b
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.