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RRB JE ECE (CBT II) Mock Test- 2 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test - RRB JE ECE (CBT II) Mock Test- 2

RRB JE ECE (CBT II) Mock Test- 2 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The RRB JE ECE (CBT II) Mock Test- 2 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The RRB JE ECE (CBT II) Mock Test- 2 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE ECE (CBT II) Mock Test- 2 below.
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RRB JE ECE (CBT II) Mock Test- 2 - Question 1

In the SPWM, the modulating signal is

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 1

  • In the SPWM, the modulating signal is sinusoidal
  • SPWM output is generated by intersection between sine signal and triangle signal
  • Sine signal is the reference waveform and triangle waveform is the carrier waveform
  • When value sine signal is large than triangle signal, the pulse will start produce to high

RRB JE ECE (CBT II) Mock Test- 2 - Question 2

Which of the following statements are false related to a typical Op-Amp?

1) High Bandwidth

2) High output impedance

3) High CMMR

4) Low slew rate

5) High input impedance

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 2

An Op-Amp should have

i) High Bandwidth

ii) High input impedance

iii) Low output impedance

iv) High Gain

v) High CMRR

vi) High slew rate

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RRB JE ECE (CBT II) Mock Test- 2 - Question 3

Which among the following indicates the orientation of sub-shells in an atom?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 3
  • The quantum number defines an electron completely in an atom. 
  • There are four types of quantum number - Principal Quantum Number, Azimuthal Quantum Number, Magnetic Quantum Number and Spin Quantum Number. 
  • Principal Quantum Number describes the size, shape and energy of the shell to which an electron belongs whereas Azimuthal Quantum Number determines the shape of the electron cloud and the number of sub-shells in a shell. 
  • Magnetic Quantum Number determines the orientation of sub-shells whereas the spin quantum number represents the direction of the electron spin. 
RRB JE ECE (CBT II) Mock Test- 2 - Question 4

Which of the following statement is true regarding avalanche and Zener breakdown?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 4

Zener Breakdown:

In a heavily doped PN junction, under high reverse voltage (around 5 V), the electrons tunnel though the depletion region due to extremely small depletion region width.

Even though the electric field is very high the electrons are able to tunnel through due to very thin depletion regions.

This causes high reverse current to flow even in reverse bias. The junction is not destroyed can be used as normal PN diode in forward bias.

Avalanche Breakdown:

In thinly doped PN junction diodes Zener effect is not possible, but under very high reverse voltage (< 8-10 V) the high electric fields (lower than Zener diode) cause the electrons to accelerate.

The accelerated electrons break free more electrons (called impact ionization) this newly free electrons in turn create more charge carriers causing a kind of avalanche effect.

This results in huge reverse current, but the junction is destroyed to this effect.

RRB JE ECE (CBT II) Mock Test- 2 - Question 5

Which one of the following layer of the atmosphere is responsible for the deflection of the radio waves?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 5

In the lower part of the thermosphere region, between 100 and 400 km, the ionisation of the atmospheric gases takes place and the layer is called ionosphere.

The ionosphere is ionized by solar radiation.

There is a peak concentration of ionized particles at 250 kms which is responsible for the deflection of radio waves.

RRB JE ECE (CBT II) Mock Test- 2 - Question 6

Delay problem is encountered in:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 6

Delay Problem:

In asynchronous counters the output of the previous stage serves as the clock of the next stage. As the number of stages increases the propagation delay of each flip flop stage adds up and this propagation delay becomes significant.

Remedy of Propagation delay:

To eliminate the propagation delay encountered in different stages, all the flip flops are provided with a common clock. Thus, the output of each stage does not depend on the clock from the previous stage but only on the common clock signal and propagation delay does not adds.

RRB JE ECE (CBT II) Mock Test- 2 - Question 7

Where was the Western Presidency situated in the early period of the East India Company?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 7

After defeating the Portuguese, Jahangir was influenced by the British,As a result, a royal decree issued in 1613 AD gave the British permission to establish a trading kothi in Surat. Thus, in the initial period of the East India Company, the Western Presidency was in Surat.
Hence, the correct option is (A)

RRB JE ECE (CBT II) Mock Test- 2 - Question 8

Five years ago, Ram was three times as old as Shyam. Four years from now, Ram will be only twice as old as Shyam. What is the present age of Ram?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 8

Let present age of Ram be x and Shyam be y 

From question, 

(x - 5) = 3(y – 5 ) => x – 3y = -10 ……. (i) 

