Law of Radioactive Decay question from MHT CET 2026 (19 May, Shift 2)

Answer: (b)

1. $N_X = N_0 e^{-\lambda t}$, $N_Y = N_0(1 - e^{-\lambda t})$.
2. $\frac{N_Y}{N_X} = e^{\lambda t} - 1 = 0.4 \Rightarrow e^{\lambda t} = 1.4$.
3. $t = \frac{\ln 1.4}{\lambda} = \frac{T \ln 1.4}{\ln 2} = \frac{T\log(1.4)}{\log(2)}$.
4. So the answer is (b).