Key Idea If a radioactive sample has more than one radioactive material then the effective half life.
$\frac{1}{T_{\text {eff T/2 }}}=\frac{1}{T_{1 \frac{1}{2}}}+\frac{1}{T_{2 \frac{1}{2}}}+\frac{1}{T_{3 \frac{1}{3}}}+\ldots$
Where, $T_{n \frac{1}{2}}=$ half life of $n^{\text {th }}$ material.
Here, effective half life of the net material,
$\begin{aligned}
& T_{\text {eff } \frac{1}{2}}=\frac{0.7 \times 0.3}{0.7+0.3}=0.21 \mathrm{hr} \\
&\left(\because T_{1 \frac{1}{2}}=0.7 \mathrm{hr}, \mathrm{T}_{2 \frac{1}{2}}=0.3 \mathrm{hr}\right)
\end{aligned}$
As we know, $T_{1 / 2}=0.693 \tau_{\text {aveg. mean life }}$
$\begin{aligned}
\Rightarrow \quad \tau_{\text {aveg. mean life }} & =\frac{T_{\text {eff } \frac{1}{2}}}{0.693} \\
& =\frac{0.21}{0.693}=0.3 \mathrm{hr} . \\
\Rightarrow \quad \tau_{\text {aveg. mean life }} & =0.3 \times 60=18 \mathrm{~min} .
\end{aligned}$
Hence, the correct option is (b).