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

MPPGCL JE Electronics Mock Test - 10 - Electronics and Communication Engineering (ECE) MCQ


Test Description

30 Questions MCQ Test - MPPGCL JE Electronics Mock Test - 10

MPPGCL JE Electronics Mock Test - 10 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The MPPGCL JE Electronics Mock Test - 10 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The MPPGCL JE Electronics Mock Test - 10 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 - 10 below.
Solutions of MPPGCL JE Electronics Mock Test - 10 questions in English are available as part of our course for Electronics and Communication Engineering (ECE) & MPPGCL JE Electronics Mock Test - 10 solutions in Hindi for Electronics and Communication Engineering (ECE) course. Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free. Attempt MPPGCL JE Electronics Mock Test - 10 | 100 questions in 120 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
MPPGCL JE Electronics Mock Test - 10 - Question 1

Which of the following options is true?

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 1

Explanation:-

The effect of noise in FM signal is determined by the extent to which it changes the frequency of the modulated signal.

Noise in FM

PSD of the noise at the detector output is directly proportional to the square of the frequency.

So we can say the effect of noise is more at higher frequencies in FM.

Discussion:-

Figure of Merit (FOM): It is the ratio of output SNR to the input SNR. FOM depends only on the demodulator.

For AM:

In noise performance, the synchronous detector is better than an envelope detector.

In AM noise has nothing to do with frequency, it depends on the amplitude of modulating signal.

 (for envelope detector).

For PM:

Noise in PM increases exponentially for the entire audio range is a wrong statement.

For FM:

The figure of merit of FM is far better than the other two. So we can say the noise performance of FM is better than AM and PM.

Important Points

1. To improve noise performance in FM we use the concept of pre-emphasis and de-emphasis.

2. WBFM is preferred over NBFM because the earlier has a better figure of merit.

MPPGCL JE Electronics Mock Test - 10 - Question 2

Which information is displayed in the fire control room?

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 2

Fire Control Room

  • For all buildings 15 m in height and above and apartment buildings with a height of 30 m and above, there shall be a Control Room on the entrance floor of the building with a communication system (suitable public address system) to all floors and facilities for receiving the message from different floors.
  • Details of all floor plans along with the details of fire fighting equipment and installations shall be displayed in the fire control room.
  • The fire control room shall also have facilities to detect the fire on any floor through indicator boards connection; fire detection and alarm systems on all floors.
  • The fire staff in charge of the fire control room shall be responsible for maintenance of the various services and the fire fighting equipment and installations in coordination with security, electrical and civil staff of the building.
1 Crore+ students have signed up on EduRev. Have you? Download the App
MPPGCL JE Electronics Mock Test - 10 - Question 3

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

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

Concept:

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

1/r term is the most dominant. 

Calculation:

Given,

E = 3 V/m for r = R

For r2 = 3R,

Electric field E2 will be 

MPPGCL JE Electronics Mock Test - 10 - Question 4

What is maximum power delivered to the load ?

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 4

Concept:

Maximum power is given as 

Where

Vth → Thevenin voltage, Rth → Thevenin resistance

Calculation:

Calculate Vth

Apply kvl in a bigger loop 

Vth + 68 + 60 = 0 

Vth = -128

Calculate Rth

For Rth 

The current source is replaced by an open circuit

The voltage source is replaced by a short circuit

Rth = 3 + 10 + 2 = 15 Ω 

MPPGCL JE Electronics Mock Test - 10 - Question 5

What will be the maximum power that can be distributed in the load in the given circuit?

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 5

Concept:

For maximum power to transfer across the load resistance RL,

RL = Rth

Under this condition, power transfer to the load is

Calculation:

First, we need to calculate equivalent resistance (Rth):

By short-circuiting the DC Source,

Rth = (14||5) + 3

Rth = 3.68 + 3 = 6.68 Ω

Now we have to calculate Vth:

PL = 0.377 W

MPPGCL JE Electronics Mock Test - 10 - Question 6

Moving iron instruments are scaled at ______.

