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(C) The modulation index $M$ is given by the formula $M = \frac{V_m}{V_c}$, where $V_m$ is the peak voltage of the message signal and $V_c$ is the pea

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decreased by $20 \%$

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(C) The modulation index $M$ is given by the formula $M = \frac{V_m}{V_c}$, where $V_m$ is the peak voltage of the message signal and $V_c$ is the peak voltage of the carrier wave.Given $V_m = 12 V$ and initial modulation index $M_1 = 0.6$.For the first case: $0.6 = \frac{12}{V_c} \Rightarrow V_c = \frac{12}{0.6} = 20 V$.For the second case, we want the modulation index $M_2 = 0.75$ with the same message signal $V_m = 12 V$.$0.75 = \frac{12}{V_c'} \Rightarrow V_c' = \frac{12}{0.75} = 16 V$.The change in carrier peak voltage is $\Delta V = V_c - V_c' = 20 V - 16 V = 4 V$.The percentage change is $\frac{\Delta V}{V_c} 	imes 100 \% = \frac{4}{20} 	imes 100 \% = 20 \%$.Since the voltage decreased from $20 V$ to $16 V$, the carrier peak voltage should be decreased by $20 \%$.

List $I$	List $II$
$A$. Attenuation	$I$. Combination of a receiver and transmitter.
$B$. Transducer	$II$. Process of retrieval of information from the carrier wave at receiver.
$C$. Demodulation	$III$. Converts one form of energy into another.
$D$. Repeater	$IV$. Loss of strength of a signal while propagating through a medium.

If a message signal of frequency $10 kHz$ and peak voltage $12 V$ is used to modulate a carrier wave of frequency $1 MHz$, the modulation index is $0.6$. To make the modulation index $0.75$, the carrier peak voltage should be

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