signal x(t) = (2 + sin t)²

A. 4 + sin²t, 4 sin t
B. 2 + sin²t, 2 sin t
C. 4, 4 sin t + sin² t
D. 2, sin t

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 74

Concept:

Any given signal x(t) can be written as the sum of its even part and odd part, i.e.

x(t) = xe(t) + xo(t)

xe(t) = Even part of g(t) calculated as:

xo(t) = Odd part of x(t) calculated as:

Calculation:

Given:

x(t) = (2 + sin t)²

x(t) = 4 + sin²t + 4 sin t

x(-t) =4 + sin2t-4 sin t

Even part of g(t) calculated as:

= 4 +sin2t

Odd part of x(t) calculated as:

= 4 sin t

MPPGCL JE Electronics Mock Test - 3 - Question 75

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Types of MOSFET

(P) P – Channel Enhancement MOSFET

(Q) P – Channel Depletion MOSFET

(R) N – Channel Enhancement MOSFET

(S) N – Channel Depletion MOSFET

A.
P – 4, Q – 2, R – 1, S – 3
B.
P – 3, Q – 2, R – 4, S – 1
C.
P – 1, Q – 3, R – 2, S – 4
D.
P – 2, Q – 4, R – 3, S - 1

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 75

TheEnhancement mode MOSFET's are equivalent to a "Normally Open" switch that requires gate-source voltage to switch ON the device.
If the positive voltage (+V_GS) is applied to an n-channel gate terminal, then only the channel will conduct and the drain current starts to flow through the channel.
If thebias voltage is zero or negative(- V_GS)then the transistorswitches OFFand the channel stays in the non-conductive state resulting in the drain current to be zero.

Transfer characteristics of an n-channel depletion MOSFET is as shown:

We observe that for V_GS= 0, drain current is flowing.

Transfer characteristics of p-channel depletion MOSFET is as shown:

Transfer characteristics of n-channelEnhancement MOSFET is as shown:

We observe that for V_GS= 0, there is no drain current flowing as there is no conduction region at V_GS= 0, in enhancement mode MOSFET.

Transfer characteristics of p-channelEnhancement MOSFET is as shown:

MPPGCL JE Electronics Mock Test - 3 - Question 76

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If 1 is added to the denominator of a fraction it becomes1/2. If 1 is added to the numerator it becomes 1. The product of numerator and denominator of the fraction is

A.
14
B.
12
C.
10
D.
6

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 76

Let the numerator = x and denominator = y

y + 3 = 2y ⇒ y = 3

x +1= 3 ⇒ x = 2

∴ xy = 2 × 3 = 6

MPPGCL JE Electronics Mock Test - 3 - Question 77

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Directions: Study the following question carefully and choose the right answer.

In a certain code ‘TERMINAL’ is written as ‘SDQLJOBM’. How is CREDIBLE written in that code?

A.
BQDCJCMF
B.
BQDCHAKD
C.
DSFEJCMF
D.
DSFEHAKD

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 77

Hence, option A is correct.

MPPGCL JE Electronics Mock Test - 3 - Question 78

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Sunny can complete a work in 15 days. If Bobby is 60% as efficient as Sunny, in how many days can Bobby complete the work?

A.
20
B.
25
C.
30
D.
40

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 78

MPPGCL JE Electronics Mock Test - 3 - Question 79

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A round punched paper is given as shown in the question figure. Figure out from the four alternatives as to how it will appear when folded.
Question Figure

Answer figure

A.
1
B.
2
C.
3
D.
4

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 79

MPPGCL JE Electronics Mock Test - 3 - Question 80

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The shadow of a building is 10 m long when the point of rise of the sun is 60°. Discover the building's stature.

A.
16.32m
B.
17.32 m
C.
18.32 m
D.
19.32m

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 80

MPPGCL JE Electronics Mock Test - 3 - Question 81

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Direction: Study the following information carefully and answer the given questions besides.

A.
Aarti
B.
Manu
C.
Kirti
D.
Sona

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 81

From the common explanation, we get Sona sits opposite Bobby.

Hence, Option E is correct.

Final Arrangement:

Common Explanation:

References:

Eight students Kirti, Ayushi, Manu, Sona, Aarti, Bobby, Ankita and Divya are sitting around the square table.