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 6

Moving Iron Instrument

  • MI instruments are used for the measurement of the RMS value of current and voltage.
  • These instruments have a non-uniform scale.
  • It works on the principle of attraction of a single piece of soft iron into the magnetic field.
  • In the MI instrument, when the current is passed through the solenoid, a magnetic field is set up inside the solenoid. Then the iron piece gets magnetized which in turn deflects the pointer. 
  • The deflecting torque produced is directly proportional to the square of the current.
  • The controlling torque is provided by the spring control.
  • The damping torque is provided by air friction damping.
MPPGCL JE Electronics Mock Test - 10 - Question 7

IRS series satellite are used for -

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 7

The correct answer is Remote sensing.

  • IRS series satellite is used for Remote sensing.

Key Points

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

Additional Information

  • ​Following are the remote sensing satellites planned by ISRO to be launched:
    • RESOURCESAT-3
      • A follow on to Resourcesat-2, it will carry a more advanced LISS-III-WS (Wide Swath) Sensor having similar swath and revisit capability as Advanced Wide Field Sensor (AWiFS).
      • The satellite would also carry Atmospheric Correction Sensor (ACS) for quantitative interpretation and geophysical parameter retrieval.
      • It slated to be launched in 2021.
    • OCEANSAT-3
      • Oceansat-3 would carry Thermal IR Sensor, 12 channel Ocean Color Monitor, Scatterometer, and Passive Microwave Radiometer.
      • IR Sensor and Ocean Color Monitor would be used in the analysis for operational Potential Fishing Zones.
      • Satellite is mainly for Ocean biology and sea state applications.
      • It is slated to the launched aboard PSLV in January 2020.
    • GISATs
      • GISAT-1 is planned for launch in May 2020 and GISAT-2 is planned for launch in August 2020. 
      • They are expected to provide images from geostationary orbit during disasters. 
MPPGCL JE Electronics Mock Test - 10 - Question 8

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

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

The correct answer is option "4"
Concept:

The  power consumption is given by 

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

where 

V is voltage in Volts

R is the resistance

Calculation:

R1 = 2000 Ω

P1 =  2 mW

P = V2/R1

V2 = PR1 = 2 × 10-3 × 2000 = 4 V

V = 2 V

for R2 = 4000 Ω

P = V2/R2

= 22 / 4000 = 1 mW

MPPGCL JE Electronics Mock Test - 10 - Question 9
A Si n-type semiconductor is doped with donor impurity of 1016 cm-3 at room temperature if the intrinsic concentration is 1010 cm-3 then the gap between Fermi energy level and intrinsic energy level is
Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 9

Concept:

In an N-type semiconductor, the doping done is of pentavalent atoms.

As the doping density increases the Fermi level moves towards the conduction band.

If ND is the doping density, then the upward shift in the Fermi level is given by

From the expression

As ND increases  increases and the shift increases.

Calculation:

The electrostatic potential of n type semiconductor

EF - Ei = qνn

EF - Ei = 0.359 eV

MPPGCL JE Electronics Mock Test - 10 - Question 10

How many roots of characteristic equation P(s) = s4 + s3 + 2s2 + 2s + 3 have (+)ve real part?

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

NOTE:

1:If all the elements of a row is zero in a Routh Hurwitz table then we consider it as a row of zeros(ROZ) , for eliminating this ROZ we find the derivative of the above row and write its coefficient

2:Note that ROZ occurs in odd power of S rows only

3:If the first element is only zero  in any row then in that place we consider an arbitrary constant whose value is tending to zero

 4:Order of the algebraic equation gives the number of poles 

 5:Number of sign changes in first column of a row gives the number of poles at the right side of the origin, or on the positive side.

Now, forming Routh Hurwitz table fore the given equation 

so as we see at  we get negative infinite, which means we get two significant changes on the first column means two poles at the right-hand side of the origin.
or We get two poles on the positive hand side

MPPGCL JE Electronics Mock Test - 10 - Question 11

Consider the purely resistive circuit as shown in Fig. where, VS is the source voltage, RS is the source resistance, RL is the load resistance and IL is the load current. The current at maximum power is equal to _______ of the maximum current.

