The rate of growth of bacteria is proportional to number present. If initially there were 1000 bacteria and…
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 $\sqrt{2}=1.414$ )
4646 approximately
5056 approximately
5656 approximately
$400 \sqrt{2}$ approximately
Solution
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
$\mathrm{N}=1000 \times 2^{\frac{5}{2}}=1000 \times 4 \sqrt{2}$
$=1000 \times 4 \times 1.414=5656$