If the population grows at the rate of $8 \%$ per year, then the time taken for the population to be doubled…
If the population grows at the rate of $8 \%$ per year, then the time taken for the
population to be doubled, is (Given $\log 2=0 \cdot 6912$ )
- $6 \cdot 8$ years
- $10 \cdot 27$ years
- $8 \cdot 64$ years
- $4 \cdot 3$ years
Solution
(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 $P=2 P_{0}$
$t=\frac{25}{2} \cdot \log \left(\frac{2 P_{0}}{P_{0}}\right)=\frac{25}{2} \log 2=8.64$
Asked in: MHT CET 2020 (19 Oct Shift 2)
Practice more Differential Equations questions on Aicharya