Given $: \mathrm{N}_{1}=2 \mathrm{N}_{2}$

Activity of radioactive substance $=\lambda \mathrm{N}$

Half life period $\mathrm{t}=\frac{\ln 2}{\lambda}$ or, $\lambda=\frac{\ln 2}{\mathrm{t}}$

$\lambda_{1} \mathrm{N}_{1}=\frac{\ln 2}{\mathrm{t}_{1}} \times \mathrm{N}_{1}=5 \mu \mathrm{C}_{\mathrm{i}}$       …… $(i)$

$\lambda_{2} \mathrm{N}_{2}=\frac{\ln 2}{\mathrm{t}_{2}} \times \mathrm{N}_{2}=10 \mu \mathrm{C}_{\mathrm{i}}$       …… $(ii)$

Dividing equation $(ii)$ by $(i)$

$\frac{t_{2}}{t_{1}} \times \frac{N_{1}}{N_{2}}=\frac{1}{2}$

$\frac{t_{2}}{t_{1}}=\frac{1}{4} \Rightarrow t_{1}=4 t_{2}$

i.e., Half life of $S_{1}$ is four times of sample $S_{2}$ Hence $5$ years and $20$ years.