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Full Test 1 - EKT CSE - AFCAT MCQ


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30 Questions MCQ Test IAF AFCAT Mock Test Series and EKT for (ME, EEE, CSE) 2025 - Full Test 1 - EKT CSE

Full Test 1 - EKT CSE for AFCAT 2024 is part of IAF AFCAT Mock Test Series and EKT for (ME, EEE, CSE) 2025 preparation. The Full Test 1 - EKT CSE questions and answers have been prepared according to the AFCAT exam syllabus.The Full Test 1 - EKT CSE MCQs are made for AFCAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Full Test 1 - EKT CSE below.
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Full Test 1 - EKT CSE - Question 1

Two balls each of radius R, equal mass and density are placed in contact, then the force of gravitation between them is proportional to –

Detailed Solution for Full Test 1 - EKT CSE - Question 1


Gravitational force,

Full Test 1 - EKT CSE - Question 2

Special theory of relativity treats problems involving –

Detailed Solution for Full Test 1 - EKT CSE - Question 2

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

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Full Test 1 - EKT CSE - Question 3

A total charge Q is broken in two parts Q1 and Q​2 and they are placed at a distance R from each other. The maximum force of repulsion between them will occur, when

Detailed Solution for Full Test 1 - EKT CSE - Question 3

The force exerted by charged particle is given by:

For F to be maximum, 

Full Test 1 - EKT CSE - Question 4

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 Full Test 1 - EKT CSE - Question 4

Heat released by sun Q1 = σAT14

As the temperature is doubled, T2 = 2T1

So Q2 = σAT24 = σA(2T1)4 = 16Q1

Full Test 1 - EKT CSE - Question 5

Dynamo is a device for converting

Detailed Solution for Full Test 1 - EKT CSE - Question 5

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

Full Test 1 - EKT CSE - Question 6

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 Full Test 1 - EKT CSE - Question 6


By conservation of momentum

Full Test 1 - EKT CSE - Question 7

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

Detailed Solution for Full Test 1 - EKT CSE - Question 7

 for constant G and M

= +2%

∴ The value of g increases by 2%

Full Test 1 - EKT CSE - Question 8

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 Full Test 1 - EKT CSE - Question 8

Here, aμw = 1.33

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

wμa = 1/ aμw = 1/1.33 = 0.7518

Full Test 1 - EKT CSE - Question 9

A disc is in rotation with angular speed ω. If a child sits on it, what is conserved –

Detailed Solution for Full Test 1 - EKT CSE - Question 9

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.

Full Test 1 - EKT CSE - Question 10

Tesla is a unit of _______.

Detailed Solution for Full Test 1 - EKT CSE - Question 10

The Tesla (symbolized T) is the standard unit of magnetic flux density.

Full Test 1 - EKT CSE - Question 11

Zero address instruction format is used for

Detailed Solution for Full Test 1 - EKT CSE - Question 11

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.

Full Test 1 - EKT CSE - Question 12

In IEEE-754 double precision number representation, number of exponent and mantissa bits respectively are:

Detailed Solution for Full Test 1 - EKT CSE - Question 12

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.

Full Test 1 - EKT CSE - Question 13

Which of the following is not true about process schedulers?

Detailed Solution for Full Test 1 - EKT CSE - Question 13

Option (A) is TRUE. Short Term Scheduler is faster than Long Term Scheduler.

Option (B) is TRUE. Medium term scheduling is part of the swapping. It removes the processes from the memory. It reduces the degree of multiprogramming. The medium term scheduler is in-charge of handling the swapped out-processes.

Option (C) is FALSE. Long Term Scheduler controls the degree of Multiprogramming. Short Term Scheduler provides lesser control over degree of Multiprogramming. Medium Scheduler reduces the Degree of Multiprogramming.

Option D is TRUE. Long Term Scheduler selects processes from the pool and loads them into memory for execution.

Full Test 1 - EKT CSE - Question 14

Which of the following main memory will be capable of supporting 80386 processor which has 32 bit address (32 pins) capacity?

Detailed Solution for Full Test 1 - EKT CSE - Question 14

80386 processor has 32-bit address capacity hence it is capable of accessing 232 = 22 × 230 = 4 GB = 4096 MB of main memory.

