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RRB JE ECE (CBT II) Mock Test- 6 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test - RRB JE ECE (CBT II) Mock Test- 6

RRB JE ECE (CBT II) Mock Test- 6 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) preparation. The RRB JE ECE (CBT II) Mock Test- 6 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The RRB JE ECE (CBT II) Mock Test- 6 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- 6 below.
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RRB JE ECE (CBT II) Mock Test- 6 - Question 1

The output stage of an op-amp amplifier is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 1

The various stages of op-amp are shown in the figure:
The last stage/output stage is the Push Pull amplifier

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

The brightest spot, on a cathode ray screen, occurs at

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

When no voltage is applied to these deflection plates, the electron beam produces a spot of light at the center as shown by point O in fig below on the screen.

The intensity of the light spot on the screen can be controlled by changing the negative potential on the control grid.

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

While transmitting data using Frequency Division Multiplexing (FDM) the filter used at the receiving end is:

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

FDM transmitter:

  • The available bandwidth of a single physical medium is divided into a number of smaller, independent frequency channels
  • Using modulation, independent message signals are translated into different frequency bands

FDM Receiver:

  • At the receiving end, the signal is applied to a bank of band-pass filters, which separates individual frequency channels
  • The bandpass filter outputs are then demodulated and distributed to different output channels

RRB JE ECE (CBT II) Mock Test- 6 - Question 4

In an AC circuit, the ratio of kW/kVA represents

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 4

Power factor: It is the ratio between true power and apparent power

Power factor = kW/kVA

Form factor: It is the ratio of the RMS (root mean square) value to the average value

Load factor: It is defined as the ratio of the average load over a given period to the maximum demand (peak load) occurring in that period.

Diversity factor: It is the ratio of sum of individual maximum demand to maximum demand on power station

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

Generation in computer terminology is

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

The generations of computers are characterized by a major technological development that fundamentally changed the way computers operate, resulting in smaller, cheaper, more powerful and more efficient and reliable devices.

The various generations of computers are listed below:

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

The virtual ground concept in op-amp is valid for

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 6
  • Virtual ground concept is valid for only Negative feedback systems
  • Negative feedback systems decrease the differential input voltage between and non-inverting terminals

When difference becomes zero, the voltage at two terminals become equal, which is a virtual ground concept

RRB JE ECE (CBT II) Mock Test- 6 - Question 7

Azhar is 4 times as good a workman as Balaraj and therefore is able to finish a job in 45 days less than Balaraj. Working together, they can do it in:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 7

Azhar takes 45 days less to finish a job and is 4 times more efficient than Balraj.
Therefore, let Balraj takes x days then Azhar takes x – 45 days.
Now, as per the question if Balraj takes 4D days then Azhar should take 4D days to
complete the job.
X/X - 45 = 4D/D
⇒ x =4x - 180
⇒ x = 60 days
Thus, Balraj takes 60days and Azhar takes (60 – 45) = 15 days to finish the job.
Working together, they need

RRB JE ECE (CBT II) Mock Test- 6 - Question 8

The power supply has a full load voltage of 24 V. The value of its no-load voltage for 5% regulation is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 8

percentage regulation =  X 100%
5/100 = 
5/100 . (24) = (VNL - 24)
VNL = 25.2

RRB JE ECE (CBT II) Mock Test- 6 - Question 9

Which of the following is true regarding an LC oscillator in resonance?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 9


At resonance
Xc = XL

RRB JE ECE (CBT II) Mock Test- 6 - Question 10

The upward shift in the position of fermi level with respect to intrinsic fermi level in n type semiconductor is given by:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 10

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
shift = kTln (ND/ni)
From the expression
As ND increases ND/nincreases and the shift increases.

RRB JE ECE (CBT II) Mock Test- 6 - Question 11

Medium frequency waves travel mainly as

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 11

Ground Waves:

Middle frequency contains frequencies in range 300 kHz to 3 MHz.

These waves travel mainly by reflection from the ground surface and are called Ground waves.

MF is mostly used for AM radio broadcasting, navigational radio beacons, maritime ship-to-shore communication, and transoceanic air traffic control.

Sky Waves:

Skywave refers to the propagation of radio waves reflected or refracted back toward Earth from the ionosphere, an electrically charged layer of the upper atmosphere.

