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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 ECE (CBT II) Mock Test- 3

RRB JE ECE (CBT II) Mock Test- 3 for Electronics and Communication Engineering (ECE) 2024 is part of Electronics and Communication Engineering (ECE) 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) 2024 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

The gain of the RC low pass filter having time constant ‘τ' and frequency ‘ω' is:

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

The Low pass filter is shown below

The transfer function is given by

Time constant RC = τ

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

A data communication system has ______ components.

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

  • 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

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

The shunt used in milliammeter

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

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 4

For an open circuited transmission line with no load impedance connected to its end, the voltage reflection coefficient is:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 4

Concept:
The reflection coefficient gives the percentage of incident voltage reflected from the load.
Mathematically
reflection coefficient = 
Where ZL is the load impedance and Zo is the characteristic impedance.
Application of Concept:
For open circuit transmission line
ZL = infinite
reflection coefficient = 
As ZL → ∞
Reflection coefficient = 1

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

A repeater has ____________ capacity.

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

An amplifier cannot discriminate between the intended signal and noise; it amplifies equally everything fed into it. A repeater dots not amplify the signal; it regenerates the signal. When it receives a weakened or corrupted signal, it creates a copy, bit for bit, at the original strength.

A repeater forwards every frame; it has no filtering capacity.

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

In a J-K flip flop, when Jn = 0 and Kn = 1, the output Qn + 1 will have a value of:

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

The Truth table of JK flip flop is

From the truth table when J = 0, K = 1 the output is reset, i.e. Qn + 1 = 0

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

Most harmful environmental pollutants are

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

Nonbiodegradable chemicals are most harmful environmental pollutants as these cannot be broken down into simpler, harmless substances in nature. Whereas, biodegradable pollutants can be rapidly decomposed into simpler and harmless substances by natural processes.

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

In star to delta transformation

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

In star to delta transformation
Ra = R2 + R3 + R2R3/R1
Rb = R1 + R3 + R1R3/R2
Rc = R1 + R2 + R1R2/R3

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

The coldest and warmest layer of atmosphere are_________ respectively.

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

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 10

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 10

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 11

The idea of “Welfare State” in the Indian Constitution is enshrined in its

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

 Self Explanatory.
Hence, the correct option is (B)

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

A microprocessor having 8-bit address line has _____ memory locations?

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

If there is n-bit address bus, then number of memory locations are 2N
Thus for 8-bit address line number of memory locations are 28= 256
Hence, the correct option is (C)

RRB JE ECE (CBT II) Mock Test- 3 - Question 13
Where is Kudremukh National Park?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 13

  • 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 14
F = v + v̅ w + v̅ w̅ x + v̅ w̅ x̅ y + v̅ w̅ x̅ y̅ z, where minimized Boolean function F is
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 14

F = v + v̅ w + v̅ w̅ x + v̅ w̅ x̅ y + v̅ w̅ x̅ y̅ z

F = v + v̅ w + v̅ w̅ x + v̅ w̅ x̅ (y + y̅ z)                 

using the relation (y + y̅ z) = (y + y̅ ) (y+z) = (y+z)

F = v + v̅ w + v̅ w̅ x + v̅ w̅ x̅ (y+z)

F = v + v̅ w + v̅ w̅ (x + x̅ (y+z))

F = v + v̅ w + v̅ w̅ ( x + y + z )

F = v + v̅ ( w + w̅ ( x + y + z ) )

F = v + v̅ (w + x + y + z )

F = v + w + x + y + z

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

The radius of curvature of a concave mirror is 30 cm. Following Cartesian Sign Convention, its focal length is expressed as ________.

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

  • In a concave mirror, the light rays from the object converge after reflection.
  • Light rays from infinity, parallel to the principal axis, thus meet at the point on the principal axis on the side of the mirror which is towards the object.
  • This distance is considered as negative.
  • Hence the focal length of a concave mirror is negative.

⇒ Focal length, f = R/2

⇒ f = 30/2 = 15 cm

For the Concave mirror, the focal length is negative.

Therefore, f = -15 cm.

RRB JE ECE (CBT II) Mock Test- 3 - Question 16
Which of the following device receive data message?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 16

A data communication system has five components:

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 17
Which of the following statement is true?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 17

Concept:

SRAM can retain stored data indefinitely and it can be implemented as a flip flop circuit to store one bit of information.

DRAM uses a single transistor to access the capacitor charge. Since stored charge is gradually lost because of leakage capacitance associated with them, the cells require periodic refreshing.

Difference between SRAM and DRAM:

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

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 19
Compound with a formula C3H6O2 is:
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 19

Concept:

The compounds are characterised by their functional group:

  1. Aldehyde:

These compounds have R -CHO as their functional group.

  1. Ketone:

These compounds are characterised by  R-CO- R type of functional groups where R is alkyl chain.

  1. Acid:

Compounds containing R-COOH group at the end are called carboxylic acids

  1. Alcohols:

Alcohols have -OH group attached to its end.

Application:

From the above discussion only Carboxylic Acid group have 2 oxygen atoms.

The given compounds can be written as:

CH3 – CH2 – COOH

Which is chemical formula for Propanoic acid.

RRB JE ECE (CBT II) Mock Test- 3 - Question 20
The pH of acid rain should always be less than
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 20

When the quantity of acids in the raining water is more than the average, then such rain is called 'Acid rain'.

