The maximum and minimum voltage of an amplitude modulated signal are $$60 \text{ V}$$ and $$20 \text{ V}$$ respectively. The percentage modulation index will be

We are given the maximum voltage $$V_{max} = 60 \text{ V}$$ and minimum voltage $$V_{min} = 20 \text{ V}$$ of an amplitude modulated signal.

The modulation index is defined as:

$$\mu = \frac{V_{max} - V_{min}}{V_{max} + V_{min}}$$

Substituting the given values:

$$\mu = \frac{60 - 20}{60 + 20} = \frac{40}{80} = 0.5$$

The percentage modulation index is:

$$\mu \times 100 = 0.5 \times 100 = 50\%$$

The correct answer is Option B.