Q 12-13-117JEE MainJEE Main 2020 (3 Sep, Shift 1)Easy
In a radioactive material, a fraction of active material remaining after the time $t$ is $\dfrac{9}{16}$. The fraction that was remaining after the time $\dfrac t2$ is:
Answer: (C) $\dfrac34$
$\dfrac{N}{N_0} = e^{-\lambda t} = \dfrac{9}{16}$, so $e^{-\lambda t/2} = \sqrt{\dfrac{9}{16}} = \dfrac34$.
Solution by Sreeraj P, M.Sc Physics