AFCAT Exam  >  AFCAT Tests  >  IAF AFCAT Mock Test Series and EKT for (ME, EEE, CSE) 2025  >  Full Test 5 - EKT EEE - AFCAT MCQ

Full Test 5 - EKT EEE - AFCAT MCQ


Test Description

30 Questions MCQ Test IAF AFCAT Mock Test Series and EKT for (ME, EEE, CSE) 2025 - Full Test 5 - EKT EEE

Full Test 5 - EKT EEE for AFCAT 2024 is part of IAF AFCAT Mock Test Series and EKT for (ME, EEE, CSE) 2025 preparation. The Full Test 5 - EKT EEE questions and answers have been prepared according to the AFCAT exam syllabus.The Full Test 5 - EKT EEE MCQs are made for AFCAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Full Test 5 - EKT EEE below.
Solutions of Full Test 5 - EKT EEE questions in English are available as part of our IAF AFCAT Mock Test Series and EKT for (ME, EEE, CSE) 2025 for AFCAT & Full Test 5 - EKT EEE solutions in Hindi for IAF AFCAT Mock Test Series and EKT for (ME, EEE, CSE) 2025 course. Download more important topics, notes, lectures and mock test series for AFCAT Exam by signing up for free. Attempt Full Test 5 - EKT EEE | 50 questions in 45 minutes | Mock test for AFCAT preparation | Free important questions MCQ to study IAF AFCAT Mock Test Series and EKT for (ME, EEE, CSE) 2025 for AFCAT Exam | Download free PDF with solutions
Full Test 5 - EKT EEE - Question 1

A bag contains 4 white, 5 red and 6 blue balls. Three balls are drawn at random form the bag. The probability that all of them are red is:

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

Total balls in the bag are = 15
Sample space =15c3
The probability that all the balls are red is = 

Full Test 5 - EKT EEE - Question 2

If the area of the square is increased by 69% the side of the square increase by

Detailed Solution for Full Test 5 - EKT EEE - Question 2

Let the side of the square is x

Now, the area of square is, A = x2

Given that area of the square increased by 69%, the area will be 1.69 A

Let the increased side of the square is nx 

⇒ (nx)= 1.69x2
⇒ n = 1.3
Hence the side of the square increases by 30%

1 Crore+ students have signed up on EduRev. Have you? Download the App
Full Test 5 - EKT EEE - Question 3

If x = a(cos t + t sin t), y = a(sin t – t cos t). The value of dy/dx is

Detailed Solution for Full Test 5 - EKT EEE - Question 3

Full Test 5 - EKT EEE - Question 4

The inverse Laplace transform of  is

Detailed Solution for Full Test 5 - EKT EEE - Question 4

The inverse Laplace transform, 

Full Test 5 - EKT EEE - Question 5

If  then A12 is

Detailed Solution for Full Test 5 - EKT EEE - Question 5

Multiplying matrix A by itself, we get A2

Similarly, 

Full Test 5 - EKT EEE - Question 6

In an AC circuit, resonance occurs when

Detailed Solution for Full Test 5 - EKT EEE - Question 6

In an AC circuit, resonance occurs when capacitive reactance equals reactive reactance.

At resonance, XL=X, Z=R

Full Test 5 - EKT EEE - Question 7

One Coulomb passing a point in one second is one ______.

Detailed Solution for Full Test 5 - EKT EEE - Question 7

The current at a point in the circuit is the amount of charge that passes that point in one second. A current of 1 A is flowing in a circuit if a charge of 1 coulomb passes any point in the circuit every second.

Full Test 5 - EKT EEE - Question 8

Tesla is a measure of

Detailed Solution for Full Test 5 - EKT EEE - Question 8

Tesla (T) is a measure of magnetic flux density.

Electrical potential measures in volts (V).

Magnetic potential measures in Amperes (A).

