**Question:**

The rate of growth of bacteria in a culture is proportional to the number of bacteris present and the bacteria count is 1000 at initial time $\mathrm{t}=0$. The number of bacteria is increased by $20 \%$ in 2 hours. If the population of bacteria is 2000 after

$\frac{\mathrm{k}}{\log _{\mathrm{e}}\left(\frac{6}{5}\right)}$ hours, then $\left(\frac{\mathrm{k}}{\log _{\mathrm{e}} 2}\right)^{2}$ is equal to

Correct Option: 1

**Solution:**

$\frac{\mathrm{dB}}{\mathrm{dt}}=\lambda \mathrm{B} \Rightarrow \int_{1000}^{1200} \frac{\mathrm{dB}}{\mathrm{B}}=\lambda \int_{0}^{2} \mathrm{dt} \Rightarrow \lambda=\frac{1}{2} \ell \mathrm{n}\left(\frac{6}{5}\right)$

$\int_{1000}^{2000} \frac{\mathrm{dB}}{\mathrm{B}}=\frac{1}{2} \ell \mathrm{n}\left(\frac{6}{5}\right)_{0}^{\mathrm{T}} \mathrm{dt} \Rightarrow \mathrm{T}=\frac{2 \ell \mathrm{n} 2}{\ell \mathrm{n}\left(\frac{6}{5}\right)}$

$\Rightarrow \mathrm{k}=2 \ell \mathrm{n} 2$