The correct option is (C) \(12 \mathrm{~h}\).
Explanation:
$$
\mathrm{T}_{(1 / 2)}=2 \mathrm{hr} .
$$
The radioactive sample decreases to \(\left(1 / 2^{n}\right)\) in \(n\) half lives.
\(\because\) initial activity is 64 times permissible level, it must become \((1 / 64)^{\text {th }}\) of its initial value.
$$
\begin{aligned}
&\therefore(1 / 64)=\left(1 / 2^{n}\right) \\
&\therefore n=6
\end{aligned}
$$
\(\therefore\) In 6 half lives activity will reduced itself to permissible level.
Given : half life \(=2 \mathrm{hrs}\) hence total time taken is \(6 \times 2=12 \mathrm{hrs}\).
