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RRB JE ECE (CBT II) Mock Test- 5 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test - RRB JE ECE (CBT II) Mock Test- 5

RRB JE ECE (CBT II) Mock Test- 5 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The RRB JE ECE (CBT II) Mock Test- 5 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The RRB JE ECE (CBT II) Mock Test- 5 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- 5 below.
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RRB JE ECE (CBT II) Mock Test- 5 - Question 1

The Laboratory apparatus are made of ________.

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 1

  • The Laboratory apparatus are made of pyrex glass. Pyrex glasses are borosilicate glasses.
  • Some of the basic laboratory apparatus comprise of a volumetric flask, graduated cylinder, funnel, and beaker.
  • Flint glasses are durable and heavy characterized by their high refractive quality, clarity, and brilliance.
  • Crookes glasses are those which are designed to diminish the transmission of UV rays.

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

Path’s name is used to locate file in directory structure is

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

A file is identified by its path through the file system beginning from the root node.

Relative Path:

Absolute Path: It always contains the root element and the complete directory list required to locate the file.

Relative Path: it needs to be combined with another path in order to access a file. This path does not start from the root node.

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

Inert gases exhibit _______.

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

Different magnetic materials and their properties are shown in the Table:

RRB JE ECE (CBT II) Mock Test- 5 - Question 4
When the USB is connected to a system, its root hub is connected to the _________.
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 4

USB root hub is the software driver that connects multiple USB peripherals to the computer. Root hub is connected to the processor through Bus.

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

Ramon Magsaysay Award is named after the former President of ________.

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 5
  • The Ramon Magsaysay Award is an annual award established to perpetuate former Philippine President Ramon Magsaysay's example of integrity in governance, courageous service to the people, and pragmatic idealism within a democratic society.
  • The prize was established in April 1957 by the trustees of the Rockefeller Brothers Fund based in New York City with the concurrence of the Philippine government.
RRB JE ECE (CBT II) Mock Test- 5 - Question 6

In the following figure, When S = R = 1, then the output P and Q are:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 6

Truth Table of NAND gate is shown

If one of the inputs is ‘1’ then the output is an inversion of other input.
The NAND based latch truth table is shown

The output will retain its previous state

RRB JE ECE (CBT II) Mock Test- 5 - Question 7

Programs stored in which of the following memories can’t be erased?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 7

ROM is a non-volatile memory. Its contents are retained even after the power is switched off. ROM stores the program required to initially boot the computer. It only allows for reading.

RRB JE ECE (CBT II) Mock Test- 5 - Question 8

A transmission line is distortionless if,

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 8

A transmission line is distortionless if R/L = G/C
Rewriting this we have, LG = RC
Note:
If R/L = G/C then phase velocity is given by
Vp = ω/β = 1/√LC
the phase velocity is independent of frequency hence the transmission line is distortionless.

RRB JE ECE (CBT II) Mock Test- 5 - Question 9

A 3-bit full adder has:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 9

A 3-bit full adder is a combinational circuit

Output depends only on the present input bits

Hence there is no memory in a 3-bit full adder

RRB JE ECE (CBT II) Mock Test- 5 - Question 10

Bitwise operators can operate upon?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 10

→ At the language level, there's no such thing as "bitwise operation on floating-point numbers". Bitwise operations in C/C++ work on value-representation of a number. And the value-representation of floating-point numbers is not defined in C/C++.
→ Floating point numbers don't have bits at the level of value-representation, which is why it is not possible apply bit wise operations to them
Program:
#include<stdio.h>
int main()
{
float a = 1.2;
a = 1 | a;
return 0;
}
Output: invalid operands of type int and float
Similarly, it will give error or double and char and float and double
Program:
#include<stdio.h>
int main()
{
char a = 2;
a = 1 | a;
printf("%d", a);
return 0;
}
Output: 3
This shows that the bit wise operator can operate on int and char.

RRB JE ECE (CBT II) Mock Test- 5 - Question 11

In balancing the following chemical reaction, what is the value of “b”?
a H2S + b O2 → c H2O + d SO2

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 11

Hydrogen and Sulphur are already balanced
Start with the balancing of oxygen
Oxygen on RHS = 3
Oxygen on LHS = 2
H2S + 3/2 O2 → H2O + SO2
To get an integer number of moles, multiply the equation by 2
2  H2S + 3 O→ 2 H2O + 2 SO2

RRB JE ECE (CBT II) Mock Test- 5 - Question 12

Which is used to store critical pieces of data during subroutines and interrupts:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 12

A stack is a place in memory for storing data sequentially.

