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RRB JE ECE (CBT II) Mock Test- 3 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test RRB JE Mock Test Series for ECE 2026 - RRB JE ECE (CBT II) Mock Test- 3

RRB JE ECE (CBT II) Mock Test- 3 for Electronics and Communication Engineering (ECE) 2025 is part of RRB JE Mock Test Series for ECE 2026 preparation. The RRB JE ECE (CBT II) Mock Test- 3 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The RRB JE ECE (CBT II) Mock Test- 3 MCQs are made for Electronics and Communication Engineering (ECE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE ECE (CBT II) Mock Test- 3 below.
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RRB JE ECE (CBT II) Mock Test- 3 - Question 1

To obtain 30 % feedback factor, the correct relation between R1 and R2 is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 1

The feedback factor is given by β

100 R2 = 30 R1 + 30 R2
70 R2 = 30 R1
7 R2 = 3 R1

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

Which one of the following is NOT a type of error of ADC?

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

Offset error:

It is defined as a constant difference, over the whole range of the ADC, between the actual output value and the ideal output value.

Gain error:

It is defined as the difference of the slope of the actual output values and the ideal output values.

Non-linearity error

For an ideal ADC, the output is divided into 2n uniform steps each with the width ∆. Any deviation from the ideal step width is the Differential Non-Linearity. (DNL)

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

By which properties, the orientation of molecules in a layer of liquid crystals can be changed?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 3

  • The optical properties of liquid crystals depend on the direction of light travels through a layer of the material
  • In LCD, the electric field is induced by a small electric voltage applied across it; Due to which the orientation of molecules in a layer of liquid crystals can be changed

RRB JE ECE (CBT II) Mock Test- 3 - Question 4
A data communication system has ______ components.
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 4

  • Message: The message is the information (data) to be communicated; Popular forms of information include text, numbers, pictures, audio, and video
  • Sender: The sender is the device that sends the data message; It can be a computer, workstation, telephone handset, video camera, and so on
  • Receiver: The receiver is the device that receives the message; It can be a computer, workstation, telephone handset, television, and so on
  • Transmission medium: The transmission medium is the physical path by which a message travels from sender to receiver; Some examples of transmission media include twisted-pair wire, coaxial cable, fiber-optic cable and radio waves
  • Protocol: A protocol is a set of rules that govern data communication

RRB JE ECE (CBT II) Mock Test- 3 - Question 5

Residual resistivity of a conductor is measured at

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 5
  • Residual resistivity is the resistivity of material at 0 kelvin
  • It depends on the imperfections in the lattice
  • It is independent of temperature
RRB JE ECE (CBT II) Mock Test- 3 - Question 6

The disadvantage of constant current battery charging method is

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 6

In constant current charging method, the battery is supplied with constant current throughout its charging period. Through this method, the charging is fast initially but slows down near high values of battery charge.

RRB JE ECE (CBT II) Mock Test- 3 - Question 7

Counter type ADC: A

Dual slope ADC: B

SAR type ADC: C

Flash ADC: D

The correct order of conversion time (T) of the ADCs shown above is –

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 7

For n-bit ADC of different type the conversion times are as follows –

Counter type (TA) = 2n TCLK

Dual slop (TB) = 2n+1 TCLK

SAR type (TC) = n TCLK

Flash type (TD) = TCLK

RRB JE ECE (CBT II) Mock Test- 3 - Question 8

Which of the following sinusoidal oscillators are L-C oscillators
I.          Wien oscillator
II.         Colpitts oscillators
III.        Hartly oscillators
IV.        Phase Shift oscillators

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 8

Different types of oscillators are:
1. LC Oscillators:

  • Colpitts Oscillator
  • Clapp's Oscillator
  • Hartley Oscillator

2. RC Oscillators

  • Wein Bridge Oscillator
  • RC Phase shift oscillator
RRB JE ECE (CBT II) Mock Test- 3 - Question 9

The shunt used in milliammeter

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 9

In ammeters, the range can be extended by using low shunt resistance. As low shunt resistance is connected in parallel, the meter resistance will get reduce.

