VITEEE 2016 Paper

Let y denote the number of bacteria at any instant t • then according to the question $\frac{dy}{dx}$ αy ⇒ $\frac{dy}{y}$ = kdt ... (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 $y_0$ ⇒ log $\frac{y}{y_0}$ = kt ... (iii) y = $\left(y_0 + \frac{10}{100}y\_\right)$ = $\frac{11y_0}{10}$ , when t = 2 So, from (iii), we get log $\frac{\frac{11y_0}{10}}{y_0}$ = k (2) ⇒ k = $\frac{1}{2}$ log $\frac{11}{10}$ ... (iv) Using (iv) in (iii) log $\frac{y}{y_0}$ = $\frac{1}{2}$ $\left(\frac{\log 11}{10}\right)$ t ... (v) let the number of bacteria become 1, 00, 000 to 2,00,000 in $t_1$ hours, i.e., y = $2y_0$ when t = $t_1$ hours, from (v) log $\frac{2y_0}{y_0}$ = $\frac{1}{2}\left(\frac{\log 11}{10}\right)t_1$ ⇒ $t_1$ = $\frac{2\log 2}{\frac{\log 11}{10}}$ Hence, the reqd. no. of hours = $\frac{2\log 2}{\log\left(\frac{11}{10}\right)}$