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

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 \(4 \mathrm{~N}\)