(x+ 4 ) = 2(y + 4) => x – 2y = 4 …….. (ii) 

After solving equation (i) and (ii) we get 

x = 32 

Hence present age of Ram be 32 years 
Hence, the correct option is (B)

RRB JE ECE (CBT II) Mock Test- 2 - Question 9

The number of memory segments in 8086 is

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 9

Segmentation is the process in which the main memory of the computer is divided into different segments and each segment has its own base address.
It is used to enhance the speed of execution of the computer system, so that processor is able to fetch and execute the data from the memory easily and fast.
Memory segmentation is shown below

RRB JE ECE (CBT II) Mock Test- 2 - Question 10
The continue statement cannot be used with
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 10

Concept:

The continue statement is used inside loops.

When a continue statement is encountered inside a loop, control jumps to the beginning of the loop for next iteration, skipping the execution of statements inside the body of loop for the current iteration.

Application:

Loop can be constructed using For, while, do while statements.

Switch is not used for looping and hence continue statement is not used with Switch case.

RRB JE ECE (CBT II) Mock Test- 2 - Question 11

The figure shown below is a network in which the diode is an ideal one. The terminal v - i characteristics of the networks is given by,

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 11

When V < 10 V I = 0

For V > 10 V, diode is forward biased

Apply kVL:

V - 4I - 10 = 0

V = 4I + 10

y = mx + c

slope = m = 4

y intercept = c = 10

 

RRB JE ECE (CBT II) Mock Test- 2 - Question 12

In the given op amp circuit if all Resistances are equal, then the magnitude of voltage gain of the op-amp is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 12

Since all resistance are equal let
R1 = R2 = R3 = R4 = R

VA = 0 (Virtual ground)
Now R2 are R4 are in parallel, connected between node voltage at B and ground.
Equivalent: R||R = R/2


V o / V i = - R/3R/2 = - 2/3 = - 0.66 V

RRB JE ECE (CBT II) Mock Test- 2 - Question 13

The power delivered by the voltage source is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 13

Current through 1Ω resistor is
I - V/R = 1V/1Ω = 1A
Apply KCL at Node A

I’ + 1 = 1A
I’ = 0A
Since I’ = 0 A, the power delivered by the voltage source = VI’ = OW

RRB JE ECE (CBT II) Mock Test- 2 - Question 14
In order to have fast, steady and accurate responses, the meters should have
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 14

  • For critically damped system ζ = 1.
  • The response of the system in this case is rapid and the system reaches its final steady state conditions smoothly without oscillations. 

RRB JE ECE (CBT II) Mock Test- 2 - Question 15

What is the range of Local area network?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 15

RRB JE ECE (CBT II) Mock Test- 2 - Question 16

Grey code for a number is 1011. The decimal equivalent for the number is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 16

Convert gray to binary:

Copy the MSB and then Exor as shown

Step 2.

Now convert the binary to decimal

1101 → (13)10

RRB JE ECE (CBT II) Mock Test- 2 - Question 17
The major portion of the alkalinity in natural water is caused by
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 17

  • Alkalinity refers to the capability of water to neutralize the acid
  • The alkalinity of natural water is determined by the soil and bedrock through which it passes
  • The main sources for natural alkalinity are rocks which contain carbonate, bicarbonate, and hydroxide compounds. Borates, silicates, and phosphates also may contribute to alkalinity

RRB JE ECE (CBT II) Mock Test- 2 - Question 18

Each cell of a static Random-Access Memory contains

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 18

SRAM has 6 MOS transistors whereas DRAM has 1 MOS transistors and 1 capacitor.

SRAM (static RAM) is random access memory (RAM) that retains data bits in its memory as long as power is being supplied

RRB JE ECE (CBT II) Mock Test- 2 - Question 19

The internal resistance of a cell is

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 19

Internal resistance of a cell: It is the resistance which is present within the battery that resists the current flow when connected to a circuit. It is the resistance offered by its electrolyte to the flow of ions. Thus, it causes a voltage drop when current flows through it.

Here E is the emf of a cell and r is the internal resistance.

Factors affecting internal resistance of a cell:

  • It is directly proportional to the distance between the electrodes
  • It is inversely proportional to the surface area of the electrodes dipped in electrolyte
  • It is inversely proportional to the temperature of electrolytes
  • It is inversely proportional to the concentration of electrolytes
  • Greater the conductivity of the electrolyte, lesser is the internal resistance of the cell. i.e. internal resistance depends on the nature of the electrolyte.
RRB JE ECE (CBT II) Mock Test- 2 - Question 20
A step-index fibre has a core index of refraction of n1 = 1.425.  The cut-off angle for light entering the fibre from air is found to be 8.50o. The numerical aperture of the fibre is:
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 20

Concept:

The numerical aperture is given by:

NA = n0 sin θ0max

Where n0 is the refractive index of the air (surrounding)

Calculation:

NA = n0sin θ0max = (1.0003) sin (8.50°) = 0.148.

