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Q1. | The requirements of carrier frequency are |

A. | precise frequency value [Wrong Answer] |

B. | low frequency drift [Wrong Answer] |

C. | both (a) and (b) [Correct Answer] |

D. | neither (a) nor (b) [Wrong Answer] |

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Explanation:-
Answer : CDiscuss it below :!! OOPS Login [Click here] is required to post your answer/resultHelp other students, write article, leave your comments |

**Also Read Similar Questions Below :**

⇒ The beam of a radar should be

very narrow

very wide

neither too narrow nor very wide

either (

*b*) or (

*c*)

⇒ If ω is the frequency of modulating voltage in FM, the modulation index is proportional to

ω

1/ω

ω

1/ω

⇒ Square law modulators utilise

linear range of V-I characteristics of a diode

linear range of V-I characteristics of a triode

non-linear region of V-I dynamic characteristics of a diode

non-linear region of V-I dynamic characteristics of a triode

⇒ In a half wave rectifier, the load current flows

only for the positive half cycle of the input signal

only for the negative half cycle of the input signal

for full cycle

for less than fourth cycle

⇒ RMS value of

*i*(

*t*) = 10[1 + sin (-

*t*)] is

10

10/2

l50

102

⇒ Directivity of antenna is defined as

maximum directive gain

average directive gain

∞

0

⇒ The Nyquist plot of a stable, transfer function

*G*(

*s*) is shown in the figure. We are interested in the stability of the closed loop system in the feedback configuration sh

G(

*s*) is an all-pass filter

G(

*s*) has a zero in the right-half plane

G(

*s*) is the impedance of a passive network

G(

*s*) is marginally stable

⇒ MTI stands for

Moving Target Indicator

Microprocessor Tracking Index

Multiplexed Telemetry Index

Maximum Transmission Index

⇒ In 8085 eight address and data buses are multiplexed.

TRUE

FALSE

⇒ Which one of the following statements regarding DSI is wrong?

It is a digital form of TASI

Though it is more efficient than TASI, it is much slower

A speaker has to wait (it at all) for only a few milliseconds for reallocation of channel

It has increased the capacity of satellite channels by a factor of 2.2 or more with out degrading speech quality

⇒ A line is excited by a 100 V dc source. If reflection coefficients at both ends are 1 each then

there will be no oscillations on line

there will be only 1 or 2 oscillations on line

there will be a finite number of oscillations on line

the oscillations will continue indefinitely

⇒ A function will have only sine terms if

*f*(

*t*) = -

*f*(

*t*)

*f*(-

*t*) =

*f*(

*t*)

*f*(-

*t*) = -

*f*(

*t*)

none of the above

⇒ The real part of complex dielectric constant and tanδ for a dielectric are 2.1 and 5 x 10

^{-4}at 100 Hz respectively. The imaginary part of dielectric constant at 100 Hz is

1.05 x 10

^{-3}

2.1 x 10

^{-3}

5 x 10

^{-3}

1.05 x 10

^{-2}

⇒ Substitution theorem applies to

Linear network

Non linear network

Linear time invariant network

Any network

⇒ In a flip-flop with a NAND latch, a low

*R*and a low

*S*produces

active condition

inactive condition

race condition

dead condition

⇒ A single mode fibre not suffer from which type of dispersion?

Waveguide dispersion

Material dispersion

Intermodal dispersion

Polarization mode dispersion

⇒

**Assertion (A):** Potentiometers can not be used as error detectors in position control systems.

**Reason (R):** The resolution of a potentiometer places an upper limit on its accuracy

Both A and R are correct and R is correct explanation of A

Both A and R are correct but R is not correct explanation of A

A is correct but R is wrong

R is correct but A is wrong

⇒ The loss of charge method can be used to measure

resistance of shunt

insulation resistance of cable

resistance of shunt winding

both (b) and (c)

⇒ In a AM wave the carrier and one of the side bands is suppressed. If

*m*= 0.5, the percentage saving in power is

0.5

0.833

0.944

1

⇒ The gain of collinear arrays is maximum when the spacing between the elements is

0.1 λ to 0.3 λ

0.3 λ to 0.9 λ

0.5 λ to 0.7 λ

0.7 λ to 0.9 λ

⇒ The sensor generally used in incremental optical encoder is

LED

LCD

photodiode

either (a) or (b)

⇒ A sine wave is passed through an amplifier which severely limits it symmetrically. If then passes to a second amplifier which is narrow band and tuned to the frequency of the original sine wave. What will be the output wave from this second amplifier?

A square wave of the same frequency as that of the original sine wave

A square wave of double the frequency of that of original sine wave

A sine wave of constant amplitude at the frequency of the original sine wave

A sine wave of constant amplitude at double frequency of the original sine wave

⇒

Match the following:

List I | List II | ||
---|---|---|---|

A. | Ratio of maximum energy stored to energy dissipated per cycle | 1. | Propagation constant |

B. | TEM mode is loss less medium | 2. | Cutoff frequency is zero |

C. | Ratio of frequency in radian to phase velocity of EM wave | 3. | Quality factor of cavity |

D. | TEM is the mode of lowest cutoff frequency | 4. | Cylindrical waveguide |

A-3, B-2, C-4, D-1

A-2, B-3, C-1, D-4

A-3, B-2, C-1, D-4

A-2, B-3, C-4, D-1

⇒ In a 3 phase full converter feeding a highly inductive load, the average load current is 150 A. The peak current through thyristor is

150 A

75 A

50 A

300 A

⇒ A CMOS amplifier when compared to an N-channel. MOSFET, has the advantage of

higher cut off frequency

higher voltage gain

higher current gain

lower current drain from the power supply, there by less dissipation

⇒ In communication receivers the fidelity is provided by

audio stage

detector range

mixer stage

all of the above

⇒ For a photo conductor with equal electron and hole mobilities and perfect ohmic contacts at the ends, an increase in the intensities of optical illumination results in

change in open circuit voltage

change in short circuit current

a reduction resistance

an increase of resistance

⇒ A dipole antenna was radiating with some excitation in free space radiating a certain amount of the power. If this antenna is immersed in a lake where water is non-dissipative but has a dielectric constant of 81, the radiated power with the same excitation will

decrease to finite non-zero value

remain the same

increase

decrease to zero

⇒ One coulomb charge is equal to the charge on

6.24 x 10

^{18}electrons

6.24 x 10

^{24}electrons

6.24 x 10

^{18}atoms

none of the above

⇒ In the circuit of figure the current through 5 Ω resistance at

*t*= &infin

0

0.416666666666667

6.67 A

5.1 A