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 $\log 2=0.6912$ )
- 6.8 year
- 4.3 years
- 10.27 years
- 8.64 years
Solution
Let the initial population be $\mathrm{P}$ and rate of increase is $8 \%$ per year.
$\begin{aligned}
& \therefore \frac{\mathrm{dP}}{\mathrm{dt}}=\frac{8}{100} \mathrm{P} \\
& \therefore \int \frac{\mathrm{dP}}{\mathrm{P}}=\int 0.08 \mathrm{t} \\
& \therefore \log |\mathrm{P}|=0.08 \mathrm{t}+\mathrm{c}
\end{aligned}$
When $\mathrm{t}=0$, we get $\mathrm{c}=\log \mathrm{P}$
$\therefore \log \mathrm{P}=0.08 \mathrm{t}+\log \mathrm{P}$
When ' $\mathrm{P}$ ' doubles, we write
$\begin{aligned}
& \log 2 p=0.08 \mathrm{t}+\log \mathrm{P} \\
& \therefore \log \left(\frac{2 \mathrm{P}}{\mathrm{P}}\right)=\log 2=0.6921=0.08 \mathrm{t} \\
& \therefore \mathrm{t}=\frac{0.6912}{0.08}=8.64 \text { years }
\end{aligned}$
Asked in: MHT CET 2021 (23 Sep Shift 1)
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