JEE Main 2021 (Online) 26th February Morning ShiftQuestion 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 $\left(\frac{k}{\log_e2}\right)^{2}$ is equal to :
Differential Equation
Solution
Concept. Growth proportional to $N$ gives $N=N_0e^{\lambda t}$. Formulas used. (1) $\dfrac{dN}{dt}=\lambda N\Rightarrow N=N_0e^{\lambda t}$Step 1 - Find $\lambda$. By (1), $N=1000e^{\lambda t}$. After $2$ hours $N=1200$, so $e^{2\lambda}=\frac65$ and $\lambda=\frac12\ln\frac65$. Step 2 - Time to reach $2000$. $e^{\lambda t}=2$, so $$t=\frac{\ln2}{\lambda}=\frac{2\ln2}{\ln\frac65}.$$ So $k=2\ln2$. Step 3 - Value. $\left(\dfrac{k}{\ln2}\right)^2=4$. Answer: (B)