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AFCAT EKT Electrical Mock Test - 2 - AFCAT MCQ


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30 Questions MCQ Test - AFCAT EKT Electrical Mock Test - 2

AFCAT EKT Electrical Mock Test - 2 for AFCAT 2025 is part of AFCAT preparation. The AFCAT EKT Electrical Mock Test - 2 questions and answers have been prepared according to the AFCAT exam syllabus.The AFCAT EKT Electrical Mock Test - 2 MCQs are made for AFCAT 2025 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for AFCAT EKT Electrical Mock Test - 2 below.
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AFCAT EKT Electrical Mock Test - 2 - Question 1

Which of the following is the most likely reason for large overshoot in a control system?

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 1

If gain is high, damping ratio is small hence overshoot is more.

AFCAT EKT Electrical Mock Test - 2 - Question 2

Conduction electrons have more mobility than holes because they

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 2

The mobility of electrons is more than that of holes because the probability of an electron having the energy required to move to an empty state in the conduction band is much greater than the probability of an electron having the energy required to move to the empty state in valance band. The mobility of hole is about half that of electrons.

AFCAT EKT Electrical Mock Test - 2 - Question 3

The principle of operation used in capacitive transducers to measure level of liquid is change of

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 3

Capacitive level transducer is an example of indirect measurement of level
The technique is frequently referred as RF as radio frequency signals applied to the capacitance circuit. The sensors can be designed to sense metrical with dielectric constants as low as 1.1 (coke and fly ash) and as high as 88 (water) or more.

AFCAT EKT Electrical Mock Test - 2 - Question 4

A pre - emphasis circuit in audio communications provides extra noise immunity by

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 4

Pre - emphasis is process of amplifying high frequency signals in spectrum before transmission of signal. It provides extra noise immunity because noise levels are higher at high frequencies there by pre-emphasis improves SNR at frequencies.
It is done before modulation of audio signal

AFCAT EKT Electrical Mock Test - 2 - Question 5

A cylindrical cavity operating in TE111 mode has a 3dB bandwidth of 2.4 MHz and its quality factor is 4000. Its resonant frequency would be

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 5

Resonant frequency f₀ = Quality factor × Bandwidth
f₀ = 4000 × 2.4 × 10⁶ = 9.6 GHz

AFCAT EKT Electrical Mock Test - 2 - Question 6

In experimentally obtaining the frequency response curve of an amplifier

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 6

The Frequency Response of an amplifier is presented in a form of a graph that shows output amplitude (or, more often, voltage gain) plotted versus frequency. Typical plot of the voltage gain of an amplifier versus frequency. The gain is null at zero frequency, then rises as frequency increases, level off for further increases in frequency, and then begins to drop again at high frequencies. The frequency response of an amplifier can be divided into three frequency regions.
The frequency response begins with the lower frequency region designated between 0 Hz and lower cutoff frequency. At lower cutoff frequency fL, the gain is equal to 0.707 Amid. Amid is a constant midband gain obtained from the midband frequency region. The third, the upper frequency region covers frequency between upper cutoff frequency and above. Similarly, at upper cutoff frequency, fH, the gain is equal to 0.707 Amid. After the upper cutoff frequency, the gain decreases with frequency increases and dies off eventually.

AFCAT EKT Electrical Mock Test - 2 - Question 7

Forbidden energy gap between valence band and conduction band is least in the case of _____

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 7

Forbidden energy gap between valence band and conduction band is least in the case of impure silicon.

AFCAT EKT Electrical Mock Test - 2 - Question 8

If f(x) = x + |x2 - 8| then the derivative of f(x) at x = 3 is

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 8

For x = 3, (x2 – 8) > 0 so |x2 - 8| = x2 – 8

f(x) = x + (x2 – 8)

f’(x) = 1 + 2x

f’(3) = 1 + 2(3) = 7

AFCAT EKT Electrical Mock Test - 2 - Question 9

For a dipole antenna

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 9

A dipole antenna is a linear antenna, usually fed in the center and producing maximum of radiation in the plane normal to the axis. It is said to be short dipole when length is less than λ / 4 and current distribution is sinusoidal. Radiation intensity is maximum along the normal to the dipole axis.