In voltmeters, the range can be extended by using high series resistance. As high series resistance is connected in series, the meter resistance will get increased.

RRB JE ECE (CBT II) Mock Test- 3 - Question 10

The colour of methyl orange in basic medium is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 10

The colour of different pH indicators in different mediums is summarized in the table below:

RRB JE ECE (CBT II) Mock Test- 3 - Question 11

The measurement of the speed of a rotating shaft by means of an electric tachometer is a:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 11

There are two basic methods of measurement:

1. Direct comparison with either a primary or a secondary standard

2. Indirect comparison using calibrated system

Direct measurements: The value of the physical parameter (measurand) is determined by comparing is directly with reference standards. The physical quantities like mass, length and time are measured by direct comparison.

Indirect measurements: The value of the physical parameter (measurand) is more generally determined by indirect comparison with secondary standards through calibration.

The measurand is converted into an analogue signal which is subsequently processed and fed to the end device that presents the result of measurement.

Primary measurements: The value of the physical parameter is determined by comparing directly with the reference standards. The required information is obtainable through senses of sight and touch.

Ex: Matching of two lengths when determining the length of an object with a ruler

Secondary and Tertiary measurements:

An indirect method may consist of developing an electrical voltage proportional to a physical value to be measured, measuring that voltage and then converting the measured voltage back to the corresponding value of the original measurand.

Electrical methods are preferred in the indirect methods due to their high speed of operation and simple processing of the measured variable.

The indirect measurements involving one conversion are called secondary instruments and those involving two conversions are called tertiary measurements.

Examples of secondary instruments:

The conversion of pressure into displacement by means of bellows

The conversion of force into displacement by means of springs

The pressure measurement by manometers and the temperature measurement by mercury in glass thermometers etc

Examples of tertiary measurements:

Measurement of pressure by a bourdon tube pressure gauge

Measurement of angular speed by an electric tachometer

RRB JE ECE (CBT II) Mock Test- 3 - Question 12

Secondary batteries are usually

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 12

Primary cells: It is a cell that once it has been discharged, cannot be recharged. Primary cells are called as dry cells because they have solid or molten electrolyte.

These are light in weight and have less cost.

Secondary cells: It is the one that can be electrically recharged after use to their original pre-discharge condition. Normally secondary cells are wet cells.

These are heavy in weight and have high initial cost.

RRB JE ECE (CBT II) Mock Test- 3 - Question 13

Turn-off time of a thyristor effects its

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 13

For turning off the SCR anode current must fall below the holding current.

Turn-off time of a thyristor effects it’s operating frequency because turn off time of thyristor is inversely proportional to frequency.

The turn – off time is divided into two intervals: reverse recovery time trr and the gate recovery time tgr

RRB JE ECE (CBT II) Mock Test- 3 - Question 14

The coldest and warmest layer of atmosphere are_________ respectively.

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 14

Structure of the atmosphere

The atmosphere is divided into five layers starting from the earth's surface:

Troposphere (0 - 18 km): This is the most important layer of the atmosphere. The air breath exists here. All the weather changes like rainfall, fog, and hailstorm occur in this layer.

Stratosphere (18 - 50 km): It is free from clouds which makes it the most ideal layer for flying Aeroplanes. It contains a layer of ozone gas which protects from the harmful effect of the sun rays.

Mesosphere (50 - 85 km): It is the third layer of the atmosphere. Temperature drops to about -95°C. The mesosphere is the coldest region of Earth’s atmosphere. 

Thermosphere (85 - 500 km): In this layer, the temperature rises very rapidly with increasing height. The ionosphere is a part of this layer. It helps in radio transmission.

Exosphere (500 - 1600 km): The uppermost layer of the atmosphere is called exosphere. It is very thin air. Temperature is very high due to direct solar radiation. Light gases like Helium and Hydrogen float into space from here.

