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The rate at which the population of a city increases varies as the population present. Within the period of 30 years, the population grew from 20 lakhs to 40 lakhs. Then, the population after a further period of 15 years will be
$($ Take $\sqrt{2}=1.41)$
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$($ Take $\sqrt{2}=1.41)$
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Verified Answer
The correct answer is:
56.4 lakhs
$\begin{aligned} & \frac{d p}{d t} \propto p \Rightarrow \frac{d p}{d t}=k p \Rightarrow \int \frac{d p}{p}=\int k d t \\ & \Rightarrow \log _e P=k t+C\end{aligned}$
Putting $\mathrm{t}=0$ year and $\mathrm{p}=20$ lakhs
We get $\mathrm{c}=\log _{\mathrm{e}} 20$
$\Rightarrow \log _e P=K t+\log _e 20$
Again putting $\mathrm{t}=30$ years and $\mathrm{P}=40$ lakhs
$\begin{aligned} & \Rightarrow \log _e 40=K \times 30+\log _e 20 \\ & \Rightarrow K=\frac{\log _e 2}{30} \\ & \Rightarrow \log _e P=\frac{\log _e 2}{30} t \log _e 20\end{aligned}$
Putting $\mathrm{t}=45$ years
We get $\log _e p=\frac{\log _e 2}{30} \times 45+\log _e 20$
$\begin{aligned} & \Rightarrow \log _e P=\frac{3}{2} \log _e 2+\log _e 20 \\ & \Rightarrow \log _e P=\log _e 2 \frac{3}{2} \times 20 \\ & \Rightarrow P=2 \sqrt{2} \times 20 \text { lakhs }\end{aligned}$
$\Rightarrow \mathrm{P}=2 \times 1.41 \times 20$ lakhs
$\Rightarrow \mathrm{P}=56.4$ lakhs
Putting $\mathrm{t}=0$ year and $\mathrm{p}=20$ lakhs
We get $\mathrm{c}=\log _{\mathrm{e}} 20$
$\Rightarrow \log _e P=K t+\log _e 20$
Again putting $\mathrm{t}=30$ years and $\mathrm{P}=40$ lakhs
$\begin{aligned} & \Rightarrow \log _e 40=K \times 30+\log _e 20 \\ & \Rightarrow K=\frac{\log _e 2}{30} \\ & \Rightarrow \log _e P=\frac{\log _e 2}{30} t \log _e 20\end{aligned}$
Putting $\mathrm{t}=45$ years
We get $\log _e p=\frac{\log _e 2}{30} \times 45+\log _e 20$
$\begin{aligned} & \Rightarrow \log _e P=\frac{3}{2} \log _e 2+\log _e 20 \\ & \Rightarrow \log _e P=\log _e 2 \frac{3}{2} \times 20 \\ & \Rightarrow P=2 \sqrt{2} \times 20 \text { lakhs }\end{aligned}$
$\Rightarrow \mathrm{P}=2 \times 1.41 \times 20$ lakhs
$\Rightarrow \mathrm{P}=56.4$ lakhs
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