AFCAT EKT Electrical Mock Test - 2 - Question 10

BCD equivalent of Gray code number 1001 is

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 10

If G0G1G2G3 is a gray code number then the BCD equivalent number B0B1B2B3 can be estimated using:

B3 = G3

B2 = B3 ⊕ G2

B1 = B2 ⊕ G1

B0 = B1 ⊕ G0

So, using the above rules, the BCD equivalent of gray code number 1001 is 1110

AFCAT EKT Electrical Mock Test - 2 - Question 11

A working diode must have

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 11

A working diode means the diode is biased or connected with the DC source. When a diode is forward biased (+ve terminal to anode and -ve terminal to cathode), it provides a low resistance path because it helps the majority carrier to crossover the barrier potential and conduct the flow of current. When a diode is reversed biased (-ve terminal to anode and +ve terminal to cathode) it provide a high resistance path because the majority carriers are flows away from the junction and no current conduction takes place.

AFCAT EKT Electrical Mock Test - 2 - Question 12

Which of the following method of biasing provides the best operating point stability

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 12

Self-bias are emitter biased circuit. Generates it's own DC bias voltage .For the emitter which has a considerable effect, on the base current .Any circuit which does generate such bias voltage without actually directly depending on an explicit power supply is a self-bias.

AFCAT EKT Electrical Mock Test - 2 - Question 13

The waves which travel close to the earth is known as

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 13

The waves that travel close to the surface of the earth are known as surface waves.

AFCAT EKT Electrical Mock Test - 2 - Question 14

Transformer oil used in transformer provides

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 14

Transformer oil is a highly refined mineral oil that is stable at high temperatures and has excellent electrical insulating properties. It is used in oil-filled transformers. Its function is to provide insulation and cooling.

AFCAT EKT Electrical Mock Test - 2 - Question 15

The sum of kinetic and potential energy of a falling body _____.

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 15

Kinetic energy (KE = 1/2 mv²) represents energy the mass possesses by virtue of its motion. Likewise, potential energy (PE = mgh) represents energy the mass possesses by virtue of its position.
E = KE + PE
E is the total energy of the mass: i.e., the sum of its kinetic and potential energies. It is clear that E is a conserved quantity: i.e., although the kinetic and potential energies of the mass vary as it falls, its total energy remains the same and one form of energy is being converted into other form of energy.

AFCAT EKT Electrical Mock Test - 2 - Question 16

A communication receiver can be used for

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 16

Communication receiver is used for signal strength measurement as it is as the receiver end and due to which if the signal is not received properly from the sender we can easily determine the strength of the signal.
It is also used for detection and display of individual components of high-frequency wave as it enables us to display the components which comes at higher frequency and which comes at lower frequency

AFCAT EKT Electrical Mock Test - 2 - Question 17

A large series motor is never started without some load on it, otherwise

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 17

In case of DC series motor, field generated depends on armature current. At no load, current drawn is very less hence the field generated is less. Since the back emf remains constant hence speed is inversely proportional to the flux. Thus the initial speed will get very high and the motor may get damaged at no load.

AFCAT EKT Electrical Mock Test - 2 - Question 18

Which of the following statements is correct for varactor diode modulator?

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 18

Varactor diodes produces considerably less noise compared to other conventional diodes.

  • These diodes are available at low costs.
  • Varactor diodes are more reliable.
  • The varactor diodes are small in size and hence, they are very light weight.
  • The diode is operated in reverse bias.
  • Increase in reverse bias of varactor diode increases the capacitance.
  • Varactor diodes can be used as frequency modulators.
  • In microwave receiver LO, varactor diodes can be used as frequency multipliers.
  • Varactor diodes can be used as RF phase shifters.
AFCAT EKT Electrical Mock Test - 2 - Question 19

The Microwave semiconductor device which cause both Microwave oscillations and amplification action is

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 19

IMPATT diode does both Microwave oscillations and amplification action due to its operation is based on supplying RF signal superimposed on the bias so as to amplify it by avalanche breakdown

AFCAT EKT Electrical Mock Test - 2 - Question 20

A Farad is defined as

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 20

1 farad = 1 coulomb/volt
One farad is defined as the capacitance of a capacitor which requires a charge of one coulomb to establish a potential difference of one volt between its plates.

