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JEE • PHYSICS • COMMUNICATION SYSTEMS

Communication systems practice · Hard

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Practice format and chapter quiz

These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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1 Hard
1 of 7
Understanding +4 | -1
Given, below are two statements
Statement I A speech signal of 2kHz2\mathrm{ }kHz is used to modulate a carrier signal of 1MHz.1MHz. The bandwidth requirement for the signal is 4kHz4\mathrm{ }kHz.
Statement II The side band frequencies are 1002kHz1002\mathrm{ }kHz and 998kHz998\mathrm{ k}Hz. In the light of the above statements, choose the correct answer from the options given below
View answer and explanation
Correct option: C
Both Statement I and Statement II are true.
Option C is correct because the bandwidth requirement for AM is twice the modulating frequency, and the sideband frequencies are the carrier frequency plus or minus the modulating frequency.
2 Hard
2 of 7
Understanding +4 | -1
Given, below are two statements
Statement I A speech signal of 2kHz2\mathrm{ }kHz is used to modulate a carrier signal of 1MHz.1MHz. The bandwidth requirement for the signal is 4kHz4\mathrm{ }kHz.
Statement II The side band frequencies are 1002kHz1002\mathrm{ }kHz and 998kHz998\mathrm{ k}Hz. In the light of the above statements, choose the correct answer from the options given below
View answer and explanation
Correct option: C
Both Statement I and Statement II are true.
Option C (Both Statement I and Statement II are true) is correct because the bandwidth requirement for AM is twice the modulating signal frequency, and the sideband frequencies are the sum and difference of carrier and modulating frequencies.
3 Hard
3 of 7
Application +4 | -1
Calculate the phase velocity of electromagnetic wave having electron density and frequency of D-layer, N=400 electron cc−1,v=200 kHzc{c^{ - 1}},v = 200\,kHz
View answer and explanation
Correct option: C
6.9×108ms−16.9 \times {10^8}\mathrm{m}{\mathrm{s}^{ - 1}}
The correct answer is C (6.9×108ms−16.9 \times {10^8}\mathrm{m}{\mathrm{s}^{ - 1}}). The phase velocity of an electromagnetic wave in a medium is calculated using the refractive index, which depends on the electron density and frequency.
4 Hard
4 of 7
Understanding +4 | -1
In a radio receiver, the short wave and medium wave stations are tuned by using the same capacitor but coils of different inductance Ls{L_s} and Lm{L_m} respectively then
View answer and explanation
Correct option: B
Ls<Lm{L_s} < {L_m}
Option B (Ls<Lm{L_s} < {L_m}) is correct because short wave stations have higher frequencies than medium wave stations, and since frequency is inversely proportional to the square root of inductance, the inductance for short waves must be smaller.
5 Hard
5 of 7
Application +4 | -1
Two electrons are moving in opposite direction with speeds 0.8 c and 0.4 c, where c the speed of light in vacuum. Then the relative speed is about
View answer and explanation
Correct option: C
0.9 c
Option C (0.9 c) is correct because it is the relative speed calculated using the relativistic velocity addition formula.
6 Hard
6 of 7
Understanding +4 | -1
Given, below are two statements
Statement I A speech signal of 2kHz2\mathrm{ }kHz is used to modulate a carrier signal of 1MHz.1MHz. The bandwidth requirement for the signal is 4kHz4\mathrm{ }kHz.
Statement II The side band frequencies are 1002kHz1002\mathrm{ }kHz and 998kHz998\mathrm{ k}Hz. In the light of the above statements, choose the correct answer from the options given below
View answer and explanation
Correct option: C
Both Statement I and Statement II are true.
Option C is correct because the bandwidth requirement for AM is twice the modulating frequency, and the sideband frequencies are the carrier frequency plus or minus the modulating frequency.
7 Hard
7 of 7
Application +4 | -1
The glass of optical fiber has refractive index 1.55 and is clad with another glass of refractive index 1.51. When the surrounding is air, the numerical aperture will be
View answer and explanation
Correct option: B
0.35
Option B (0.35) is correct because it is the result of applying the numerical aperture formula with the given refractive indices.
Revision notes and key takeaways

Review COMMUNICATION SYSTEMS in your textbook or class notes, then use this set to check your understanding.

Key Takeaways

  • ✓ This set contains 7 single-correct-answer MCQs on COMMUNICATION SYSTEMS.
  • ✓ Try each question before opening its answer. Read the explanation where one is available.
  • ✓ Sign in to save submitted answers. Revisit incorrect answers alongside your textbook or class notes.
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