Let y denote the number of bacteria at any instant $\mathrm{t} \cdot$ then according to the question
$$
\frac{\mathrm{dy}}{\mathrm{dt}} \alpha \mathrm{y} \Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}=\mathrm{k} \mathrm{dt}......(i)
$$
$\mathrm{k}$ is the constant of proportionality, taken to be + ve on integrating (i),
we get $\log y=k t+c......(ii)$
$\mathrm{c}$ is a parameter. let $\mathrm{y}_{0}$ be the initial number of bacteria
i.e., at $t=0$ using this in (ii), $c=\log y_{0}$
$$
\begin{array}{l}
\Rightarrow \log y=k t+\log y_{0} \\
\Rightarrow \log \frac{y}{y_{0}}=k t......(iii)
\end{array}
$$
$\mathrm{y}=\left(\mathrm{y}_{0}+\frac{10}{100} \mathrm{y}_{0}\right)=\frac{11 \mathrm{y}_{0}}{10}$, when $\mathrm{t}=2$
So, from (iii), we get $\log \frac{\frac{11 \mathrm{y}_{0}}{10}}{\mathrm{y}_{0}}=\mathrm{k}$ (2)
$\Rightarrow \mathrm{k}=\frac{1}{2} \log \frac{11}{10}......(iv)$
Using (iv) in (iii) $\log \frac{\mathrm{y}}{\mathrm{y}_{0}}=\frac{1}{2}\left(\log \frac{11}{10}\right) \mathrm{t}......(v)$
let the number of bacteria become 1,00 , 000 to $2,00,000$ in $\mathrm{t}_{1}$ hours. i.e., $\mathrm{y}=2 \mathrm{y}_{0}$ when $\mathrm{t}=\mathrm{t}_{1}$ hours. from (v)
$$
\log \frac{2 \mathrm{y}_{0}}{\mathrm{y}_{0}}=\frac{1}{2}\left(\log \frac{11}{10}\right) \mathrm{t}_{1} \Rightarrow \mathrm{t}_{1}=\frac{2 \log 2}{\log \frac{11}{10}}
$$
Hence, the reqd. no. of hours $=\frac{2 \log 2}{\log \frac{11}{10}}$