Q 12-13-035JEE MainJEE Main 2026 (28 Jan, Shift 1)Medium
An atom ${}^{8}_{3}X$ is bombarded by shower of fundamental particles and in 10 s this atom absorbed 10 electrons, 10 protons and 9 neutrons. The percentage growth in the surface area of the nucleons is recorded by :
Answer: (C) $125\%$
Only protons and neutrons change the mass number: $A' = 8 + 10 + 9 = 27$.
Nuclear radius $R \propto A^{1/3}$, so surface area $S \propto A^{2/3}$:
$$\frac{S'}{S} = \left(\frac{27}{8}\right)^{2/3} = \left(\frac32\right)^2 = \frac94$$
Percentage growth $= \left(\dfrac94 - 1\right)\times100 = 125\%$.
Solution by Sreeraj P, M.Sc Physics