Law of Radioactive Decay question from JEE Main 2003 (AIEEE)

Answer: (a)

1. $5000 \to 1250$ is a fall to $1/4$, i.e. 2 half-lives in 5 min, so $T = 2.5$ min.
2. $\lambda = \frac{\ln 2}{T} = \frac{\ln 2}{2.5} = 0.4\ln 2$ per minute.
3. So the answer is (a) $0.4\ln 2$.