An alternating voltage \(\mathrm{V}(\mathrm{t})=220 \sin 100 \ \pi \mathrm{t}\) volt is applied to a purely resistive load of \(50\ \Omega\). The time taken for the current to rise from half of the peak value to the peak value is _______.

Given equation of alternating voltage:
\( V(t) = 220 \sin(100 \pi t) \)

Comparing this with the standard alternating voltage equation \( V(t) = V_0 \sin(\omega t) \), we get:
Peak voltage, \( V_0 = 220\text{ V} \)
Angular frequency, \( \omega = 100 \pi\text{ rad/s} \)

For a purely resistive load, the current \( I(t) \) is in phase with the voltage and is given by:
\( I(t) = I_0 \sin(\omega t) \) (where \( I_0 \) is the peak current value)

Step 1: Find the time \( t_1 \) when the current reaches half of its peak value \( \left(\frac{I_0}{2}\right) \):
\( I(t_1) = \frac{I_0}{2} \)
\( I_0 \sin(\omega t_1) = \frac{I_0}{2} \)
\( \sin(\omega t_1) = \frac{1}{2} \)
\( \omega t_1 = \frac{\pi}{6} \implies t_1 = \frac{\pi}{6\omega} \)

Step 2: Find the time \( t_2 \) when the current reaches its full peak value \( (I_0) \):
\( I(t_2) = I_0 \)
\( I_0 \sin(\omega t_2) = I_0 \)
\( \sin(\omega t_2) = 1 \)
\( \omega t_2 = \frac{\pi}{2} \implies t_2 = \frac{\pi}{2\omega} \)

Step 3: Calculate the time taken \( t \) to rise from half of the peak value to the peak value:
\( t = t_2 - t_1 = \frac{\pi}{2\omega} - \frac{\pi}{6\omega} \)
\( t = \frac{3\pi - \pi}{6\omega} = \frac{2\pi}{6\omega} = \frac{\pi}{3\omega} \)

Explanation of your doubt (\( \frac{\pi}{3\omega} = \frac{\pi}{300\pi} \)):
Since we found \( \omega = 100\pi \) from the given voltage equation, substituting this value into the denominator gives:
\( 3\omega = 3 \times 100\pi = 300\pi \)

Now, substitute \( \omega = 100\pi \) into the time equation:
\( t = \frac{\pi}{3 \times 100\pi} = \frac{\pi}{300\pi} \)
\( t = \frac{1}{300}\text{ seconds} \)

Converting seconds into milliseconds (\( 1\text{ s} = 1000\text{ ms} \)):
\( t = \frac{1}{300} \times 1000\text{ ms} = \frac{10}{3}\text{ ms} \approx 3.33\text{ ms} \)

Therefore, the correct option is B.