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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) 2025 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) 2025 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

A set of programs which enables computer hardware and software to work together is called

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

System software is a type of computer program that is designed to run a computer's hardware and application programs. If we think of the computer system as a layered model, the system software is the interface between the hardware and user applications.

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

The Gross Domestic Product of a country is defined as which among the following?

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

  • The Gross Domestic Product is defined as the total value of the finished goods and services produced within the boundary of a country in a financial year. 
  • The GDP of India has shown a growth of 8.2% in the first quarter of the FY 2018-19. 

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

The early effect in a bipolar junction transistor is caused by

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

  • A large collector base reverse bias is the reason behind early effect manifested by BJTs
  • The depletion layer penetrates more into the base as the base is lightly doped increasing the concentration gradient in the base
  • As reverse biasing of the collector to base junction increases, depletion region penetrates more into the base, this reduces the effective base width
  • This reduction in base width causes less recombination of carriers in the base region
  • This is known as an early effect

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

Which of the following is an operating system?

I. Ubuntu

II. Linux

III. Unix

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

Ubuntu, Linux, Unix, Microsoft Windows and macOS all fall under the category of operating systems.

An Operating system is basically a low-level system software and an important program that runs on every computer. Its basic function is to perform basic tasks such as controlling peripherals, managing computer software and hardware resources, keeping track of directories and files in the storage drive, sending output to the display and more.

RRB JE ECE (CBT II) Mock Test- 2 - Question 5
An increase in which of the following gases leads to global warming?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 5

  • Carbon dioxide is responsible for trapping heat energy from the Sun within the Earth's atmosphere by preventing the rays of the Sun from escaping into the atmosphere. 
  • This leads to an increase in the overall temperature and this phenomenon is called global warming.

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

Which instrument has the highest frequency range with accuracy within reasonable limits?

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

Rectifier instrument has the highest frequency range with accuracy within reasonable limits. These instruments are nothing but permanent magnet moving coil instruments used in conjunction with rectifying device for AC measurements (current and voltage) from about 20 Hz to 20 kHz.

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

Which law of thermodynamics defines the concept temperature?

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

The Zeroth law of thermodynamics defines temperature. It states that “If two systems are in thermal equilibrium with the third system, then they are in thermal equilibrium with each other.”
Zeroth law – concept of temperature
First law – concept of internal energy
Second law – concept of entropy

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

The correct sequence of different layers of the atmosphere from the surface of the Earth upwards is

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

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

  1. 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.
  2. 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.
  3. Mesosphere (50 - 85 km): It is the third layer of the atmosphere. Temperature drops to about -95°C.
  4. 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.
  5. 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.

The Exosphere is the uppermost layer of the atmosphere.

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

Which of the following alloy is used for making clocks and scientific instruments?

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

Invar is an alloy composed of Iron, nickel, and Carbon where Iron is a major component with 63%, nickel with 36% and Carbon with 1%. Invar is an alloy used for making clocks and scientific instruments because of its property of negligible coefficient of expansion.

RRB JE ECE (CBT II) Mock Test- 2 - Question 10

Which of the following is the information retrieval service?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 10

The internet is a globally connected network system that transmits data via various types of media; Sometimes it is referred to as a ‘network of networks’

The World Wide Web, or simply web, is a way of accessing information over the medium of the internet

A web browser is a software application for accessing information on the World Wide Web

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

The chemical formula of Hematite is ________.

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 11
  • The chemical formula of Hematite is Fe2O3, called as ferric oxide. It is a reddish black mineral and an ore of Iron.
  • FeO(OH).nH2O is also an ore of Iron and is the chemical formula of Limonite.
  • FeS2 is the chemical formula of Pyrite.
RRB JE ECE (CBT II) Mock Test- 2 - Question 12

In a bag, there are coins of 25 p, 10 p and 5 p in the ratio of 2 : 3 : 4. If there is Rs. 50 in all, how many 5 p coins are there?

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

25p, 10p and 5p coins are in the ratio of 2:3:4.
So, number of 25p coins = 2 units
Its value = 25 x 2 = 50 units
number of 10 p coins = 3 units
Its value = 3 x 10 = 30 units
number of 5p coins = 4 units
Its value = 4 x 5 = 20 units
Total value = 50 + 30 + 20 = 100 units = 50 Rs.
1 unit = 1/2, 1 unit = 50p
value of 5p coins = 20 units = 20 x 50 p.
Number of 5p coins = 20 X 50/5 = 200

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

A wire of resistance R is divided in 10 equal parts. These parts are connected in parallel, the equivalent resistance of such connection will be 

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

Each part will have a resistance r = R/10
Let equivalent resistance be r R, then
1/r R = 1/r + 1/r + 1/r...........10 times
∴ 1/r R = 10/r = 10/R/10 = 100/R ⇒ r R = R/100 = 0.01 R

RRB JE ECE (CBT II) Mock Test- 2 - Question 14

Which of the following hardware device provides a way to access data flowing in a network?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 14

A network tap is a hardware device which provides a way to access the data flowing across a computer network.

In many cases, it is desirable for a third party to monitor the traffic between two points in the network.

 If the network between points A and B consists of a physical cable, a "network tap" may be the best way to accomplish this monitoring.

The network tap has (at least) three ports: a A port, a B port, and a monitor port.

A tap inserted between A and B passes all traffic through unimpeded, but also copies that same data to its monitor port, enabling a third party to listen.

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

Which of the following is not a control signal?