RRB JE ECE (CBT II) Mock Test- 2 - Question 21

When a body moves in a circular path, no work is done by the force since,

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 21

The work done on a body moving in a circular path is also zero. This is because,when a body moves in a circular path, then the centripetal force acts along the radius of the circle, and it is at right angles to the motion of the body. Thus, the work done in the case of moon moving round the earth is also zero.

RRB JE ECE (CBT II) Mock Test- 2 - Question 22
In Guided media transmission using twisted pair, the twisting of cables is done:
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 22

  • A number of these pairs are bundled together into a cable by wrapping them in a tough protective sheath
  • Over longer distances, cables may contain hundreds of pairs
  • The twisting tends to decrease the crosstalk interference between adjacent pairs in a cable
  • Neighbouring pairs in a bundle typically have somewhat different twist lengths to reduce the crosstalk interference

RRB JE ECE (CBT II) Mock Test- 2 - Question 23
A master slave flip flop has the characteristic that
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 23

Master-Slave flip flop is the cascaded combination of two flip-flops among which the first is designated as master flip-flop while the next is called slave flip-flop.

The master flip-flop is triggered by the external clock pulse train while the slave is activated at its inversion i.e. if the master is positive edge-triggered, then the slave is negative-edge triggered and vice-versa.

There will be a change in the output when the state of the slave is affected because the output is taken at the slave.

RRB JE ECE (CBT II) Mock Test- 2 - Question 24

The drain current in MOSFET is varied by:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 24

The drain current in MOSFET in saturation is given by

From the equation it is evident that MOSFET’s drain current depends on:

  1. Device parameters W/L
  2. Gate to source Voltage VGS

To realise this physically, as the Gate to source voltage is increased, the channel becomes deeper, hence there is more area through which electrons can move.

RRB JE ECE (CBT II) Mock Test- 2 - Question 25
Which of the following theorem can be applied to any network-linear or non-linear, active or passive, time variant or time invariant?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 25

Various Theorem and the circuits where they are applicable is shown below in the table

RRB JE ECE (CBT II) Mock Test- 2 - Question 26
Determine the deflection sensitivity (in m/V) of a CRO, when the value of the deflection factor is 0.5 V/m.
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 26

Given that, deflection factor (d) = 0.5 V/m

Deflection sensitivity (s) is the reciprocal of deflection factor.

s = 1/d = 1/0.5 = 2 m/V

RRB JE ECE (CBT II) Mock Test- 2 - Question 27

What is the status of Carry flag after the execution of following instruction?

MVI A, FF H

ADI 01 H

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 27

FF + 01 = 1 00 H

The accumulator contains 00 H

And 1 carry is obtained

The important point to note here that carry is obtained and this will set the carry flag.

This is the main difference between INR A and ADI 01 instruction.

ADI instruction will affect Carry flag, but INR instruction will not.

RRB JE ECE (CBT II) Mock Test- 2 - Question 28

The expected value of voltage across a resistor is 40 V. The measurement gives a value of 39 V, then the percentage accuracy is given by:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 28

percentage accuracy = measured value / expected value.100 %

percentage accuracy = 39 / 40 .100 %

= 97.5 %

RRB JE ECE (CBT II) Mock Test- 2 - Question 29

If a copper wire is stretched to make it 0.1% longer. The percentage change in its resistance is

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 29

Let wire be of length 1000 unit and cross area be also 1000 unit2

After stretching new length = 1.1% of 1000 = 1001

Volume is same (length x cross section)

1000 × 1000 = 1001 × A’

A’ = 1000 × 1000/1001

= 999

% change = -0.1%

R = ρ l/A

Taking log and differentiating

dR / r = dl / L - dA / A

Substituting value

dR / R x 100 = 0.1 -(-0.1) = 0.2%

RRB JE ECE (CBT II) Mock Test- 2 - Question 30

Two balls are dropped from heights h and 2h respectively. What would be the ratio of times taken by balls to reach the earth?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 30

Using the equation,

for the ball released from height h,

for the ball released from height 2h

The ratio of the time taken is

t / t1 = t / √2t = 1 / √2t

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