Q 12-08-073JEE MainJEE Main 2023 (29 Jan, Shift 2)Easy
The modulation index for an A.M. wave having maximum and minimum peak to peak voltages of $14\ \text{mV}$ and $6\ \text{mV}$ respectively is
Answer: (B) 0.4
$$\mu=\frac{V_{max}-V_{min}}{V_{max}+V_{min}}=\frac{14-6}{14+6}=0.4$$
Solution by Sreeraj P, M.Sc Physics