Full Test 1 - EKT CSE - Question 15

Gray code representation of 14 is___________________.

Detailed Solution for Full Test 1 - EKT CSE - Question 15

The reflected binary code (RBC), also known as Gray code after Frank Gray, is a binary numeral system where two successive values differ in only one bit. The reflected binary code was originally designed to prevent spurious output from electromechanical switches. Today, Gray codes are widely used to facilitate error correction in digital communications such as digital terrestrial television and some cable TV systems.

Conversion from Gray Code to Binary Code:

Let Gray Code be G3  G2  G1  G0 and Binary Code be B3 B2 B1 B0. Then the respective Grey Code can be obtained as follows:

Now, convert 14 in grey code:

Binary code of 14 ® 1110

Hence the grey code is: 1001.

Full Test 1 - EKT CSE - Question 16

The main purpose of using core in a transformer is to

Detailed Solution for Full Test 1 - EKT CSE - Question 16

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.

Full Test 1 - EKT CSE - Question 17

The following C declaration 

struct node {

int i ;

float j ;

} ;

struct node * S[10] ;

Define S to be

Detailed Solution for Full Test 1 - EKT CSE - Question 17

As it as an array of pointers to struct type.

Full Test 1 - EKT CSE - Question 18

LRU (Least Recently used) is almost universally used as the cache replacement policy because

Detailed Solution for Full Test 1 - EKT CSE - Question 18

OPT is not achievable, requiring perfect knowledge of future use.

Full Test 1 - EKT CSE - Question 19

BIOS stands for

Detailed Solution for Full Test 1 - EKT CSE - Question 19

BIOS (basic input/output system) is the program a personal computer's microprocessor uses to get the computer system started after we turn it on. It also manages data flow between the computer's operating system and attached devices such as the hard disk, video adapter, keyboard, mouse and printer.

Full Test 1 - EKT CSE - Question 20

A decoder is a special case of

Detailed Solution for Full Test 1 - EKT CSE - Question 20

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.

Full Test 1 - EKT CSE - Question 21

X.25 is used for

Detailed Solution for Full Test 1 - EKT CSE - Question 21

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.

Full Test 1 - EKT CSE - Question 22

Loop back IP address is

Detailed Solution for Full Test 1 - EKT CSE - Question 22

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.

Full Test 1 - EKT CSE - Question 23

The address bus is

Detailed Solution for Full Test 1 - EKT CSE - Question 23

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

Full Test 1 - EKT CSE - Question 24

Meta data is

Detailed Solution for Full Test 1 - EKT CSE - Question 24

Metadata is the data providing information about the other data.

Full Test 1 - EKT CSE - Question 25

Address line of 8086 microprocessor is

Detailed Solution for Full Test 1 - EKT CSE - Question 25

8086 has 20 bits of address lines 

4 bits = 1 nibble

thus 20 bits = 5 nibble

8086 microprocessor can address 220 memory locations or 1 MB of memory.

Full Test 1 - EKT CSE - Question 26

Shortest Remaining Time Next (SRTN) Scheduling may be implemented-

Detailed Solution for Full Test 1 - EKT CSE - Question 26

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.

Full Test 1 - EKT CSE - Question 27

What is decipherment?

Detailed Solution for Full Test 1 - EKT CSE - Question 27

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

Full Test 1 - EKT CSE - Question 28

Which of the following is true about WORM?

Detailed Solution for Full Test 1 - EKT CSE - Question 28

A worm is a self-replicating malicious software that does not alter files but duplicates itself. It is common for worms to be noticed only when their uncontrolled replication consumes system resources, slowing or halting other tasks.

Full Test 1 - EKT CSE - Question 29

Which command(s) is/are used to to add, delete, or modify columns in an existing table in SQL?

Detailed Solution for Full Test 1 - EKT CSE - Question 29

To delete a column in a table, use the following syntax (notice that some database systems don't allow deleting a column):
ALTER TABLE table_name
DROP COLUMN column_name;

Full Test 1 - EKT CSE - Question 30

The Class A has ________ network bits.

Detailed Solution for Full Test 1 - EKT CSE - Question 30

The size of network number bit field for class A is 8. It is 16 for class B; 24 for class C whereas it is not defined for class D and class E.

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