We are aware that normally rainwater has a pH of 5.6.  When the pH of the rainwater drops below 5.6, it is called acid rain.

Acid rain refers to the ways in which acid from the atmosphere is deposited on the earth’s surface. Oxides of nitrogen and sulphur which are acidic in nature can be blown by the wind along with solid particles in the atmosphere and finally settle down either on the ground as dry deposition or in water, fog and snow as wet deposition.

The bad effects of acid rain

  • When acid rain falls and flows as groundwater to reach rivers, lakes etc, it affects plants and animal life in the aquatic ecosystem
  • Acid rain is harmful to agriculture, trees and plants as it dissolves and washes away nutrients needed for their growth
  • It causes respiratory ailments in human beings and animal
  • It may also cause corrosion in many buildings bridges, monuments, fencing etc
  • It causes irritation in the eyes and skin of human beings
  • This rain reduces the lustre of the metals too
  • Acid rain damages buildings and other structures made of stone or metal
  • The Taj Mahal in India has been affected by acid rain

RRB JE ECE (CBT II) Mock Test- 3 - Question 21

A low pass circuit with large time constant can be used as:

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 21

The low pass circuit is shown below:

Vi = VR + VC

VR = iR

Current through capacitor is given by

As time constant is large

RC > 1

The output voltage is integral of input voltage and hence the low pass filter acts as integrator.

RRB JE ECE (CBT II) Mock Test- 3 - Question 22
Which of the following is not a characteristic of Data communication?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 22

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 23

What is the content of memory location 5000 after execution?

MVI H, 40 H

MVI L, 20 H

SHLD 5000

Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 23

MVI H, 40 H     // H ← 40 H

MVI L, 20 H     // L ← 20 H

SHLD means store the content of HL pair directly to memory address

Following the “little endian rule”, the data from lower order register will go to lower order address and data from higher order register will go to higher order address

SHLD 5000   

5000 ← L

5001 ← H 

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

The inductance of a long solenoid is measured as 1 mH. What will be its inductance, if the number of turns is doubled?

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

Inductance of a coil is proportional to square of number of turns

L ∝ N2

Given N ↑2 times

∴ New inductance L2 = 4 L1 = 4 × 1 = 4 mH

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

Hall effect can be used to measure

1. Carrier concentration

2. Type of semiconductor

3. Magnetic field

4. Conductivity

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

Concept:

Hall Voltage states that if a specimen (metal or semiconductor) carrying a current I is placed in transverse magnetic field B, an electric field is induced in direction perpendicular to both I and B.

Hall Voltage is given by:

V H = Ed = B I/pW

Applications of Hall effect:

Hall effect can be used to find:

1. Carrier concentration

2. Type of semiconductor

3. Conductivity

4. Mobility

It cannot be used to find magnetic field.

RRB JE ECE (CBT II) Mock Test- 3 - Question 26
While executing IN and OUT instruction the data is transferred between?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 26

8085 has 74 instruction. Out of 74 instructions 72 are used for operation between memory and microprocessor and 2 are used for the transfer of data between microprocessor and I/O device.

IN, 8bit address

The content of the input port whose address is given is read and loaded into the accumulator.

OUT, 8bit address

The content of the accumulator is copied into the output port whose address is given are read and loaded into the accumulator.

RRB JE ECE (CBT II) Mock Test- 3 - Question 27
Which of the following images are formed by the Plane mirror?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 27

  • The Images formed by the Plane Mirror are laterally inverted, virtual and erect.
  • Plane mirrors are used in looking glasses, Solar Cooker etc.
  • If an object moves with a certain velocity towards a plane mirror then its image appears to be moving with twice the velocity it was moving towards the mirror.

RRB JE ECE (CBT II) Mock Test- 3 - Question 28
Who won the Women’s British Open 2019?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 28

  • Hinako Shibuno of Japan won the Women’s British Open by one shot over Lizette Salas on 4 August 2019.
  • She played for the first time outside of her country and birdied five of the final nine holes in a 4-under 68 and 18-under 270 overall.
  • Morgan Pressel (67) finished fourth at 15 under, just ahead of former leader Ashleigh Buhai (70).

RRB JE ECE (CBT II) Mock Test- 3 - Question 29

The equation –

Cu + x HNO3 → Cu (NO3 )2 + y NO2 + 2 H2O

The values of x and y ar
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 29

Start with hydrogen since it occurs in only one compound on each side of equation

Hydrogen on LHS = x

Hydrogen on RHS = 4

Replace x by 4

Cu + 4 HNO3 → Cu(NO3 )2 + y NO2 + 2 H2O

Now balance nitrogen

Nitrogen on LHS = 4

Nitrogen on RHS = 2 + y

Replace y by 2

Cu + 4 HNO3 → Cu(NO3 )2 + 2 NO2 + 2 H2O

The equation is balanced.

RRB JE ECE (CBT II) Mock Test- 3 - Question 30
Which constellation is known as the ‘Saptarishi’?
Detailed Solution for RRB JE ECE (CBT II) Mock Test- 3 - Question 30

  • The Indian name of Ursa Major constellation or Great Bear constellation is 'Saptarishi'.
  • The Ursa Major constellation consists of 7 bright rishis (stars) and these rishis become the 7 stars of the BIG DIPPER constellation.
  • Ursa Major is visible throughout the year from the northern hemisphere. 

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