Four students sit on each side while rest on each corner.

Students at the side face away from the centre and students at the corner faces towards the centre.

Inferences:

So, we need to arrange as follows to accommodate the information as stated in the question,

References:

Divya sits third to the left of Ayushi.

Kirti sits opposite to Manu and both faces away from the centre.

Inferences:

From the above references, we get the following two different cases:

Case 1:

Case 2:

References:

Neither Kirti nor Manu are beside Ayushi.

Sona sits immediate left of Ankita, who sits on the side of the table.

Bobby is not an immediate right of Ayushi.

Bobby is not adjacent to Manu.

Inferences:

So, from this case 2 will be eliminated as neither Kirti nor Manu are beside Ayushi.

Now, from case 1 we get the following final arrangements:

MPPGCL JE Electronics Mock Test - 3 - Question 82

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The HCF of two numbers is 15 and their LCM is 300. If one of the number is 60, the other is :

A.
75
B.
50
C.
100
D.
65

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 82

Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
First number × Second number = HCF × LCM

MPPGCL JE Electronics Mock Test - 3 - Question 83

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Directions: Read the given information carefully and answer the questions given beside:

Q.Who sits two places away from Y?

A.
X
B.
Z
C.
C
D.
either B or C option

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 83

Following the common explanation, we get “C and Z sits two places away from Y”.
Hence, option D is correct.
Final Arrangement:

MPPGCL JE Electronics Mock Test - 3 - Question 84

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Which revolution in the world inspired the Indians to set up a socialist economy?

A.
French Revolution
B.
Turkish Revolution
C.
Russian Revolution
D.
American War of Independence

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 84

During that time Lenin introduced the idea of a socialist, cooperative economy and emphasized collectivisation .Hence, these values inspired Indians to set up a socialist economy.

MPPGCL JE Electronics Mock Test - 3 - Question 85

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When did Zimbabwe attain independence and from whom?

A.
1970, from Black minority rule
B.
1880, from White minority rule
C.
1980, from Americans
D.
1980, from White minority rule

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 85

The country gained official independence as Zimbabwe on18 April 1980. The government held independence celebrations in Rufaro stadium in Salisbury, the capital.

MPPGCL JE Electronics Mock Test - 3 - Question 86

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Which of the following does not include election procedure?

A.
Voting
B.
Nomination of Candidate
C.
Booth capturing
D.
Canvassing

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 86

Booth capturing is a type of electoral fraud in which party loyalists "capture" a polling booth and vote in place of legitimate voters to ensure that their candidate wins.

MPPGCL JE Electronics Mock Test - 3 - Question 87

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Which of the following is not an agent of erosion and deposition

A.
Running water
B.
Glaciers
C.
Earthquake
D.
Wind

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 87

MPPGCL JE Electronics Mock Test - 3 - Question 88

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State animal of Haryana is:

A. Black buck
B. Asiatic Lion
C. Chinkara
D. Tiger

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 88

The correct answer is theBlackbuck.

Important Points

The Blackbuck is the state animal of Haryana. In India hunting of Blackbuck is ban under theWildlife Protection Act of 1972, due to excessive hunting andhabitat degradation of Blackbucks during the 20th century.
The Blackbuck or an Indian antelope are found in India, Nepal, and Pakistan. They have a lifespan of approximately 10 to 15 years.

Additional Information

Black Francolin of thepheasant family is the State bird of Haryana.
There are 2 National parks located in Haryana viz,Sultanpur National Park located in Gurugram andKalesar National Park located in Yamunanagar.
Chinkara alsoknown as Indian Gazelleis thestate animal of Rajasthan.

MPPGCL JE Electronics Mock Test - 3 - Question 89

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While working with 75% of its efficiency a pipe can fill a tank in 80 minutes. In how many minutes can the pipe fill the tank while working with 100% efficiency?

A.
50
B.
60
C.
75
D.
90

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 89

Hence, Option B is correct.

MPPGCL JE Electronics Mock Test - 3 - Question 90

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Seven persons B, C, D, E, F, G and H are born on seven different months of a year starting from January, but not necessarily in the same order. Who was born in July?