The address of the stack top is given by stack pointer.

Whenever there is an interrupt or a subroutine call, microprocessor stores the address of next instruction on stack and then executes the subroutine.

After the execution of subroutine, microprocessor fetches the return address from the stack and executes the instructions sequentially.

RRB JE ECE (CBT II) Mock Test- 5 - Question 13

In which of the following cases, work is done by the force?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 13

Option a) While the man is moving upon a staircase, he has to do work against the gravity. Hence work is done here.

Option b) When the main is carrying a bucket of water walking on a level road with a uniform velocity then the direction of the displacement is at right angle with the direction of the force (i.e., gravity). Hence the dot product of the force and displacement is zero => Work done is zero.

Option c) When a drop of rain is falling vertically with a constant velocity then there is no net force acting hence the work done is zero.

Option d) Here also, the centripetal force is perpendicular to the direction of the displacement hence the work done is zero.

RRB JE ECE (CBT II) Mock Test- 5 - Question 14

Gate whose output is zero, only when both the inputs are opposite

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 14

The truth table for EXOR and EXNOR is:

From the truth table. It is clear the required gate is EXNOR

Trick:

The questions can be solved without drawing the truth table.

Since both the inputs are opposite

The output is 1 bar either A'B' or AB.

i.e. Y = AB' + A'B

This is the expression for EXNOR gate.

RRB JE ECE (CBT II) Mock Test- 5 - Question 15

When is the World Population Day observed?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 15

The United Nations observes 11 July every year as World Population Day.

This year’s World Population Day calls for global attention to the unfinished business of the 1994 International Conference on Population and Development.

United Nations Headquarters - New York, USA .

Founded - 24 October 1945.

Secretary General - Antonio Guterres.

RRB JE ECE (CBT II) Mock Test- 5 - Question 16

The closed loop voltage gain of the op-amp shown is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 16

The given op-amp is configured in non inverting mode
The gain is given by
 Av = 1 + Rf/R1 = 1 + 100/4.7 = 22.3
Hence, the correct option is (D)

RRB JE ECE (CBT II) Mock Test- 5 - Question 17

In a 8085 microprocessor the first machine cycle of an instruction is

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 17

The time required by the microprocessor to complete an operation of accessing memory or input/output devices is called Machine Cycle

The 4 steps involved in machine cycle in order are:

Fetch 

decode 

execute 

store

In execution of any instruction first cycle is op-code fetch cycle.

After that depending upon the given instruction the memory read/write and I/O read/ write cycle occurs.
Hence, the correct option is (D)

RRB JE ECE (CBT II) Mock Test- 5 - Question 18
Which of the following radars have highest Pulse Repetition Frequency (PRF):
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 18

MTI Radar:

MTI radar operates on low pulse repetition frequencies thus causing ambiguous Doppler measurements (blind speeds) but unambiguous range measurements (no second-time-around echoes).

Pulse Doppler Radar:

Pulse-Doppler radar operates on high pulse repetition frequency thus causing unambiguous doppler measurements (no blind speeds) but ambiguous range measurements (second-time-around echoes).

RRB JE ECE (CBT II) Mock Test- 5 - Question 19
Devices on one network can communicate with devices on another network via a-
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 19

  • There are many networks that exist often with different hardware and software
  • It is needed to connected devices on one network with devices on another network
  • To connect such incompatible networks, necessary translation in terms of hardware and software is required, the gateway does this task
  • A file server is a computer attached to a network that provides a location for shared disk access that can be accessed by the workstations through a shared link
  • Printer server is a device that connects printers to client computers over a network
  • The Utility Server runs a number of utility applications that support or enhance the component applications facilitating a complete single box solution

RRB JE ECE (CBT II) Mock Test- 5 - Question 20
A tunnel-diode is best suited for
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 20

According to quantum mechanics tunneling occurs at high frequencies ideally at the speed of light hence tunnel diode can be used as high frequency (microwave range) oscillator.

RRB JE ECE (CBT II) Mock Test- 5 - Question 21
Which one of the following four metals would be displaced from the solution of its salts by other three metals?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 21

Concept:

A metal of high reactivity displaces a metal of low reactivity from its salt solution.

The reactivity of metal is decided by its reactivity series.