AFCAT EKT Electrical Mock Test - 2 - Question 21

Which of the following internet services is operate at network layer ?

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 21

Ethernet--→data link layer

Internet Protocol--→ network layer

Hyper text protocol--→application layer

AFCAT EKT Electrical Mock Test - 2 - Question 22

Lorentz force is given by

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 22

Lorentz force is the combination of electric force  and magnetic force . Thus the Lorentz force is given by .

AFCAT EKT Electrical Mock Test - 2 - Question 23

A glass optical fiber has refractive indices of core and cladding as 1.5 and 1 respectively. Assuming c = 3 × 108 m/s the multipath time dispersion per unit length will be 

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 23

In an optical fiber with core and cladding refractive indices n₁ and n₂, the time dispersion Δt is given by:

Δt = (L n₁) / (c n₂ (n₁ - n₂))

L = length of the optic fiber, hence multipath time dispersion per unit length of fiber is:

Δt / L = (1.5) / (1 × 3 × 10⁸ (1.5 - 1)) = 2.5 ns/m

AFCAT EKT Electrical Mock Test - 2 - Question 24

Equalizer rings can be used by

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 24

Equalizer rings are used in lap wounded armatures only to remove any circulating currents in parallel paths.

AFCAT EKT Electrical Mock Test - 2 - Question 25

Given P(A) = 1/4, P(B) = 1/3 and P(AUB) = 1/2. Value of P(A/B) is

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 25

Conditional probability P(A|B):

P(A|B) = P(A ∩ B) / P(B) = [P(A) + P(B) – P(A ∩ B)] / P(B) = [1/4 + 1/3 – 1/2] / (1/3) = 1/4

AFCAT EKT Electrical Mock Test - 2 - Question 26

A transmission line having 50 Ω impedance is terminated in a load of (40 + j30) Ω. The VSWR is

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 26

Given,
Z₀ = 50 Ω and ZL = 40 + j30

Reflection co-efficient:
P = (ZL - Z) / (ZL + Z)
= (40 + j30 - 50) / (40 + j30 + 50)
= (-10 + j30) / (90 + j30)

|P| = √[(10)² + (30)²] / √[(90)² + (30)²]
= √(10) / 90

V.W.W.R = (1 + 1/3) / (1 - 1/3)
= 2

AFCAT EKT Electrical Mock Test - 2 - Question 27

In the circuit below, Vo is (in V)

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 27

In positive feedback op-Amp acts in its saturation region

⇒ Vo = ±Vsat
Vo = 12 V

AFCAT EKT Electrical Mock Test - 2 - Question 28

The angle between two vectors a = i + 2j – k and b = 2i + j + k is

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 28

Angle between two vectors:

cos θ =  = 1/2

AFCAT EKT Electrical Mock Test - 2 - Question 29

The function of backward wave oscillator is similar to that of

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 29

It is somewhat similar to TWT and can deliver microwave power over a wide frequency band. It has an electron gun and a helix structure. However the interaction between electron beam and RF wave is different than in TWT. The growing RF wave travels in opposite direction to the electron beam.The frequency of wave can be changed by changing the voltage which controls the beam velocity. Moreover the amplitude of oscillations can be decreased continuously to zero by changing the beam current.

AFCAT EKT Electrical Mock Test - 2 - Question 30

The order and degree of the differential equation xy (d²y/dx²) + x (dy/dx)² - y (dy/dx) = 0 are _____.

Detailed Solution for AFCAT EKT Electrical Mock Test - 2 - Question 30

Order of a differential equation is defined as the order of the highest order derivative and the degree of a differential equation, when it is a polynomial equation in derivatives, is defined as the highest power (positive integral index) of the highest order derivative.
The highest order derivative present in the given differential equation is d²y/dx², so it’s order is two. It is a polynomial equation in d²y/dx² and dy/dx and the highest power raised to d²y/dx² is one, so its degree is one.

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