JEE Main · 2022 · 24 Jun 2022 · Shift 2 · Modern Physics

Both substances start with the same mass, $10^{-2}\mathrm{kg}$. Reading the 'amount' as the mass remaining, the atomic weights do not enter: in $16\mathrm{s}$ substance A (half-life $4\mathrm{s}$) passes through four half-lives and is left at $\frac{1}{16}$ of its mass, while B (half-life $8\mathrm{s}$) passes through two half-lives and is left at $\frac{1}{4}$. The ratio of the surviving masses is then $\frac{1/16}{1/4}=\frac{1}{4}=\frac{25}{100}$, which fixes $x=25$. Had the comparison instead been of the number of atoms, the stated atomic-weight ratio $M_A:M_B=1:2$ would make the initial atom ratio $2:1$ and the same decay would give $\frac{2\times 1/16}{1\times 1/4}=\frac{1}{2}=\frac{50}{100}$; the official key corresponds to the mass reading, $x=25$.