Bacteria increases at the rate proportional to the number of bacteria present. If the original number N…
Bacteria increases at the rate proportional to the number of bacteria present. If the
original number N doubles in 4 hours, then the number of bacteria will be $4 \mathrm{~N}$ in
2 hours
4 hours
6 hours
8 hours
Solution
Let $\mathrm{N}$ be the number of bacteria present at time $t_{0}$. Let $\mathrm{N}_{0}$ be the initial number of bacteria. Here $\frac{\mathrm{dN}}{\mathrm{dt}} \alpha \mathrm{N} \Rightarrow \frac{\mathrm{dN}}{\mathrm{dt}}=\mathrm{KN} \Rightarrow \frac{\mathrm{dN}}{\mathrm{N}}=\mathrm{K} \mathrm{dt}$
$\therefore \int \frac{\mathrm{dN}}{\mathrm{N}}=\mathrm{K} \int \mathrm{dt} \Rightarrow \log \mathrm{N}=\mathrm{Kt}+\mathrm{C}$
When $\mathrm{t}=0, \mathrm{~N}=\mathrm{N}_{0}$
$\therefore \log N_{0}=C \Rightarrow \log \left(\frac{N}{N_{0}}\right)=K t$
When $t=4, N=2 N_{0} \Rightarrow 4 K=\log 2$
$\begin{array}{l}
\mathrm{K}=\frac{1}{4} \log 2 \\
\therefore \log \left(\frac{\mathrm{N}}{\mathrm{N}_{0}}\right)=\frac{\mathrm{t}}{4} \log 2
\end{array}$
When $\mathrm{N}=4 \mathrm{~N}_{0}$, we get
$\log 4=\frac{t}{4} \log 2 \Rightarrow 2(\log 2)=\frac{t}{4}(\log 2) \Rightarrow t=8 \text { hours }$
This problem can also be saved as follows :
Number of bacteria doubles in 4 hrs.
$\therefore$ If initial number of bacteria are $N$, then
After 4 hours number of bacteria become $2 \mathrm{~N}$.
After 8 hours, number of bacteria becomes $4 \mathrm{~N}$