Question

The rate of growth of bacteria in culture is proportional to the number of bacteria present and the bacteria count is $1000$ at the initial time $t=0$.

The number of bacteria has increased by $20\%$ in $2hours$. If the population of bacteria is $2000$ after $\frac{k}{{\log }_e\left({\displaystyle \frac{6}{5}}\right)}hours$, 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

$4$

Find the given value

Given, $x=2000$at $\frac{k}{{\log }_e\left({\displaystyle \frac{6}{5}}\right)}hours$

Let the number of bacteria at the time $t$ is $x$

$$\begin{gathered} \Rightarrow \frac{dx}{dt}\alpha x \\ \Rightarrow \frac{dx}{dt}=\lambda x \end{gathered}$$ [ By variable separable ]

Now, integrating both sides

$$\begin{gathered} \Rightarrow {\int }_{1000}^x\frac{1}{x}dx=\lambda \int dt \\ \Rightarrow \ln \left(x\right)-\ln \left(1000\right)=\lambda t \\ \Rightarrow \ln \left(\frac{x}{1000}\right)=\lambda t...\left(1\right) \end{gathered}$$

Put $t=2,x=1200$

$\lambda =\frac{1}{2}\ln \left(\frac{6}{5}\right)....\left(2\right)$

From $\left(1\right)$ and $\left(2\right)$

$x=1000e^{\frac{t}{2}\ln \left(\frac{6}{5}\right)}$

Put $x=2000$and $t=\frac{k}{{\log }_e\left({\displaystyle \frac{6}{5}}\right)}hours$

$\Rightarrow {\left(\frac{k}{\ln \left(2\right)}\right)}^2=4$

Hence, the correct option is $A$.