The bacteria increases at the rate proportional to the number of bacteria present. If the original number…
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
4N
3N
8N
6N
Solution
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 $=8 \mathrm{~N}$