Law of Radioactive Decay question from JEE Main 2021 (27 Jul, Shift 1)

Answer: 150

1. Discharging capacitor: $q = q_0 e^{-t/RC}$. Activity: $A = A_0 e^{-t/\tau}$, where $\tau$ is the mean life.
2. $\dfrac{q}{A} = \dfrac{q_0}{A_0} e^{-t\left(\frac{1}{RC} - \frac{1}{\tau}\right)}$ is constant only if $RC = \tau$.
3. $R = \dfrac{\tau}{C} = \dfrac{30 \times 10^{-3}}{200 \times 10^{-6}} = 150\ \Omega$
4. So $R = 150\ \Omega$.