Space Waves:

The high-frequency electromagnetic wave is not reflected back by the ionosphere, so to use high-frequency electromagnetic wave in communication we used space wave propagation.

  • Space waves are used in two types of communication −
  • Line-of-sight (LOS) propagation.
  • Satellite communication
RRB JE ECE (CBT II) Mock Test- 6 - Question 12

The process of increasing fertility of soil by earthworms is known as

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 12

Vermicompost is the product of the composting process using various species of worms, usually red wigglers, white worms, and other earthworms, to create a mixture of decomposing vegetable or food waste, bedding materials, and vermicast. This process of increasing fertility of soil by earthworms is known as vermicomposting.

Important Points:

Eutrophication: It is an enrichment of water by nutrient salts that causes structural changes to the ecosystem.

Compounds that contain nitrogen and phosphorus (fertilizers) drives the eutrophication. These nutrients are generally not toxic.

RRB JE ECE (CBT II) Mock Test- 6 - Question 13

In which year was the Battle of Waterloo fought?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 13

The Battle of Waterloo was fought between the French army commanded by Napoleon Bonaparte and the British army led by the Duke of Wellington in 1815.

The Battle of Waterloo is considered one of the most dreadful battles in history. In this war, the Duke of Wellington defeated Napoleon Bonaparte.

Napoleon surrendered in front of British forces after a few months. He was sent to St. Helena where he died in 1921.

RRB JE ECE (CBT II) Mock Test- 6 - Question 14

Calculate the fastest rise time (in ms) a sine wave can have to be reproduced by a CRO, if the bandwidth ranges from 0 Hz to 10 Hz.

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 14

Given that, bandwidth (B.W.) = 10 Hz
Raise time is, t r = 0.35/B.W. = 0.35/10 = 0.035 = 35 ms

RRB JE ECE (CBT II) Mock Test- 6 - Question 15

A concave lens has a focal length of 20 cm. At what distance the image be formed from the lens if the object is placed at a distance of 40 cm from the lens.

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 15

Using,
1/f = 1/v - 1/u
Where,
f = focal length = -20 cm
v = Image distance = v
u = Object distance = -40 cm
1/-20 = 1/v - (1/-40)
v = -13.33 cm

RRB JE ECE (CBT II) Mock Test- 6 - Question 16
Which of the following rivers originates outside India?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 16

  • River Sutlej originates outside India, at Mount Kailash, Tibet.
  • It enters India through Shipki la Pass.
  • Yamuna River originates at Yamunotri Glacier, Uttarakhand.
  • Ganga River originates at Gaumukh, Gangotri Glacier, Uttarakhand.
  • Luni is the only river that flows in Rajasthan.

RRB JE ECE (CBT II) Mock Test- 6 - Question 17
Which of the following gain unauthorized access to the user system and demand ransom to regain access?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 17

Ransomware is a type of malicious software that threatens to publish the victim's data or perpetually block access to it unless a ransom is paid

  • In May 2017, WannaCry ransomware targeted computers running the Microsoft Windows operating system by encrypting data and demanding ransom payments in the Bitcoin cryptocurrency
  • Around 37 incidents of ransomware attacks were reported to the Indian Computer Emergency Response Team (CERT-In), according to Minister of State for Electronics and IT

Adware is any software application in which advertising banners are displayed while a program is running

Spyware is mostly used for the purposes of tracking and storing internet user movements on the web and provides pop-up ads to the internet user

Trojan horse is a program that appears harmless but in fact malicious which misleads user of its true intent

RRB JE ECE (CBT II) Mock Test- 6 - Question 18
A galvanometer (G) measures up to 100 mA current. It is to be converted to a voltmeter to measure up to 100 volts. What is required to be done?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 18

In order to convert a Galvanometer into voltmeter, a very high resistance known as "series resistance" is connected in series with the galvanometer.

Required resistance = 100/100 mA = 1000 Ω

RRB JE ECE (CBT II) Mock Test- 6 - Question 19
What will happen if in a C program you assign a value to an array element whose subscript exceeds the size of array?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 19

Consider an example:

{

int a[10]; \ the system reserves memory for 10 elements of array

a[11] = 12; \ enter the value into the 11 the element of array which was not defined

printf(“%d”, a[11]);

}

Now if the 11th element tries to overwrite the space which is occupied by some other important data, the program may crash.

If the system does not tries to override important data, then the program gives output as 12.

