The 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 in 12 hours will be

108. (C)
Let \(x\) be the number of bacteria at time \(t\).
\(\frac{\mathrm{dx}}{\mathrm{dt}} \propto \mathrm{x} \Rightarrow \frac{\mathrm{dx}}{\mathrm{dt}}=\mathrm{kx} \Rightarrow \int \frac{\mathrm{dx}}{\mathrm{x}}=\int \mathrm{k} \cdot \mathrm{dt}\)
\(\log \mathrm{x}=\mathrm{kt}+\mathrm{c}\)
Initially, i.e. when \(t=0\), let \(x=x_{0}\)
\(\begin{array}{l}\log x_{0}=k \times 0+c \\ \log x=k t+\log x_{0}\end{array} \quad \Rightarrow c=\log x_{0}\)
\(\log x-\log x_{0}=k t \quad \Rightarrow \log \left(\frac{x}{x_{0}}\right)=k t\)...(1)
Since the number doubles in \(4 \mathrm{hrs}\), i.e.
\(\begin{aligned}
t &=4 \text { and } x=2 x_{0} \\
\therefore \log \left(\frac{2 x_{0}}{x_{0}}\right) &=4 k \quad \Rightarrow k \quad=\frac{1}{4} \log 2
\end{aligned}\)
\(\begin{array}{l}
\text { When } t=12, \log \left(\frac{x}{x_{0}}\right)=\frac{12}{4} \log 2=3 \log 2 \\
\therefore \log \left(\frac{x}{x_{0}}\right)=\log 8 \quad \Rightarrow x=8 x_{0}
\end{array}\)
This problem can be alternatively solved as follows :
Original number \(=\mathrm{N}\).
After four hours, number \(=2 \mathrm{~N}\)
After eight hours, number \(=4 \mathrm{~N}\)
After twelve hours, number \(=8 \mathrm{~N}\)