Question

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. $\frac{2}{\log \frac{11}{10}}$
• B. $\frac{2\log 2}{\log \left(\frac{11}{10}\right)}$
• C. $\frac{\log 2}{\log 11 }$
• D. $\frac{\log 2}{\log \left(\frac{11}{10}\right)}$

Answer 1

Answer: (B)

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

Answer 2

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), $c = \log \, y_0$ $\Rightarrow \log \, y = kt + \log \, y_{0}$ $\Rightarrow\log \frac{y}{y_{0}} = kt$ .....(iii) $y = \left(y_{0} + \frac{10}{100} y_{0}\right) = \frac{11y_{0}}{10}, t = 2$ So, from (iii) , we get $\log \frac{\frac{11y_{0}}{10}}{y_{0}} = k \left(2\right)$ $\Rightarrow k = \frac{1}{2} \log \frac{11}{10}$ .........(iv) Using (iv) in (iii) $\log \frac{y}{y_{0} } = \frac{1}{2} \left(\log \frac{11}{10}\right)t$ 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) $\log \frac{2y_{0}}{y_{0}} = \frac{1}{2} \left(\log \frac{11}{10}\right)t_{1} \Rightarrow t_{1} = \frac{2\log 2}{\log \frac{11}{10}}$ Hence, the reqd. no. of hours $= \frac{ 2\log 2}{\log \frac{11}{10}}$