VITEEE 2021 Paper

Let y denote the number of bacteria at any instant t . then according to the question $\frac{dy}{dt} \alpha y \Rightarrow \frac{dy}{y}=k dt$ ... (i) $k$ is the constant of proportionality, taken to be $+$ ve on integrating (i), we get $\log y = kt + c$ ...(ii) $c$ is a parameter. let $y_0$ be the initial number of bacteria i.e., at $t=0$ using this in (ii), $c = \log y_0$ ⇒ $\log y = kt + \log y_0$ $\Rightarrow \log \frac{y}{y_0} = kt$ ... (iii) $y = \left(y_0 + \frac{10}{100} y_0\right) = \frac{11 y_0}{10}$ , when $t=2$, So, from (iii), we get $\log \frac{\frac{11 y_0}{10}}{y_0} = k$ (2) ⇒ $k = \frac{1}{2} \log \frac{11}{10}$ ...(iv) Using (iv) in (iii) $\log \frac{y}{y_0} = \frac{1}{2}\left(\log \frac{11}{10}\right) t$ ... (v) let the number of bacteria become $1,00,000$ to $2,00,000$ in $t_1$ hours. i.e., $y = 2 y_0$ when $t = t_1$ hours, from (v) $\log \frac{2 y_0}{y_0} = \frac{1}{2}\left(\log \frac{11}{10}\right) t_1 \Rightarrow t_1 = \frac{2 \log 2}{\log \frac{11}{10}}$ Hence, the reqd. no. of hours $= \frac{2 \log 2}{\log \frac{11}{10}}$