Question

Bacteria multiply at a rate proportional to the number present. If the original number $N$ double in $3$ hours, the number of the bacteria will be $4N$ is (in hours).

• A. $6$
• B. $4$
• C. $5$
• D. $8$

Answer 1

The correct option is A $6$

Let $N\left(t\right)$ be the number of bactiria at $t$ hours.

As $\frac{dN\left(t\right)}{dt}\propto N\left(t\right)$,

$\frac{dN\left(t\right)}{dt}=kN\left(t\right)\Rightarrow \frac{dN\left(t\right)}{N\left(t\right)}=kdt$

On integrating both sides,

$\ln N\left(t\right)=kt+C$ where C is the constant of Integration.

when $t=0hours\Rightarrow \ln N\left(0\right)=\ln N=C$

when $t=3hours\Rightarrow \ln N\left(3\right)=\ln 2N=3k+\ln N\Rightarrow k=\frac{\ln 2}{3}$

$\ln \frac{N\left(t\right)}{N}=\frac{\ln 2}{3}t\Rightarrow t=3\frac{\ln \frac{N\left(t\right)}{N}}{\ln 2}=3{\log }_2\frac{N\left(t\right)}{N}$

When $N\left(t\right)=4N$, $t=3{\log }_{2}4=6$

So, Option A is correct.