The modulation index for an A.M. wave having maximum and minimum peak to peak voltages of $$14$$ mV and $$6$$ mV respectively is:

We need to find the modulation index for an AM wave.

The modulation index is given by $$ m = \frac{V_{max} - V_{min}}{V_{max} + V_{min}} $$. Here, $$V_{max}$$ and $$V_{min}$$ are the maximum and minimum peak-to-peak voltages respectively.

Given that $$V_{max} = 14$$ mV and $$V_{min} = 6$$ mV, substituting these values yields $$ m = \frac{14 - 6}{14 + 6} = \frac{8}{20} = 0.4 $$.

The correct answer is Option B: 0.4.

The recorded answer code is 7, which corresponds to Option B, matching our calculation.