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 11

Maximum power transfer theorem:

Maximum power transfer theorem states that " In a linear bilateral network if the entire network is represented by its Thevenin's equivalent circuit then the maximum power transferred from source to the load when the load resistance is equal to the Thevenin's resistance.

1)  If source impedance is complex then load impedance has to be a complex conjugate of source impedance for maximum power transfer to occur.

2) Maximum efficiency is not related to maximum power transfer.

3) If the source and load impedances are matched then there are no reflections but maximum power transfer.

Proof: 

Let's consider variable resistive load and Thevenin's equivalent network as shown below,

 

Apply KVL, for the above circuit

VS - IL RS - IL RL = 0

Where

VS is the source or Thevenin's voltage, IL is the load current, RL is the load resistance, RS is the source or Thevenin's resistance

V= I(RS + RL)

IL = Vs / (Rs + RL)

P = ILRL

For maximum power transfer, RL = RS

Then the maximum power transferred is given by

Properties of maximum power transfer theorem:

  • This theorem is applicable only for linear networks i.e networks with R, L, C, transformer, and linear controlled sources as elements.
  • The presence of dependent sources makes the network active and hence, MPPT is used for both active as well as passive networks.
  • This theorem is applicable when the load is variable.

Calculation:

For the given circuit, maximum power transfers at RL = Rs

The current at this condition is, 

The maximum value of current occurs at R = 0 and it is given by, 

Therefore, the current at maximum power is equal to 50% of the maximum current.

MPPGCL JE Electronics Mock Test - 10 - Question 12

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

I. Mobility of carriers

II. Threshold voltage

III. Drain current

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 12

Short-channel effect in MOSFET:

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

So, based on the above points all statements are true

MPPGCL JE Electronics Mock Test - 10 - Question 13

Equipotential surface is a

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 13

Concept:

Equipotential surface:

  • The locus of all the points on the surface which has the same electric potential is called the equipotential surface. Since the electric field lines are imaginary lines points over the field lines would also be imaginary.
  • The electric field is always perpendicular to the equipotential surface.
  • Two equipotential surfaces can never intersect.
  • The direction of the equipotential surface is from high potential to low potential.
  • The potential difference between any two points on the equipotential surface is always zero.
  •  So, zero work is to be done to move a charge from one point to another point on the equipotential surface. For example surface of the charged conductor is an equipotential surface.

For point charge, the equipotential surface would be a concentric circle.

MPPGCL JE Electronics Mock Test - 10 - Question 14

Tunnel diode is a pn diode with

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 14
  • Tunnel diode is a highly doped semiconductor diode.
  • The p-type and n-type semiconductor is heavily doped in a tunnel diode due to a greater number of impurities. Heavy doping results in a narrow depletion region.
  • When compared to a normal p-n junction diode, tunnel diode has a narrow depletion width.
  • The Fermi level moves in the conduction band on the n-side and inside the valence band on the p-side.
  • Below the Fermi level, all states are filled and above the Fermi level all states are empty

Step 1: At zero bias there is no current flow as the Fermi level are aligned

Step 2: A small forward bias is applied. The potential barrier is still very high-no noticeable injection and forward current through the junction.

However, electrons in the conduction band of the n region will tunnel electrons to the empty states of the valence band in p region.

This will create a forward bias tunnel current

Step 3. With a larger voltage, the energy of the majority of electrons in the n-region is equal to that of the empty states (holes) in the valence band of p-region; this will produce maximum tunneling current

Step 4 : As the forward bias continues to increase, the number of electrons in the n side that is directly opposite to the empty states in the valence band (in terms of their energy) decreases.

Therefore, the decrease in the tunneling current will start

Step 5.

As more forward voltage is applied, the tunneling current drops to zero.

But the regular diode forward current due to electron-hole injection increases due to a lower potential barrier.

Step 6: With further voltage increase, the tunnel diode I-V characteristic is similar to that of a regular p-n diode.

Under Reverse Bias

In this case the, electrons in the valence band of the p side tunnel directly towards the empty states present in the conduction band of the n side creating large tunneling current which increases with the application of reverse voltage.

