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AFCAT EKT Computer Science Mock Test - 4 - Computer Science Engineering (CSE) MCQ


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30 Questions MCQ Test - AFCAT EKT Computer Science Mock Test - 4

AFCAT EKT Computer Science Mock Test - 4 for Computer Science Engineering (CSE) 2025 is part of Computer Science Engineering (CSE) preparation. The AFCAT EKT Computer Science Mock Test - 4 questions and answers have been prepared according to the Computer Science Engineering (CSE) exam syllabus.The AFCAT EKT Computer Science Mock Test - 4 MCQs are made for Computer Science Engineering (CSE) 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for AFCAT EKT Computer Science Mock Test - 4 below.
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AFCAT EKT Computer Science Mock Test - 4 - Question 1

An array element is accessed using

Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 1

An array is collection of variables that are accessed with an index number. The first element of the array is at index 0.

AFCAT EKT Computer Science Mock Test - 4 - Question 2

8 bits put together make up a

Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 2

A byte is a unit of data that is eight binary digits (bits) long.

AFCAT EKT Computer Science Mock Test - 4 - Question 3

Belady’s Anomaly is a behaviour of which page replacement Algorithm?

Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 3

In computer storage, Belady’s anomaly proves that it is possible to have more page faults when increasing the no. of page frames while using the first in the first out (FIFO) page replacement algorithm.

AFCAT EKT Computer Science Mock Test - 4 - Question 4
The refractive index of water with respect to air is 1.33, find the refractive index of air with respect to water?
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 4

Here, aμw = 1.33

∴ Refractive index of air w.r. to water,

wμa = 1/ aμw = 1/1.33 = 0.7518

AFCAT EKT Computer Science Mock Test - 4 - Question 5

If the radius of the earth decreases by 1% and its mass remains same, then the acceleration due to gravity _______

Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 5

g = GM/R², for constant G and M

∴ (Δg / g) × 100 = -2 × (ΔR / R) × 100 = -2 × (-1)%

= +2%

∴ The value of g increases by 2%

AFCAT EKT Computer Science Mock Test - 4 - Question 6
X.25 is used for
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 6

X.25 is a standard for WAN communications that defines how connections between user devices and network devices are established and maintained. X.25 is designed to operate effectively regardless of the type of systems connected to the network. It is typically used in the packet-switched networks (PSNs) of common carriers, such as the telephone companies. Subscribers are charged based on their use of the network.

AFCAT EKT Computer Science Mock Test - 4 - Question 7

What is the value of lim (x→∞) x² sin(1/x²)?

Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 7

Since x → ∞, t → 0
∴ lim (x→∞) x² sin(1/x²) = lim (t→0) (sin t / t) = 1

AFCAT EKT Computer Science Mock Test - 4 - Question 8
A network that spreads over cities is
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 8

MAN (Metropolitan Area Network) is a computer network that usually spans a city or a large campus.

Whereas WAN (wide area network) is a large network that interconnects many other networks spanning cities, countries or continents.

LAN stands for Local Area Network.

AFCAT EKT Computer Science Mock Test - 4 - Question 9
The main purpose of using core in a transformer is to
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 9

In an electrical power transformer, there are primary, secondary and may be tertiary windings. The performance of a transformer mainly depends upon the flux linkages between these windings. For efficient flux linking between these windings, one low reluctance magnetic path common to all windings should be provided in the transformer. This low reluctance magnetic path in transformer is known as core of transformer.

AFCAT EKT Computer Science Mock Test - 4 - Question 10
Loop back IP address is
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 10

Loopback address is a special IP number (127.0.0.1) that is designated for the software loopback interface of a machine. The loopback interface has no hardware associated with it, and it is not physically connected to a network.

AFCAT EKT Computer Science Mock Test - 4 - Question 11

A body of mass M moving with a velocity V explodes into two equal parts. If one comes to rest and the other body moves with velocity v, what would be the value of v?

Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 11


By conservation of momentum:
MV = (M/2) × 0 + (M/2) × v
⇒ v = 2V

AFCAT EKT Computer Science Mock Test - 4 - Question 12
Meta data is
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 12

Metadata is the data providing information about the other data.

AFCAT EKT Computer Science Mock Test - 4 - Question 13
A disc is in rotation with angular speed ω. If a child sits on it, what is conserved –
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 13

When child sits on the rotating disc, his weight acts downwards and does not produce any torque. Hence the angular momentum of the system is conserved.

AFCAT EKT Computer Science Mock Test - 4 - Question 14
A decoder is a special case of
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 14

If we set input value of de-multiplexer to 1, it works as a decoder.

Hence decoder is a special case of a de-multiplexer.

AFCAT EKT Computer Science Mock Test - 4 - Question 15
In C++, by default, the members of a class are –
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 15

By default , all members of a class declared with the “class” Keyword have private access for all its members. 

