If the population grows at the rate of \(8 \%\) per year, then the time taken for the population to be doubled, is (Given \(\log 2=0 \cdot 6912\) )

(D) \(\mathrm{P}_{0}\) be the initial population and let the population after \(\mathrm{t}\) years be \(2 \mathrm{P}_{0}\). then,
\(\begin{aligned}
\frac{\mathrm{dP}}{\mathrm{dt}} &=\frac{8 \mathrm{P}}{100} \Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}}=\frac{2 \mathrm{P}}{25} \\
\therefore \frac{\mathrm{dP}}{\mathrm{P}} &=\frac{2}{25} \mathrm{dt} \Rightarrow \int \frac{1}{\mathrm{P}} \mathrm{dP}=\frac{2}{25} \int \mathrm{dt}
\end{aligned}\)
\(\log P=\frac{2}{25} t+C\)...(1)
At, \(t=0, P=P_{0}\)
\(\log P_{0}=\frac{2 \times 0}{25}+C \Rightarrow C=\log P_{0}\)
\(\therefore \log P=\frac{2}{25} t+\log P_{0} \Rightarrow \log \frac{P}{P_{0}}=\frac{2}{25} t\)
\(\therefore t=\frac{25}{2} \cdot \log \left(\frac{P}{P_{0}}\right)\)
When \(P=2 P_{0}\)
\(t=\frac{25}{2} \cdot \log \left(\frac{2 P_{0}}{P_{0}}\right)=\frac{25}{2} \log 2=8.64\)