RRB JE ECE (CBT II) Mock Test- 3 - Question 15

In ARPANET, Internet Message Processor (IMP) was used to connect

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 15

The Internet Message Processor (IMP) was the packet switching node used to interconnect participant networks to the ARPANET from the late 1960s to 1989. It was the first generation of gateways, which are known today as routers.

RRB JE ECE (CBT II) Mock Test- 3 - Question 16

In a uniformly doped PN junction, the doping level of the n side is twice the doping level of the P side. Find the ratio of depletion layer width of n side and P-side.

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 16

Total space charges per unit area in P-side must be equal to n-side as per the space charge neutrality.

NA xpo = ND xno

Hence, the correct option is (D)

RRB JE ECE (CBT II) Mock Test- 3 - Question 17

A carrier is simultaneously amplitude modulated by two sine waves with individual modulation of 30% and 40%. The overall modulation index is

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 17

Effective modulation index, 
We have μ1 = 0.3 and μ2 = 0.4
Thus, 
In percentage,μ = 50%
Hence, the correct option is (A)

RRB JE ECE (CBT II) Mock Test- 3 - Question 18

Which of the following is true about below given statement in C++;
signed float 2const;

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 18
  • In signed float 2const, signed is modifier and float data type 2const is identifier
  • An identifier should not start with a number but identifier 2const is starting with 2(number) hence it will give an error

Hence, the correct option is (C)

RRB JE ECE (CBT II) Mock Test- 3 - Question 19

The demand for the Tebhaga Peasant Movement in Bengal was for

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 19

Tebhaga movement in Bengal was militant campaign initiated by Kisan Sabha in 1946. During that period, share-cropping peasants had to give half of their share of harvest to Jotedars who were the owners of land. They demanded reduction in distribution of harvest share given to landlords to only one-third the price, which seems to be option A.
Hence, the correct option is (A)

RRB JE ECE (CBT II) Mock Test- 3 - Question 20

The voltage difference (VAB) between terminal A and B is.

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 20

Applying source transformation

Apply KVL:
(since the loop is not closed there will not be any current flow)
VA - 1 - 4 + 9 = VB
VA + 4 = VB
VAB = VA - VB = - 4V

RRB JE ECE (CBT II) Mock Test- 3 - Question 21
Where is Kudremukh National Park?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 21

  • Kudremukh National park is located in Chikkamagaluru district of Karnataka.
  • It was declared as a National Park in 1987.
  • Other famous National parks in Karnataka are Bandipur, Nagarhole, and Anshi National Park.
  • Famous National Parks in Kerala are Periyar, Silent Valley, and Anamudi Shola National Park.
  • Famous National Parks in Tamil Nadu are Mudumalai, Mukurthi, and Indira Gandhi National Park.
  • Famous National Parks in Odisha are Simlipal, and Bhitarkanika National park.

RRB JE ECE (CBT II) Mock Test- 3 - Question 22

Find the value of built in voltage of the PN diode when N A = 1013 cm-3
And N D = 25 x 1011 cm-3 assume intrinsic concentration is 5 x 1012 cm-3

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 22

Concept:
V bi = V 0 = Built in Potensial 
Application:
NA ND = 25 x 1014
 = 25 x 1014
Here NA ND  hence we have
V 0 = KTLn(1) = 0 V

RRB JE ECE (CBT II) Mock Test- 3 - Question 23
Which of the following data structure can’t store the non-homogeneous data elements?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 23

Array:

  • An array is a group (or collection) of same data types.
  • For example, an int array holds the elements of int types while a float array holds the elements of float types.
  • Non-homogenous data elements cannot be stored in array.

Records:

  • In database management systems, records are composed of fields, each of which contains one item of information.
  • A set of records constitutes a file.
  • For example, a personnel file might contain records that have three fields: a name field, an address field, and a phone number field.

Records is not a data structure.

Pointers:

A pointer is a variable that stores the address of another variable. Unlike other variables that hold values of a certain type, pointer holds the address of a variable.

Pointer is a primitive data type and is homogenous data structure.

Stack:

Stack can be implemented using array or linked list.

