JEE Main 1-Sep-2021 Shift 2 Solved Paper

Given, the amplitude of the carrier wave, Ac = 250 V
The amplitude of the message wave, Am = 150 V
We know that,
The maximum amplitude, Amax = Ac+Am
Substituting the values in the above equation, we get
Amax = 250 + 150 = 400 V
We know that,
The minimum amplitude, Amin=Ac−Am
Substituting the values in the above equation, we get
Amin = 250 - 150 = 100 V
Thus, the ratio of the minimum amplitude to the maximumamplitude of the modulated wave is

Amin

Amax

=

100

400

=

1

4

Comparing with, 1: x
The value of the x = 4.