## Lakshya Education MCQs

Question: In frequency modulation
Options:
 A. The amplitude of modulated wave varies as frequency of carrier wave B. The frequency of modulated wave varies as amplitude of modulating wave C. The amplitude of modulated wave varies as amplitude of carrier wave D. The frequency of modulated wave varies as frequency of modulating wave E. The frequency of modulated wave varies as frequency of carrier wave
: B

The process of changing the frequency of a carrier wave (modulated wave) in accordance with the audio frequency signal (modulating wave) is known as frequency modulation (FM).

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## More Questions on This Topic :

Question 1. A 500 Hz modulating voltage fed into an FM generator produces a frequency deviation of 2.25 kHz. If amplitude of the voltage is kept constant but frequency is raised to 6 kHz then the new deviation will be
1.    4.5 kHz
2.    54 kHz
3.    27 kHz
4.    15 kHz
5.    The frequency of modulated wave varies as frequency of carrier wave
: B

mf = δfm=2250500 = 4.5
New deviation = 2(mffm) = 2 × 4.5 × 6 = 54 kHz.
Question 2. Broadcasting antennas are generally
1.    Omni directional type
2.    Vertical type
3.    Horizontal type
4.    None of these
: B

Broadcasting antennas are generally vertical type.
Question 3. In AM, the centpercent modulation is achieved when
1.    Carrier amplitude = signal amplitude
2.    Carrier amplitude−1 = signal amplitude
3.    Carrier frequency = signal frequency
4.    Carrier frequency−1 = signal frequency
5.    The frequency of modulated wave varies as frequency of carrier wave
: A

When signal amplitude is qual to the carrier amplitude, the amplitude of carrier ware varies between 2A and zero.

ma=AmplitudechargeofcarrierAmplitudeofnormalcarrier=2AAA×100 = 100%
Question 4. A photodetector is made from a semiconductor In 0.53Ga0.47As with Eg = 0.73 eV. What is the maximum wavelength, which it can detect
1.    1000 nm
2.    1703 nm
3.    500 nm
4.    173 nm
5.    The frequency of modulated wave varies as frequency of carrier wave
: B

Limiting value of hv is Eg, such that hv = hcλ = Eg
or λ=hcEg=6.63×1034Js×3×108ms10.73×1.6×1019J
= 1703 nm
Question 5. The attenuation in optical fiber is mainly due to
1.    Absorption
2.    Scattering
3.    Neither absorption nor scattering
4.    Both (a) and (b)
: D

A very small part of light energy is lost from an optical fiber due to absorption or due to light leaving the fiber as a result of scattering of light sideways by impurities in the glass fiber.
Question 6. Consider telecommunication through optical fibres. Which of the following statements is not true.
1.    Optical fibres may have homogeneous core with a suitable cladding.
2.    Optical fibres are subject to electromagnetic interference from outside.
3.    Optical fibres can be of graded refractive index
4.    Optical fibres have extremely low transmission loss