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The growth of population is proportional to the number present. If the population of a colony doubles is 50…
The growth of population is proportional to the number present. If the population of a colony doubles is 50 years, then the population will become triple in _____ years
$5\left(\frac{\ell \circ g 2}{\ell \operatorname{og} 3}\right) \mathrm{yrs}$ $50\left(\frac{\log 3}{\ell \mathrm{og} 2}\right) \mathrm{yrs}$ $5\left(\frac{\log 3}{\log 2}\right) \mathrm{yrs}$ $50\left(\frac{\log 2}{\ell \mathrm{og} 3}\right) \mathrm{yrs}$ Solution
Let $\mathrm{P}_{0}=$ Initial population
Given $\frac{\mathrm{dP}}{\mathrm{dt}} \propto \mathrm{P} \Rightarrow \frac{\mathrm{dP}}{\mathrm{dt}}=\lambda \mathrm{P} \Rightarrow \int \frac{\mathrm{dP}}{\mathrm{P}}=\int \lambda \mathrm{dt}$
$\log |\mathrm{P}|=\lambda \mathrm{t}+\mathrm{C}\ldots(1)$
At $t=0, \quad P=P_{0}$ we get $\log P_{0}=0+C \Rightarrow C=\log P_{0}$
$\log P=\lambda t+\log P_{0} \Rightarrow \log \left(\frac{P}{P_{0}}\right)=\lambda t$
When $P=2 P_{0}, t=50 \Rightarrow \log \left(\frac{2 P_{0}}{P_{0}}\right)=50 \lambda$
$\therefore \log 2=50 \lambda \Rightarrow \lambda=\frac{1}{50} \log 2$
$\therefore \log \left(\frac{\mathrm{P}}{\mathrm{P}_{0}}\right)=\frac{\mathrm{t}}{50}(\log 2)$
When $\mathrm{P}=3 \mathrm{P}_{0}$ we get
$\log 3=\frac{t}{50}(\log 2) \Rightarrow t=50\left(\frac{\log 3}{\log 2}\right)$ yrs.
Asked in: MHT CET 2020 (15 Oct Shift 2)
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