JEE PYQ: Differential Equation Question 13
Question 13 - 2021 (26 Feb Shift 1)
The rate of growth of bacteria in a culture is proportional to the number of bacteria present and the bacteria count is 1000 at initial time $t = 0$. The number of bacteria is increased by 20% in 2 hours. If the population of bacteria is 2000 after $\frac{k}{\log_e\left(\frac{6}{5}\right)}$ hours, then $\left(\frac{k}{\log_e 2}\right)^2$ is equal to:
(1) 4
(2) 2
(3) 16
(4) 8
Show Answer
Answer: (1)
Solution
$\frac{dN}{dt} = \lambda N$. $N = 1000e^{\lambda t}$. At $t = 2$: $1200 = 1000e^{2\lambda}$, so $\lambda = \frac{1}{2}\ln\frac{6}{5}$. For $N = 2000$: $2 = e^{\lambda t}$, $t = \frac{\ln 2}{\lambda} = \frac{2\ln 2}{\ln(6/5)} = \frac{k}{\ln(6/5)}$. So $k = 2\ln 2$, and $\left(\frac{k}{\ln 2}\right)^2 = 4$.
BETA