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

Three control signals are RD, WR & ALE.
RD − This signal indicates that the selected IO or memory device is to be read and is ready for accepting data available on the data bus
WR − This signal indicates that the data on the data bus is to be written into a selected memory or IO location
ALE − It is a positive going pulse generated when a new operation is started by the microprocessor; When the pulse goes high, it indicates address. When the pulse goes down it indicates data
Three status signals are IO/M, S0 & S1.

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

perror( ) function used to ?

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

The C library function void perror(const char *str) prints a descriptive error message to stderr.

Program:

#include<stdio.h>

int main()

{

perror("ERROR");

return 0;

}

Output: ERROR: No error

Concept:

Since the above code contains no error, compiler throws a message No error and output is ERROR: No error.

Instead of perror("ERROR") if printf("ERROR") was used then output would have been ERROR. Therefore perror() doesn’t work same as printf().

RRB JE ECE (CBT II) Mock Test- 2 - Question 17

Quantization noise occurs in –

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 17

PAM and PPM are pulse analogue modulation schemes.
There is no quantization involved in them.
Of the three modulation schemes, quantization noise occurs only in delta modulation.
Quantization error:
The quantisation error is the difference between the sampled value and the quantized value.
The corresponding noise is called Quantization noise and is given by Δ2/12 , where Δ  is the step size 
Hence, the correct option is (C)

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

Which interrupt in 8085 Microprocessor is non-maskable?

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

A non-maskable is an interrupt which can not be disabled.
RST 7.5, RST 5.5 are maskable interrupts but TRAP is a non-maskable interrupt.

Hence, the correct option is (C)

RRB JE ECE (CBT II) Mock Test- 2 - Question 19
Automatic switching OFF function is accomplished by a _________ in an MCB
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 19

A miniature circuit breaker (MCB) automatically switches off power supply during overload and faults. This function of automatic switching is accomplished by a bimetallic strip.

Whenever continuous over current flows through MCB, the bimetallic strip is heated and deflects by bending. This deflection of bimetallic strip releases mechanical latch. As this mechanical latch is attached with operating mechanism, it causes to open the miniature circuit breaker contacts, and the MCB turns off thereby stopping the current to flow in the circuit.

RRB JE ECE (CBT II) Mock Test- 2 - Question 20
Which of these is a universal register?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 20

A shift register that can operate in any of the four modes. i.e.

  1. Serial In Serial out (SISO)
  2. Serial In Parallel Out (SIPO)
  3. Parallel In Serial Out (PISO)
  4. Parallel In Parallel Out (PIPO)

Common IC’s

SISO → 74 HC 595

SIPO → 74LS164 and 74LS594

PISO → 74HC166

Universal Shift Register → TTL 74LS194, 74LS195 or the CMOS 4035

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

The number of memory segments in 8086 is

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

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 22
The maximum number of I/O devices that can be accessed using IN and OUT instructions is?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 22

Concept:

The format of IN and OUT instruction is

IN 8 bit address

OUT 8 bit address

This address is the port address of I/O device.

Calculation:

Since port the address is of 8 bits , there are 28 port address and 256 maximum I/O devices that can be connected to these ports.

RRB JE ECE (CBT II) Mock Test- 2 - Question 23

The power delivered by the voltage source is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 23

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 24
Where are the Buddhas of Bamiyan located?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 24

  • The Buddhas of Bamiyan is located in Hazarat region of Afghanistan.
  • Buddhas of Bamiyan are 2 statues of Gautama Buddha made in the 6th century.
  • The statues are made in Gandharan Style of Architecture.
  • They were destroyed by the Taliban in the year 2001.

RRB JE ECE (CBT II) Mock Test- 2 - Question 25
Two or more computers connected to each other for sharing information form a ________.
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 25

A computer network is a set of computers connected together for the purpose of sharing resources. Computers and devices that allocate resources for a network are called servers. Computer networks support an enormous number of applications and services such as access to the World Wide Web and digital video. The best - known computer network is the Internet.

RRB JE ECE (CBT II) Mock Test- 2 - Question 26
A man is standing on a boat in still water. If he walks towards the shore, the boat will
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 26

A man is standing on a boat in still water. If he walks towards the shore, the boat will move away from the shore. This is according to Newton’s third law of motion - to every action there is equal and opposite reaction.

RRB JE ECE (CBT II) Mock Test- 2 - Question 27
Which of the following is not an anti-virus?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 27

Python is not an antivirus software, it is a computer language.

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

The Characteristic impedance of transmission line

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

The Characteristic impedance of the transmission line is given as 
Hence it is independent of the length of the transmission line.

RRB JE ECE (CBT II) Mock Test- 2 - Question 29
In a microprocessor-based system, a DMA facility is required to increase the speed of the data transfer between the
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 29

  • Direct memory access (DMA) is a feature of computer systems that allows certain hardware subsystems to access main system memory (random-access memory), independent of the central processing unit (CPU).
  • DMA avoids using the CPU thus allowing the CPU to attend another job
  • It increases the speed of the data transfer between the memory and the I/O devices
  • DMA is used when a large amount of data is to be printed out from the memory of a computer
  • The problem of slow data transfer between input-output port and memory or between two memory is avoided by implementing the Direct Memory Access (DMA) technique
  • It is implemented by using 8257 DMA controller

RRB JE ECE (CBT II) Mock Test- 2 - Question 30
Minimum number of JK flip flops required to construct a BCD counter is:
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 2 - Question 30

Concept:

A counter having N states requires m counters where

m ≥ log2 N

Application:

A BCD counter has 10 states and counts from 0 to 9

m ≥ log2 10

m = 4

Hence BCD counter requires 4 flip flops

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