An alternating voltage $V(t)=220 \sin 100\, pt$ is applied to a purely resistive load of 50 W. The time taken for the current to rise from half of the peak value to the peak value is:

$3.3\, ms$

$ V = 220 sin 100\pi t$

$ I = \frac{V}{R} = 44 sin 100\pi t$

$ I = \frac{I_0}{2} $

$\Rightarrow 100\pi t = \frac{\pi}{6}$

$\Rightarrow t_1 = \frac{1}{600}$

$ I = I_0 \text{at } 100\pi t =\frac{\pi}{2}$

$\Rightarrow t_2= \frac{1}{200}$

$\Rightarrow \text{Time taken } = t_2 - t_1 = 3.3 ms$