JEE Main 2019PhysicsCommunication SystemEasyMCQ

JEE Main 2019 — Communication System Question with Solution

JEE Main 2019 (10 Apr Shift 1)

Question

A message signal of frequency 100 MHz and peak voltage 100 V is used execute amplitude modulation on a carrier wave of frequency 300 GHz and peak voltage  400 V . The modulation index and difference between the two side band frequencies are:

Choose an option

Show full solutionCorrect option: B
Correct answer
B0.25;2×108 Hz

Step-by-step explanation

Modulating index is given by,
μ=AmAe
=100 V400 V
=0.25
The difference between the maximum frequency of modulated wave and the minimum frequency of modulated wave is given by
Δf=fmax-fmin
=fc+fm-fc-fm
=2fm
=2(100 MHz)
=2×108 Hz

Practice this on the real CBT interface

Solve this JEE Main question (and the rest of the Communication System chapter) on PrepSharp's TCS iON-style CBT player — with timer, bookmarks and session analytics.

Solve interactively →

About this question

This is a previous-year question from JEE Main 2019, covering the Communication System chapter of Physics. PrepSharp catalogues every PYQ from JEE Main with a verified answer key and step-by-step solution prepared by IIT alumni — so you can search by chapter, topic or year and revise efficiently.