JEE Main 2026 — Jan 28, Shift 1 — Physics solution

JEE Main2026Jan 28, Shift 1Physics

An atom $_{3}^{8} X$ 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 :

A
250%
B
150%
C
125%
D
900%

Solution

Nuclear surface area scales as $A^{2/3}$ , where $A$ is the mass number (electrons do not contribute to the nucleus's size).

Initial: $A_{i} = 8$ , so $S_{i} \propto 8^{2/3} = 4$ .

Final: $A_{f} = 8 + 10 + 9 = 27$ (10 protons and 9 neutrons added to the nucleus), so $S_{f} \propto 2 7^{2/3} = 9$ .

Percentage growth $= \frac{9 - 4}{4} \times 100 = 125%$ .

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