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

Let ' \(x\) ' be the number of bacteria present at time ' \(t\) '.
\(\therefore \quad \frac{\mathrm{d} x}{\mathrm{dt}} \propto x\)
\(\therefore \quad \frac{\mathrm{d} x}{\mathrm{dt}}=\lambda x\), where \(\lambda\) is the constant of proportionality
Integrating on both sides, we get
\(\log x=\lambda t+c\)
When \(\mathrm{t}=0, x=1000\)
\(\begin{array}{ll}
\therefore \quad & \log 1000=0+c \\
& \Rightarrow \mathrm{c}=\log (1000) \\
\therefore \quad & \log x=\lambda \mathrm{t}+\log (1000)...(i) \\
& \text { When } \mathrm{t}=2 \\
& x=1000+(20 \% \text { of } 1000) \\
& =1000+200 \\
& =1200
\end{array}\)
\(\begin{array}{ll}
\therefore \quad & \log 1200=2 \lambda+\log 1000 \\
& \Rightarrow \lambda=\frac{1}{2} \log \left(\frac{1200}{1000}\right)=\frac{1}{2} \log \left(\frac{6}{5}\right) \\
\therefore \quad & \log x=\frac{\mathrm{t}}{2} \log \left(\frac{6}{5}\right)+\log (1000)
\end{array}\)
...[From (i)]
\(\begin{aligned} & \text { When } \mathrm{t}=\frac{\mathrm{k}}{\log \left(\frac{6}{5}\right)}, x=2000 \\ & \therefore \quad \log 2000=\frac{\mathrm{k}}{\log \left(\frac{6}{5}\right)} \times \frac{1}{2} \log \left(\frac{6}{5}\right)+\log (1000) \\ & \Rightarrow \log \left(\frac{2000}{1000}\right)=\frac{\mathrm{k}}{2} \Rightarrow \log 2=\frac{\mathrm{k}}{2} \\ & \Rightarrow \\ & \Rightarrow \frac{\mathrm{k}}{\log 2}=2 \\ & \Rightarrow\left(\frac{\mathrm{k}}{\log 2}\right)^2=2^2=4\end{aligned}\)