Q59 / 75 · JEE Main 2026 (28 January Shift 1) · Physics
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$900 \%$
- Correct answerD$125 \%$
Solution
Surface area $x \propto A^{2/3}$
$X_i = 8^{2/3} K = 4K$
$X_f = (8 + 10 + 9)^{2/3} K = 9K$
% increase in surface area of nucleus
$x_i = \dfrac{9K - 4K}{4K} \times 100 = 125\%$
$X_i = 8^{2/3} K = 4K$
$X_f = (8 + 10 + 9)^{2/3} K = 9K$
% increase in surface area of nucleus
$x_i = \dfrac{9K - 4K}{4K} \times 100 = 125\%$
Hence, the correct answer is option: 'D'.