AFCAT EKT Computer Science Mock Test - 4 - Question 16
Time sharing system is
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 16

The round robin scheduling algorithm is designed especially for time-sharing systems. It gives each job a time slot or quantum (its allowance of CPU time) and interrupting the job if it is not completed by then. The job is resumed next time a time slot is assigned to that process.

AFCAT EKT Computer Science Mock Test - 4 - Question 17
Zero address instruction format is used for
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 17

In stack organized architecture push and pop instruction is needs a address field to specify the location of data for pushing into the stack and destination location during pop operation but for logic and arithmetic operation the instruction does not need any address field as it operates on the top two data available in the stack.

AFCAT EKT Computer Science Mock Test - 4 - Question 18

Let f : Z → R be a function such that f(x) = |x + 1|. Which one of the following statements is TRUE?

Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 18

To disprove that f is surjective (by counter-example), we find a y ∈ R such that there

does not exist x ∈ Z with f(x) = y.

Let y = 0.5. Then, |x + 1| = 0.5 ⇒ x = - 0.5 or x = - 1.5; neither of these is an integer. Hence, there does not exist x 2 Z with f(x) = 0:5 and so f is not surjective.

To disprove that f is injective (by counter-example), we find an x, y ∈ Z such that

f(x) = f(y) but x ≠ y.

Let x = 1, y = - 3. Then f(1) = 2 = f(-3) but 1 ≠ -3. Thus, f is not injective.

AFCAT EKT Computer Science Mock Test - 4 - Question 19
What is decipherment?
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 19

Decipherment in cryptography refers to decryption which converts ciphertext (Secret or coded text) into plaintext (original text).

AFCAT EKT Computer Science Mock Test - 4 - Question 20
A modem performs ________.
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 20

Modem (Modulator-Demodulator) performs modulation & Demodulation i.e. formats the computer data so it can travel over phone or cable lines. Also, Modem perform data compression. The bits are grouped together by modem, in order to compress them.

AFCAT EKT Computer Science Mock Test - 4 - Question 21
Cookie in parlance to internet is:
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 21

Cookie is a small piece of data sent from a website and stored in a user’s web browser while a user is browsing a website. It is used to maintain a session.

AFCAT EKT Computer Science Mock Test - 4 - Question 22
In IEEE-754 double precision number representation, number of exponent and mantissa bits respectively are:
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 22

In IEEE-754 double precision number representation, number of exponent and mantissa bits respectively are 11 and 52, and 1 bit is used to represent sign.

AFCAT EKT Computer Science Mock Test - 4 - Question 23
Special theory of relativity treats problems involving –
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 23

Special theory of relativity treats problems involving inertial frame of reference.

AFCAT EKT Computer Science Mock Test - 4 - Question 24
If the temperature of the sun is doubled, the rate of energy received on earth will be increased by a factor of
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 24

Heat released by sun Q1 = σAT14

As the temperature is doubled, T2 = 2T1

So Q2 = σAT24 = σA(2T1)4 = 16Q1

AFCAT EKT Computer Science Mock Test - 4 - Question 25
Dynamo is a device for converting
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 25

A dynamo converts mechanical energy to electrical energy, whereas an motor converts electrical energy into mechanical energy.

AFCAT EKT Computer Science Mock Test - 4 - Question 26
________ is used for high speed telecommunications.
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 26

Ethernet cabling is most commonly used in high-speed wired computer networks. Data can be transmitted over wireless access points, twisted pair coaxial or fiber optic cable at a speed of 10mbps to 1000mbps. Standardization of Internet connections (based on Ethernet communication) with speeds upto 100 Gigabit per second has just begun in 2009.

AFCAT EKT Computer Science Mock Test - 4 - Question 27
Paging is a –
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 27

is a memory-management scheme that permits the physical address space of a process to be noncontiguous. Paging avoids external fragmentation and the need for compaction. It also solves the considerable problem of fitting memory chunks of varying sizes onto the backing store.

AFCAT EKT Computer Science Mock Test - 4 - Question 28
The address bus is
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 28

Address bus is unidirectional out of microprocessor it carries address which maps to a particular memory location.

AFCAT EKT Computer Science Mock Test - 4 - Question 29
The task of the lexical analysis phase is
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 29

Lexical analysis is the process of converting a sequence of characters into a sequence of tokens. Lexical analysis phase, builds literal table (created to describe all literals used in the source program); Identifier table (created to describe all identifiers used in the source program) and uniform symbol table ( created to represent the program as a string of tokens rather than of individual characters).

AFCAT EKT Computer Science Mock Test - 4 - Question 30
Shortest Remaining Time Next (SRTN) Scheduling may be implemented-
Detailed Solution for AFCAT EKT Computer Science Mock Test - 4 - Question 30

Shortest job first (SJF) can be implemented in pre-emptive or non- preemptive manner But SRTN is the preemptive version of SJF in which the scheduler always dispatches the ready process which was the shortest expected time to completion.

Note: SRTF(Shortest Remaining Time First) is same as SRTN.

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