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The rate of growth of bacteria is proportional to number present. If initially there
were 1000 bacteria and the number doubles in 1 hour then the number of bacteria
after $2 \frac{1}{2}$ hours are (Given $\sqrt{2}=1.414$ )
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were 1000 bacteria and the number doubles in 1 hour then the number of bacteria
after $2 \frac{1}{2}$ hours are (Given $\sqrt{2}=1.414$ )
Solution:
2779 Upvotes
Verified Answer
The correct answer is:
5656 approximately
The rate of growth is proportional to the number present
$\therefore \frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}} \propto \mathrm{N} \Rightarrow \frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\mathrm{kN} \Rightarrow \frac{\mathrm{d} \mathrm{N}}{\mathrm{N}}=\mathrm{k} \mathrm{dt}$
$\therefore$ On integrating we get
$\int \frac{\mathrm{dN}}{\mathrm{N}}=\mathrm{k} \int \mathrm{dt}$
$\therefore \log \mathrm{N}=\mathrm{kt}+\mathrm{C}$
When $\mathrm{t}=0, \mathrm{~N}=1000 \Rightarrow \mathrm{C}=\log 1000$
$\therefore \log \mathrm{N}=\mathrm{kt}+\log 1000$
$\therefore \quad \log \left(\frac{\mathrm{N}}{1000}\right)=\mathrm{kt}$
$\mathrm{N}=1000 \mathrm{e}^{\mathrm{kt}}$ ...(1)
When $\mathrm{t}=1, \mathrm{~N}=2000$
$\therefore \mathrm{e}^{\mathrm{k}}=2 \quad \Rightarrow \mathrm{N}=1000 \times 2^{1} \quad \ldots[$ from (1) $]$
When $t=2 \frac{1}{2}$, we get
$\mathrm{N}=1000 \times 2^{\frac{5}{2}}=1000 \times 4 \sqrt{2}$
$=1000 \times 4 \times 1.414=5656$
$\therefore \frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}} \propto \mathrm{N} \Rightarrow \frac{\mathrm{d} \mathrm{N}}{\mathrm{dt}}=\mathrm{kN} \Rightarrow \frac{\mathrm{d} \mathrm{N}}{\mathrm{N}}=\mathrm{k} \mathrm{dt}$
$\therefore$ On integrating we get
$\int \frac{\mathrm{dN}}{\mathrm{N}}=\mathrm{k} \int \mathrm{dt}$
$\therefore \log \mathrm{N}=\mathrm{kt}+\mathrm{C}$
When $\mathrm{t}=0, \mathrm{~N}=1000 \Rightarrow \mathrm{C}=\log 1000$
$\therefore \log \mathrm{N}=\mathrm{kt}+\log 1000$
$\therefore \quad \log \left(\frac{\mathrm{N}}{1000}\right)=\mathrm{kt}$
$\mathrm{N}=1000 \mathrm{e}^{\mathrm{kt}}$ ...(1)
When $\mathrm{t}=1, \mathrm{~N}=2000$
$\therefore \mathrm{e}^{\mathrm{k}}=2 \quad \Rightarrow \mathrm{N}=1000 \times 2^{1} \quad \ldots[$ from (1) $]$
When $t=2 \frac{1}{2}$, we get
$\mathrm{N}=1000 \times 2^{\frac{5}{2}}=1000 \times 4 \sqrt{2}$
$=1000 \times 4 \times 1.414=5656$
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