An alternating voltage is represented by $V = 120\sin(100\pi t)$ volts. What is the root mean square (r.m.s.) value of this voltage, correct to one decimal place?

From the expression $V = V_0 \sin(\omega t)$, the peak voltage $V_0 = 120\,\text{V}$. The r.m.s. voltage is $V_{rms} = \frac{V_0}{\sqrt{2}} = \frac{120}{\sqrt{2}} = \frac{120}{1.4142} \approx 84.9\,\text{V}$. Option A (60 V) results from dividing by 2 instead of $\sqrt{2}$, and option D (169.7 V) results from multiplying the peak value by $\sqrt{2}$ instead of dividing.