Reactivity series:

K > Na > Ca > Mg > Al > Zn > Fe > Sn > Pb > H > Cu > Hg > Ag > Au.

Application:

Since Mg, Zn, Cu are more reactive than Silver Ag, they can replace silver from its salt solution.

RRB JE ECE (CBT II) Mock Test- 5 - Question 22
A labeled statement consist of an identifier followed by
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 22

The label is an identifier.

 When goto statement is encountered, control of the program jumps to label: and starts executing the code.

The name of the label is followed by a colon (:)

Example of Label:

int main()

{

 printf (“Hello Worldn”);

 goto Label;

 printf(“How are you?”);

 printf(“Are you Okey?”);

 Label:

 printf(“Hope you are fine”);

 getch();

}

RRB JE ECE (CBT II) Mock Test- 5 - Question 23
The earth tester works on the principles of _________
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 23

  • The earth tester works on the principles of fall of potential method
  • The instrument used for measuring the resistance of the earth is known as earth tester
  • The inherent function of a fall-of-potential ground tester is to input a constant current into the earth and measure the voltage drop by means of auxiliary electrodes.

RRB JE ECE (CBT II) Mock Test- 5 - Question 24
The real part of the propagation constant shows:
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 24

Concept:

  • The propagation constant of a wave is complex quantity and is given by γ.
  • The propagation constant of a sinusoidal electromagnetic wave is a measure of the change undergone by the amplitude and phase of the wave as it propagates in a given direction.
  • The quantity being measured can be the voltage, the current in a circuit, or a field vector such as electric field strength or flux density

Application:

Mathematically propagation constant γ is written by

γ = α + j β

γ = propagation constant

α = real part = Attenuation constant

β = Imaginary part = Phase constant

Attenuation constant:

  • Attenuation constant gives attenuation of an electromagnetic wave propagating through a medium per unit distance from the source.
  • It is the real part of the propagation constant and is measured in nepers per metre.

Phase Constant:

  • The phase constant is the imaginary component of the propagation constant for a plane wave.
  • It represents the change in phase per unit length along the path travelled by the wave at any instant.
  • Phase constant is measured in radians per unit length.

RRB JE ECE (CBT II) Mock Test- 5 - Question 25

A file pointer predefined in stdio.h that refers to standard input stream is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 25

RRB JE ECE (CBT II) Mock Test- 5 - Question 26

The value of Rs. such that output voltage Vo is Zero is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 26

Concept:
Output voltage is sum of output voltage due to the different inputs
Vo = V01 + V02

Calculations

Vo = 0
Vo = 0
2/12 = Rs / Rs + 20
12 Rs = 2 Rs + 40
10 Rs = 40
Rs = 4 kΩ

RRB JE ECE (CBT II) Mock Test- 5 - Question 27

The most common observational error is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 27

RRB JE ECE (CBT II) Mock Test- 5 - Question 28
The point where control returns after a subprogram is completed is known as the:
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 28

The address where the microprocessor returns after the execution of instruction Is known as a return address.

During any subroutine call, whenever call instruction is executed, microprocessor stores the current value of Program counter (PC) in the stack.

This value of the program counter serves as the return address for the microprocessor.

RRB JE ECE (CBT II) Mock Test- 5 - Question 29
An expression contains relational, assignment and arithmetic operators. If Parenthesis are not present, the order will be
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 29

Concept:

Operator is a programming element that specifies what kind of an operation need to perform on operands or variables.

Arithmetic Operators:

Arithmetic Operators are used to perform basic arithmetic calculations like addition, subtraction, division, etc. based on our requirements.

Relational Operators

Relational Operators are used to check the relation between two operands like we can determine whether two operand values equal or not.

Assignment Operators

Assignment Operators are used to assign a new value to the operand and these operators will work with only one operand.

Application:

The priority order of operators is Arithmetic > Relational > Assignment

RRB JE ECE (CBT II) Mock Test- 5 - Question 30

A mod-3 counter is constructed using J-K flip flops each having a propagation delay of 50 nsec. The maximum frequency of clock that can be applied to the input signal is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 5 - Question 30

If n-flip-flops are required to construct a counter each having propagation delay of tpd sec, then clock frequency 

Calculation:

For mod 3 counter number of flip flops required are

n ≥ log230

n = 5

Propagation delay of 1 flip flop = 50 nsec

Total propagation delay = 5 × 50 = 250 nsec

Clock frequency = 1/250 nsec

1000 x 106 / 250

= 4 MHZ

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