The V-I characteristic of a tunnel diode are best represented by:

MPPGCL JE Electronics Mock Test - 10 - Question 15
A ________ is a reverse biased silicon or germanium pn junction in which reverse current increase when the junction is exposed to light.
Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 15

Photo-Diode:

  • A photodiode is a semiconductor p–n junction device that converts light into an electrical current
  • The current is generated when photons are absorbed in the photodiode.
  • The photodiode is a special type of diode which operates in reverse bias conditions.
  • When the incident light on the photodiode increases, the reverse current in a photo-diode also increases.
  • Diode current flows from n to p.
  • Net current flowing is due to only minority charge carriers.
  • The current flowing in the photodiode is due to diffusion current only.
MPPGCL JE Electronics Mock Test - 10 - Question 16

The steady-state error for a system is 0.1. Steady-state error for the previously mentioned system being closed loop with a unity negative feedback and pulse input for 1 sec having a magnitude of 10 is?

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 16

Concept:

The steady-state error for a system is defined as:

R(s) = Input

G(s) = open loop transfer function

H(s) = feedback gain = 1 for unity feedback system

Analysis:

For a unit step input, we have:

The steady-state error becomes:

1 + G(0) = 10

G(0) = 9

Again we have the input as:

r(t) = 10 [u(t) – u(t - 1)]

∴ The steady-state error for the input will be:

Important Points

The shifting in input signal does not affect the steady-state error.  

MPPGCL JE Electronics Mock Test - 10 - Question 17
Spread-spectrum signaling technique used in the following mobile communication system
Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 17
CDMA (Code Division Multiple Access) uses a digital modulation technique called spread spectrum which spreads voice data over a very wide spectrum using a user or cell specific pseudo-random codes.
MPPGCL JE Electronics Mock Test - 10 - Question 18

The above circuit acts as:

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 18

Inverter

This is equal to NOT gate in the digital circuitry.

The output of this is a compliment of the input.

This can be built from the transistors like BJT, MOSFET, etc…

The CMOS inverter consists of the NMOS and the PMOS field-effect transistors connected in one below the other.

When In = Low

PMOS will be shorted and output will be High.

When In = High

NMOS will be shorted and output will be Low.

Hence it acts as an inverter.

Important Points

The structure with the BJT when acting as inverter is:

MPPGCL JE Electronics Mock Test - 10 - Question 19

A Wheatstone bridge with resistors as shown in figure is connected to a 0.2 A current source. Find the voltage across the 0.2 A current source, if the 140 Ω resistor is shorted.

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 19

Concept

The equivalent resistance of two resistance connected in parallel is given by:

Calculation

If the 140 Ω resistors is shorted, then the modified diagram is as follows:

80 Ω and 240 Ω are in parallel.

Their equivalent resistance is given by:

40 Ω and 120 Ω are in parallel.

Their equivalent resistance is given by:

The voltage across the current source is given by:

V = 18 volts

MPPGCL JE Electronics Mock Test - 10 - Question 20

Which of the following is the principle of operation of fibre-optic cable?

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 20

The principle of operation of fibre-optic cable is Reflection.

Key Points

  • In an optical fiber, the information is passed through light, which must not escape outside of it.
  • This phenomenon of confining the light inside the optical fiber is termed as Total internal reflection.
  • For this, the construction and material used to ensure that the total internal reflection of light takes place to prevent the escape of it.

Principle:

  • When light travels from a high refractive index medium to a low-refractive-index medium, it is refracted away from the normal as shown:

          

  • At a certain angle θi, there is no refracted wave and the wave is totally internally reflected  (θ= 90°).
  • This angle is called a critical angle.
  • Inside an optical fiber, we have a high refractive index core (n1) and low refractive index cladding (n2).
  • This results in the propagation of waves inside a fiber through total internal reflection phenomenon.
MPPGCL JE Electronics Mock Test - 10 - Question 21

What is the inverse z-transform of  with ROC |z| < 2

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 21

Concept:

If x(n) = anu(n)

And when x(n) = -anu(-n - 1)

If x(n) ↔ X(z)