When stack is implemented using link list then it can store non homogeneous data when it is implemented using array it stores homogeneous data structure.

RRB JE ECE (CBT II) Mock Test- 3 - Question 24

Which of the following hardware interrupt of 8085 can be disabled by DI instruction but cannot be masked?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 24

The following table shows the masking of hardware interrupts

From the above table INTR can be disabled by DI but cannot be masked using SIM.

RRB JE ECE (CBT II) Mock Test- 3 - Question 25

Assertion (A): Two wires of same length with different cross-sectional areas are connected in series. The heat produced by the current is more for the thicker wire.
Reason (R): The thicker wire has low resistance.

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 25

We know that, R = pl / A
Cross sectional area is more for thicker wire. As area is inversely proportional to resistance, the resistance of a thicker wire is low.
As wires are connected in series, same current flows through both the wires.
Heat produced by the current I is given by I­2R.
As same current flows through both the wires, more heat produced in the high resistive wire. Thus, heat produced by the current is more for the thinner wire.
Hence, A is false, and R is true.

RRB JE ECE (CBT II) Mock Test- 3 - Question 26
Which of the following is not a characteristic of Data communication?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 26

Characteristics of Data Communication:

The effectiveness of any data communications system depends upon the

following four fundamental characteristics:

Delivery:

The data should be delivered to the correct destination and correct user.

Accuracy:

The communication system should deliver the data accurately, without introducing any errors. The data may get corrupted during transmission affecting the accuracy of the delivered data.

Timeliness:

Audio and Video data has to be delivered in a timely manner without any delay; such a data delivery is called real time transmission of data.

Jitter:

It is the variation in the packet arrival time. Uneven Jitter may affect the timeliness of data being transmitted.

Transmission Channel is a component of Communication system and not its characteristic.

RRB JE ECE (CBT II) Mock Test- 3 - Question 27
On which day was the 20th Anniversary of Kargil Vijay Diwas celebrated?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 27

  • The 20th Anniversary of Kargil Vijay Diwas was celebrated on 26 July 2019
  • The day is named after the successful Operation Vijay in 1999, when India regained the control of the high outposts which had been stealthily taken over by Pakistan.
  • The celebrations of 20th Anniversary of Kargil Vijay Diwas began in Delhi on 14 July 2019.

RRB JE ECE (CBT II) Mock Test- 3 - Question 28

The reaction in which ester is treated with an alkali to obtain carboxylic acid and alcohol is called:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 28

Concept:

  • Esters are sweet-smelling substances.
  • These are used in making perfumes and as flavouring agents.
  • Esters react in the presence of an acid or a base to give the alcohol and salt.
  • This reaction is known as saponification because it is used in the preparation of soap

Underlying reaction:

CH3COOC2H5 + N aOH → C 25OH + C H 3COON a

RRB JE ECE (CBT II) Mock Test- 3 - Question 29
A metal reacts with oxygen to form metallic oxide. This oxide is dissolved in water to form X. Which of the following is true about the solution X:
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 29

Concept:

Metals react with oxygen to form metallic oxides.

These metallic oxides are basic in nature

Metal + Oxygen → Metallic oxide

Example:

When magnesium burns in air it combines with the oxygen of the air to form magnesium oxide

2 Mg + O2 → 2 MgO

Magnesium oxide dissolves in water to form magnesium hydroxide solution

MgO + H2O → Mg(OH)2

Magnesium hydroxide turns red litmus blue which shows its basic nature.

RRB JE ECE (CBT II) Mock Test- 3 - Question 30
The prescribed permissible noise level, Leq for residential area at night-time is
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 30

The prescribed permissible noise level for the residential area is 55 and 45 decibels during daytime and night-time respectively.

Important Points:

  • The prescribed permissible noise level for silence zones is 50 and 40 decibels during daytime and night-time respectively
  • The prescribed permissible noise level for the industrial area is 75 and 70 decibels during daytime and night-time respectively
  • The prescribed permissible noise level for the commercial area is 65 and 55 decibels during daytime and night-time respectively

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