RRB JE ECE (CBT II) Mock Test- 6 - Question 20

Which of the following is the shortcut to save and print a web page in windows?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 20

Click on the webpage and use the appropriate keyboard shortcut:

  • For Chrome OS and Windows, use Ctrl + S.
  • For Mac OS X and macOS Sierra, use Command + S.

Clicking the Print button on the toolbar instantly prints the web page.

  • To work the Print dialogue box, choose Print from the Print button’s menu or press Ctrl + p
  • Ctrl + F4 and Ctrl + w both closes current window.

Alt + F6 toggles windows. Alt + F8 toggles find result windows.

RRB JE ECE (CBT II) Mock Test- 6 - Question 21
The Imaginary part of the propagation constant shows:
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 21

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- 6 - Question 22

Which of the following flag is not affected by INX instruction:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 22

The various instructions and flags they affect is given by:

For INX and DCX instructions no flags affected

RRB JE ECE (CBT II) Mock Test- 6 - Question 23
A  device which feeds data to the computer system for processing is known as
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 23

Input is any data or information is sent to computer for processing it into meaningful information. Input devices feed data to the computer system for processing. Eg.Keyboard.

RRB JE ECE (CBT II) Mock Test- 6 - Question 24
Memory with highest storage capacity is:
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 24

Primary Memory:

Primary memory is the main memory of the computer system.

Accessing data from primary memory is faster because it is an internal memory of the computer.

The primary memory is most volatile which means data in primary memory does not exist if it is not saved when a power failure occurs.

The primary memory is a semiconductor memory.

It is costlier compared with secondary memory.

The capacity of primary memory is very much limited and is always smaller compares to secondary memory.

E.g. RAM, ROM        

Secondary Memory:

All secondary storage devices which are capable of storing high volume data is referred to secondary memory.

It's slower than primary memory.

It can save a substantial amount of data, in the range of gigabytes to terabytes.

This memory is also called backup storage or mass storage media.

E.g. Hard Disk, USB, Floppy, CD, DVD, Magnetic disks, Magnetic tapes.

RRB JE ECE (CBT II) Mock Test- 6 - Question 25

What is the content of HL pair after the execution of instructions?

MVI H, 40 H

MVI L, 20 H

MVI M, 10 H 

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 25

The first two MVI instruction stores the data in H and L registers respectively

H ← 40

L ← 20

The last instruction MVI M, 10 H will store the data 10 in the memory location pointed by HL pair

[4020] ← 10

Thus, the content of HL pair is unaffected by last MVI instruction

HL pair contains 4020

RRB JE ECE (CBT II) Mock Test- 6 - Question 26
Which one of the following metals do not react with cold as well as hot water?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 26

Na, Ca and Mg are reactive metals and quickly react with water.

At room temperature, the reaction of Fe is very slow but red-hot iron reacts with steam to form Fe3O4 with hydrogen gas.

3Fe + 4H2O → Fe3O4 + 4 H2

RRB JE ECE (CBT II) Mock Test- 6 - Question 27

The input impedance of a line of infinite length is equal to ___________

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 27

A transmission line of finite length that is terminated at one end with an impedance equal to the characteristic impedance appears to the source like an infinitely long transmission line and produces no reflections.

The behaviour of transmission line due to variation in length is tabulated below:

RRB JE ECE (CBT II) Mock Test- 6 - Question 28
In a negative feedback amplifier, current sampling
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 28

RRB JE ECE (CBT II) Mock Test- 6 - Question 29

Refer to the given circuit. Which of the following equations is correct about the current in this network, according to Kirchhoff's Current Law (KCL)?

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 29

Concept:

According to Kirchhoff’s current law (KCL), the algebraic sum of the electric currents meeting at a common point is zero. I.e. the sum of currents entering a node is equal to the sum of currents leaving the node. It is based on the conservation of charge.

Calculation:

By applying KCL,

i1 + i5 = i2 + i3 + i4

RRB JE ECE (CBT II) Mock Test- 6 - Question 30
What does FTP stand for in computer language?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 6 - Question 30

The File Transfer Protocol (FTP) is a standard network protocol used to transfer data from one medium to another over a TCP-based network, such as the Internet. It is usually built on a client-server format and uses control and data connections between the client and the server. It is one of most widely used systems to transfer data from one computer to another.

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