Then x(n – n0) ↔ X(z).z-no

Calculation:

Using partial fraction method,

2z = A(z - 3) + B(z + 2)

2(3) = 5B (Putting z = 3)

B = 6/5

2(-2) = -5A

A = 4/5 (Putting z = -2)

Which can be written as;

The IZT of X(z) is, therefore,

The above can now be manipulated as;

MPPGCL JE Electronics Mock Test - 10 - Question 22
Which of the following types of standard input signals has the Laplace transformation as  ?
Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 22

Concept:

Some important Laplace transforms are:

MPPGCL JE Electronics Mock Test - 10 - Question 23
In moving coil instruments the scale used is:
Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 23

The correct answer is 'option 2'

Solution:

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

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

Additional Information

  • In thermocouple instruments, the deflection is directly proportional to the square of the current. i.e. θ ∝ I2
  • In moving iron instruments, the deflection is directly proportional to the square of the current. i.e. θ ∝ I2
  • In hotwire instruments, the deflection is directly proportional to the square of the current. i.e. θ ∝ I2
MPPGCL JE Electronics Mock Test - 10 - Question 24

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.

If a paper is folded in a particular manner and punch is made, when, unfolded this paper appears as given below in the question figure. Find out the manner in which the paper is folded and the punch is made from the answer figures given.
Question Figure

Answer figure

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 24

MPPGCL JE Electronics Mock Test - 10 - Question 25

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

Eight person A, B, C, D, E, F, G and H sit around the circular table facing towards the centre but not necessarily in the same order.
G sits opposite to B. One person sits between A and C, who sits second to the right of H. Both G and B do not sit immediate left of H. D sits second to the left of B. A doesn’t sit beside D. E doesn’t face C.
Q. Who sits opposite to F?

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 25

From the common explanation, we get C sits opposite to F.

Hence, Option C is correct.

Common Explanation:

References:

One person sits between A and C, who sits second to the right of H.

G sits opposite to B.

Both G and B do not sit immediate left of H.

Inferences:

From the above reference, we get two different cases:

References:

D sits second to the left of B.

A doesn’t sit beside D.

E doesn’t face C.

Inferences:

From the above reference, case 1 will be eliminated because A doesn’t sit beside of D.

So, the final arrangement:

MPPGCL JE Electronics Mock Test - 10 - Question 26

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

They will have completed the work by the time we get there.

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 26

The sentence "They will have completed the work by the time we get there" in the Active Voice can be transformed into the Passive Voice as:

A. The work will have been completed by the time we get there.

This option accurately represents the original sentence in the Passive Voice, maintaining its intended meaning.

MPPGCL JE Electronics Mock Test - 10 - Question 27

Directions to Solve

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

Question -

27 : 125 :: 64 : ?

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 27

27 → 33; 125 → 53; 64 → 43

Therefore ? = 63 = 216.

MPPGCL JE Electronics Mock Test - 10 - Question 28

Direction: A word in capital letters is followed by four words. Choose the word that is most nearly opposite in meaning to the word given in capital letters.

MANSION

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 28

As we know that ,
Mansion is a very big house, while hovel is a small, dingy house, castle is a large fortified residence.
Hence , hovel is an opposite word of Mansion . 

MPPGCL JE Electronics Mock Test - 10 - Question 29

Direction: A word in capital letters is followed by four words. Choose the word that is most nearly opposite in meaning to the word given in capital letters.

TRACTABLE

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 29

As we can say that ,
Tractable is something that can be managed easily, i.e., manageable.
From above it is clear that Unmanageable is opposite word of Tractable . 

MPPGCL JE Electronics Mock Test - 10 - Question 30

Terms of credit are with respect to :

Detailed Solution for MPPGCL JE Electronics Mock Test - 10 - Question 30

The mode through which the borrower will repay the loan must be clearly mentioned. Interest rate, collateral and documentation requirement and the mode of repayment together comprise what is called the terms of credit.

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

Top Courses for Electronics and Communication Engineering (ECE)

Download as PDF

Top Courses for Electronics and Communication Engineering (ECE)