The rate of disintegration of a radio active element at time \(\mathrm{t}\) is proportional to its
mass at that time. Then the time during which the original mass of \(1 \cdot 5 \mathrm{gm}\). will
disintegrate into its mass of \(0.5 \mathrm{gm}\). is proportional to

(C)
Let \(m\) be the mass of the radioactive element at time \(t\).
Then the rate of disintegration is \(\frac{\mathrm{dm}}{\mathrm{dt}}\) which is proportional to \(\mathrm{m}\).
\(\therefore \frac{\mathrm{dm}}{\mathrm{dt}} \propto \mathrm{m} \Rightarrow \frac{\mathrm{dm}}{\mathrm{dt}}=-\mathrm{km}\), where \(\mathrm{k}>0\) \(\therefore \frac{\mathrm{dm}}{\mathrm{m}}=-\mathrm{kdt} \Rightarrow \int \frac{1}{\mathrm{~m}} \mathrm{dm}=-\mathrm{k} \int \mathrm{dt}+\mathrm{c}\)
\(\therefore \log \mathrm{m}=-\mathrm{kt}+\mathrm{c}\)
Initially, i.e. when \(\mathrm{t}=0, \mathrm{~m}=1.5\)
\(\therefore \log (1.5)=-\mathrm{k} \times 0+\mathrm{c} \Rightarrow \mathrm{c}=\log \left(\frac{3}{2}\right)\) \(\therefore \log \mathrm{m}=-\mathrm{kt}+\log \left(\frac{3}{2}\right) \Rightarrow \log \mathrm{m}-\log \frac{3}{2}=-\mathrm{kt}\)
\(\therefore \log \left(\frac{2 m}{3}\right)=-k t\)
When \(\mathrm{m}=0.5=\frac{1}{2}\), then
\(\begin{aligned}
& \log \left(\frac{2 \times \frac{1}{2}}{3}\right)=-\mathrm{kt} \Rightarrow \log \left(\frac{1}{3}\right)=-\mathrm{kt} \Rightarrow-\log 3=-\mathrm{kt} \\
\therefore & t=\frac{1}{k} \log 3
\end{aligned}\)
\(\therefore\) Thus required time is proportional to \(\log 3\).