Q45 / 75 · VITEEE 2021 · Mathematics
In a culture the bacteria count is $1,00,000$. The number is increased by $10 \%$ in 2 hours. In how many hours will the count reach 2,00,000 if the rate of growth of bacteria is proportional to the number present.
- A$\frac{2}{\log \frac{11}{10}}$
- Correct answerB$\frac{2 \log 2}{\log \left(\frac{11}{10}\right)}$
- C$\frac{\log 2}{\log 11}$
- D$\frac{\log 2}{\log \left(\frac{11}{10}\right)}$
Solution
Let $y$ denote the number of bacteria at any instant $t \cdot$ then according to the question
$$
\frac{\mathrm{dy}}{\mathrm{dt}} \alpha \mathrm{y} \Rightarrow \frac{\mathrm{dy}}{\mathrm{y}}=\mathrm{kdt}
$$
$\mathrm{k}$ is the constant of proportionality, taken to be + ve on integrating (i), we get
$$
\log y=k t+c
$$
$\mathrm{c}$ is a parameter. let $\mathrm{y}_0$ be the initial number of bacteria
i.e., at $\mathrm{t}=0$ using this in (ii), $\mathrm{c}=\log \mathrm{y}_0$
$$
\begin{aligned}
& \Rightarrow \quad \log \mathrm{y}=\mathrm{kt}+\log \mathrm{y}_0 \Rightarrow \log \frac{\mathrm{y}}{\mathrm{y}_0}=\mathrm{kt} \\
& \mathrm{y}=\left(\mathrm{y}_0+\frac{10}{100} \mathrm{y}_0\right)=\frac{11 \mathrm{y}_0}{10} \text {, when } \mathrm{t}=2 \\
& \text { So, from (iii), we get } \log \frac{\frac{11 \mathrm{y}_0}{10}}{\mathrm{y}_0}=\mathrm{k} \text { (2) } \\
& \Rightarrow \mathrm{k}=\frac{1}{2} \log \frac{11}{10}
\end{aligned}
$$
Using (iv) in (iii) $\log \frac{\mathrm{y}}{\mathrm{y}_0}=\frac{1}{2}\left(\log \frac{11}{10}\right) \mathrm{t}$
let the number of bacteria become $1,00,000$ to $2,00,000$ in $t_1$ hours. i.e., $y=2 y_0$ when $\mathrm{t}=\mathrm{t}_1$ hours. from $(\mathrm{v})$
$$
\log \frac{2 \mathrm{y}_0}{\mathrm{y}_0}=\frac{1}{2}\left(\log \frac{11}{10}\right) \mathrm{t}_1 \Rightarrow \mathrm{t}_1=\frac{2 \log 2}{\log \frac{11}{10}}
$$
Hence, the reqd. no. of hours $=\frac{2 \log 2}{\log \frac{11}{10}}$
Hence, the correct answer is option: 'B'.