A 100V carrier wave is made to vary between 160 V and 40 V by a modulating signal. What is the modulation index?

For an amplitude-modulated (A.M.) wave, the modulation index (also called the depth of modulation) is defined by the standard formula

$$m \;=\; \frac{V_{\text{max}} \;-\; V_{\text{min}}}{V_{\text{max}} \;+\; V_{\text{min}}},$$

where $$V_{\text{max}}$$ is the maximum (positive-peak) instantaneous value of the modulated carrier and $$V_{\text{min}}$$ is the minimum (negative-peak) instantaneous value.

We are told that the modulated carrier swings between $$160\text{ V}$$ and $$40\text{ V}$$. Hence

$$V_{\text{max}} \;=\; 160\text{ V}, \qquad V_{\text{min}} \;=\; 40\text{ V}.$$

Substituting these values into the formula gives

$$m \;=\; \frac{160 \;-\; 40}{160 \;+\; 40}.$$

Now we carry out the subtraction in the numerator:

$$160 \;-\; 40 \;=\; 120.$$

In the denominator we perform the addition:

$$160 \;+\; 40 \;=\; 200.$$

So the expression for $$m$$ becomes

$$m \;=\; \frac{120}{200}.$$

Dividing numerator and denominator by $$40$$ (or directly calculating with a calculator) we obtain

$$m \;=\; \frac{120 \div 40}{200 \div 40} \;=\; \frac{3}{5}.$$

Converting the fraction to decimal form,

$$\frac{3}{5} \;=\; 0.6.$$

Thus the modulation index is $$0.6$$, which is less than $$1$$ and therefore represents under-modulation, as expected.

Hence, the correct answer is Option B.