In a culture, the bacteria count is 1,00,000.

Solution:

Let y be the number of bacteria at any instant t.

It is given that the rate of growth of the bacteria is proportional to the number present.

$\therefore \frac{d y}{d t} \propto y$

$\Rightarrow \frac{d y}{d t}=k y$ (where $k$ is a constant)

$\Rightarrow \frac{d y}{y}=k d t$

Integrating both sides, we get:

$\int \frac{d y}{y}=k \int d t$

$\Rightarrow \log y=k t+\mathrm{C}$              ....(1)

Let y0 be the number of bacteria at t = 0.

⇒ log y0 = C

Substituting the value of C in equation (1), we get:

$\log y=k t+\log y_{0}$

$\Rightarrow \log y-\log y_{0}=k t$

$\Rightarrow \log \left(\frac{y}{y_{0}}\right)=k t$

$\Rightarrow k t=\log \left(\frac{y}{y_{0}}\right)$                 ...(2)

Also, it is given that the number of bacteria increases by 10% in 2 hours.

$\Rightarrow y=\frac{110}{100} y_{0}$

$\Rightarrow \frac{y}{y_{0}}=\frac{11}{10}$     ...(3)

Substituting this value in equation (2), we get:

$k \cdot 2=\log \left(\frac{11}{10}\right)$

$\Rightarrow k=\frac{1}{2} \log \left(\frac{11}{10}\right)$

Therefore, equation (2) becomes:

$\frac{1}{2} \log \left(\frac{11}{10}\right) \cdot t=\log \left(\frac{y}{y_{0}}\right)$

$\Rightarrow t=\frac{2 \log \left(\frac{y}{y_{0}}\right)}{\log \left(\frac{11}{10}\right)}$              ...(4)

Now, let the time when the number of bacteria increases from 100000 to 200000 be t1.

$\Rightarrow y=2 y_{0}$ at $t=t_{1}$

From equation (4), we get:

$t_{1}=\frac{2 \log \left(\frac{y}{y_{0}}\right)}{\log \left(\frac{11}{10}\right)}=\frac{2 \log 2}{\log \left(\frac{11}{10}\right)}$

Hence, in $\frac{2 \log 2}{\log \left(\frac{11}{10}\right)}$ hours the number of bacteria increases from 100000 to 200000 .