Question

The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is $1000$ at initial time $t=0.$ The number of bacteria is increased by $20$% in $2$ hours. If the population of bacteria is $2000$ after $\frac{k}{{\log }_e\left(\frac{6}{5}\right)}$ hours, then ${\left(\frac{k}{{\log }_{e}2}\right)}^2$ is equal to :

• A. $4$
• B. $2$
• C. $16$
• D. $8$

Answer 1

The correct option is A $4$

Let $x$ be the number of bacteria at time $t$.
$\Rightarrow \frac{dx}{dt}\propto x$
$\Rightarrow \frac{dx}{dt}=\lambda x$
$\Rightarrow \underset{1000}{\overset{x}{\int }}\frac{dx}{x}={\int }_0^t\lambda dt$
$\Rightarrow \ln x-\ln 1000=\lambda t$
$\Rightarrow \ln \left(\frac{x}{1000}\right)=\lambda t$
Put $t=2,x=1200$
$\Rightarrow \ln \left(\frac{12}{10}\right)=2\lambda \Rightarrow \lambda =\frac{1}{2}\ln \left(\frac{6}{5}\right)$

Now, $\ln \left(\frac{x}{1000}\right)=\frac{t}{2}\ln \left(\frac{6}{5}\right)$
$\Rightarrow x=1000e^{\frac{t}{2}\ln \left(\frac{6}{5}\right)}$
Given, $x=2000$ at $t=\frac{k}{{\log }_e\left(\frac{6}{5}\right)}$
$\Rightarrow 2000=1000e^{\frac{k}{2\ln \left(6/5\right)}\times \ln \left(6/5\right)}$
$\Rightarrow 2=e^{k/2}$
$\Rightarrow \ln 2=\frac{k}{2}$
$\Rightarrow \frac{k}{\ln 2}=2$
$\Rightarrow {\left(\frac{k}{\ln 2}\right)}^2=4$