Full Test 5 - EKT EEE - Question 9

A thin circular ring of mass M and radius r is rotating about its axis with a constant angular velocity ω. Two objects each of mass ‘m’ are attached gently to the opposite ends of a diameter of the ring. The ring will now rotate with an angular velocity –

Detailed Solution for Full Test 5 - EKT EEE - Question 9

By conservation of angular momentum

I’ω’ = Iω

(M + 2m) r2ω’ = Mr2ω

Full Test 5 - EKT EEE - Question 10

What happens if the reverberation time is too large?

Detailed Solution for Full Test 5 - EKT EEE - Question 10

Reverberation is the time taken for the sound to fall below the minimum audibility measured from the instant when the source stopped sounding. Therefore if the reverberation time becomes too large it produces an echo.

Full Test 5 - EKT EEE - Question 11

Optical fiber works on the principle of –

Detailed Solution for Full Test 5 - EKT EEE - Question 11

Optical fibre works on the principle of total internal reflection. Here light is used to transmit information from one place to another.

Full Test 5 - EKT EEE - Question 12

In which type of the switchings given below, entire capacity of a dedicated link is used?

Detailed Solution for Full Test 5 - EKT EEE - Question 12

In circuit switching entire channel capacity is used for single communication.
Two types of switched services are used in Wide Area Network:
(1). Circuit Switching: In circuit switching, a dedicated channel, called circuit, is set up between two stations for the duration of the transmission.The circuit must be established to make a complete end-to-end  connection before any data transmission.The whole bandwidth is dedicated to to this connection even if no data is being transferred.
(2). Packet Switching: In packet switching there is no dedicated circuit between any two end nodes, links can be dynamically shared among several communications over time. To provide the link sharing, user data must be segmented into discrete, variable length units called packets.

Full Test 5 - EKT EEE - Question 13

The diode with least reverse recovery time is

Detailed Solution for Full Test 5 - EKT EEE - Question 13

The Schottky diode is a semiconductor formed by the junction of a semiconductor with metal. It has a low forward voltage drop and near zero reverse recovery time hence used in very fast switching circuits.

Full Test 5 - EKT EEE - Question 14

The charge carriers in an n-channel FET are

Detailed Solution for Full Test 5 - EKT EEE - Question 14

The charge carriers in an n-channel FET are electrons alone

Full Test 5 - EKT EEE - Question 15

Tunnel diode

Detailed Solution for Full Test 5 - EKT EEE - Question 15

Tunnel diode owing to heaving doping(typically 100 to several thousand times normal doping semiconductor diode) resulting in extremely small depletion layer width. Due to this extremely thin depletion region width the charge carriers can "tunnel" through depletion region

Full Test 5 - EKT EEE - Question 16

In an unbiased pn junction, zero current implies that 

Detailed Solution for Full Test 5 - EKT EEE - Question 16

In an unbiased pn junction when the Total current crossing the junction from p - side to n - side equals the total current crossing the junction from n - side to p - side then the current is said to be zero current. 

Full Test 5 - EKT EEE - Question 17

Which of the following statement is false?

Detailed Solution for Full Test 5 - EKT EEE - Question 17

The Zener Diode because of its stable output characteristic, is used as a constant voltage source, gate electrode on the FET corresponds to the base in bipolar transistor and the SCR is a silicon rectifier with a gate electrode to control when current flows from anode to cathode.

Full Test 5 - EKT EEE - Question 18

Frequency of oscillation of an RC phase shift oscillator using three identical RC sections is

Detailed Solution for Full Test 5 - EKT EEE - Question 18

The oscillation frequency of an RC phase shift oscillator using three identical RC sections is 

Full Test 5 - EKT EEE - Question 19

If both inputs of J-K Flip flop are same then it acts as

Detailed Solution for Full Test 5 - EKT EEE - Question 19

If both inputs of JK flip flop are same, then it acts as T flip flop. If both the inputs are low, then the output will be same as the previous output. If both the inputs are high, then the output will be complement of the previous output.

Full Test 5 - EKT EEE - Question 20

The Logic gate that will have HIGH or “1” at its output when one of its inputs is high is

Detailed Solution for Full Test 5 - EKT EEE - Question 20

The below truth table is for OR gate.

 

From the above truth table we can say that if one of the input is high, the output will be high.

The below truth table is for AND gate.