A.
If the data in statement I alone is sufficient to answer the question.
B.
If the data in statement II alone is sufficient to answer the question.
C.
If the data in either statement I or statement II is sufficient to answer the question.
D.
If the data in both statement I and statement II is together necessary to answer the question.

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 90

Checking Statement I:

Using the above references, we get the following arrangement:

Clearly, D was born in July.

Hence data in statement I is sufficient to answer the question.

Checking Statement II:

Using the above references, we get the following arrangement:

Clearly, D was born in July.

Hence data in statement II is sufficient to answer the question.

Here, the data in either Statement I or II alone is sufficient to answer the question.

Hence, Option C is correct.

MPPGCL JE Electronics Mock Test - 3 - Question 91

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Sambhu beats Kalu by 30 metres in10 seconds. How much time was taken by Sambhu tocomplete a race 1200 meters.

A.
6 min 30 s
B.
3 min 15 s
C.
12 min 10 s
D.
2 min 5 s

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 91

Kalu’s speed = 3 m/s.
For 1200 m, Kalu would take 400 seconds and Sambhu would take 10 seconds less. Hence, 390
seconds.

MPPGCL JE Electronics Mock Test - 3 - Question 92

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A piece of paper is folded and cut as shown below in the question figures. From the given answer figures, indicate how it will appear when opened.
Question Figure

Answer figure

A.
1
B.
2
C.
3
D.
4

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 92

MPPGCL JE Electronics Mock Test - 3 - Question 93

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A.
18
B.
20
C.
22
D.
24

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 93

MPPGCL JE Electronics Mock Test - 3 - Question 94

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Directions to Solve

In each of the following questions find out the alternative which will replace the question mark.

Question -

68 : 130 :: ? : 350

A.
220
B.
224
C.
222
D.
226

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 94

As, 68 = (4)³ + 4

130 = (5)³ + 5

and 350 = (7)³ + 7

Therefore, ? = (6)³ + 6 = 222

MPPGCL JE Electronics Mock Test - 3 - Question 95

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In a school having roll strength 299, the ratio of boys and girls is 8 : 5. If 25 more girls get admitted into the school, the ratio of boys and girls becomes

A.
46: 35
B.
43: 35
C.
56: 33
D.
53: 33

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 95

MPPGCL JE Electronics Mock Test - 3 - Question 96

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'Dogs' is related to 'Bark' in the same way as 'Goats' is related to:

A.
Bleat
B.
Crow
C.
Grunt
D.
Howl

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 96

As the cry of 'Dog' is called 'Bark' in the same way the cry of 'Goat' is called 'Bleat'.

MPPGCL JE Electronics Mock Test - 3 - Question 97

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The product of two numbers is 1280 and their H.C.F. is 8. The L.C.M. of the number will be ?

A.
160
B.
150
C.
120
D.
140

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 97

Product of two numbers = HCF × LCM
1280 = 8 × LCM
160 = LCM

MPPGCL JE Electronics Mock Test - 3 - Question 98

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Select the related letters from the given alternatives.

ACFOMR : FCARMO :: MNTDEF : ?

A.
TNMFDE
B.
NMTFED
C.
TNMFED
D.
TMNFED

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 98

The logic follows here is:-

1st and 3rd letters are interchanged.

4th and 6th letters are interchanged.

And 2nd letter is placed as it is.

Similarly,

Hence, the correct answer is "TNMFED".

MPPGCL JE Electronics Mock Test - 3 - Question 99

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In a school 30% of the students play football and 50% of students play cricket. If 40% of the students play neither football nor cricket, what percentage of total students play both the games?

A.
10
B.
12
C.
14
D.
20

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 99

The number of students who play both the games = (30 + 50 + 40)% – 100% = 20%
Hence, Option D is correct.

MPPGCL JE Electronics Mock Test - 3 - Question 100

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A.
The data in Statement I alone is sufficient to answer the question while the data in Statement II is not sufficient to answer the question.
B.
The data in both the Statements I and II is not sufficient to answer the question.
C.
The data in Statement II alone is sufficient to answer the question while the data in Statement I is not sufficient to answer the question.
D.
The data in both the Statements I and II together is necessary to answer the question.

Detailed Solution for MPPGCL JE Electronics Mock Test - 3 - Question 100

From Statement 1:
A sits to the immediate left of C.
D sits second to the right of A.

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