b
Since system dissipates its energy gradually, and hence amplitude will also decreases with time according to

\(a=a_{0} \mathrm{e}^{-\mathrm{bt/m}}\)              \(...(i)\)

Energy of vibration drop to half of its in itial value \(\left(\mathrm{E}_{0}\right),\) as \(\mathrm{E} \propto \mathrm{a}^{2} \Rightarrow \mathrm{a} \propto \sqrt{\mathrm{E}}\)

\(a=\frac{a_{0}}{\sqrt{2}} \Rightarrow \frac{b t}{m}=\frac{10^{-2} t}{0.1}=\frac{t}{10}\)

From \(eq^n(i),\)

\(\frac{a_{0}}{\sqrt{2}}=a_{0} e^{-t / 10}\)

\(\frac{1}{\sqrt{2}}=e^{-t / 10}\) or \(\sqrt{2}=e^{\frac{t}{10}}\)

\(\ln \sqrt{2}=\frac{t}{10} \quad \therefore t=3.5\) seconds