From the above truth table we can say that if one of the input is low, the output will be high.

Full Test 5 - EKT EEE - Question 21

Decimal 43 is Hexadecimal and BCD number system is respectively

Detailed Solution for Full Test 5 - EKT EEE - Question 21

Full Test 5 - EKT EEE - Question 22

What is addition of (-64)10 and (80)16?

Detailed Solution for Full Test 5 - EKT EEE - Question 22

Converting in decimal we have

Full Test 5 - EKT EEE - Question 23

A full adder can be realized by using – 

Detailed Solution for Full Test 5 - EKT EEE - Question 23

A full adder can be realized by using two half adders and one OR gate.

Full Test 5 - EKT EEE - Question 24

A multiplexer has____________

Detailed Solution for Full Test 5 - EKT EEE - Question 24

A multiplexer has many inputs and one output and selection lines.

Full Test 5 - EKT EEE - Question 25

For which operation of a DC motor is generally preferred over an AC motor

Detailed Solution for Full Test 5 - EKT EEE - Question 25
  • D.C. motor is preferred over an A.C. motor when we need variable speed operation because As in DC series motor field winding and armature winding is in series. So whenever armature current changes field current also changes as they are same due to which its torque changes rapidly and we get a variable speed.
  • While in AC motor for same purpose we require frequency change which requires another complex circuitry

Advantages of DC motors:

  • The attractive feature of the dc motor is that it offers the wide range of speed control both above and below the rated speeds.
  • Dc series motors are termed as best suited drives for electrical traction applications used for driving heavy loads in starting conditions.
  • Quick starting, stopping, reversing and acceleration

  • Free from harmonics, reactive power consumption and many factors which makes dc motors more advantageous compared to A.C induction motors.
Full Test 5 - EKT EEE - Question 26

The efficiency of the transformer is maximum at

Detailed Solution for Full Test 5 - EKT EEE - Question 26

The efficiency of the transformer is maximum at when copper loss is equal to iron loss.

Full Test 5 - EKT EEE - Question 27

A dc series motor

Detailed Solution for Full Test 5 - EKT EEE - Question 27

A dc series motor should always be started on load. When the motor is connected across the supply mains without load, it draws small current from the supply mains flowing through the series field and armature, the speed tends to increase so that back emf may approach the applied voltage in magnitude. The increase in back emf weakens the armature current and hence the field current. This cause again increase in speed so in back emf. Thus the field continues to weaken and speed continues increase until the armature produced such centrifugal force that it is coming out from its shaft and gets damaged.

Full Test 5 - EKT EEE - Question 28

The stator core of an induction motor is made of

Detailed Solution for Full Test 5 - EKT EEE - Question 28

The stator frame consists of laminations of silicon steel, usually with a thickness of about 0.5 millimetre. Lamination is necessary since a voltage is induced along the axial length of the steel as well as in the stator conductors. The laminations are insulated from each other usually by a varnish layer. This breaks up the conducting path in the steel and limits the losses (known as eddy current losses) in the steel.

Full Test 5 - EKT EEE - Question 29

The reactance of a transformer is determined by its

Detailed Solution for Full Test 5 - EKT EEE - Question 29

All the flux in the transformer will not be able to link with both the primary and secondary windings. A small portion of flux will link either winding but not both. This portion of flux is called leakage flux. Due to this leakage flux in transformer, there will be a self-reactance in the transformer windings.

Full Test 5 - EKT EEE - Question 30

Movement of charge carriers from an area of high carrier concentration to an area of low carrier concentration is called

Detailed Solution for Full Test 5 - EKT EEE - Question 30

Diffusion is a physical process where molecules of a material move from an area of high concentration (where there are many molecules) to an area of low concentration (where there are fewer molecules).

View more questions
69 tests
Information about Full Test 5 - EKT EEE Page
In this test you can find the Exam questions for Full Test 5 - EKT EEE solved & explained in the simplest way possible. Besides giving Questions and answers for Full Test 5 - EKT EEE, EduRev gives you an ample number of Online tests for practice

Top Courses for AFCAT

Download as PDF

Top Courses for AFCAT