Q 11-06-070JEE MainJEE Main 2025 (24 Jan, Shift 2)Easy
A solid sphere and a hollow sphere of the same mass and of same radius are rolled on an inclined plane. Let the time taken to reach the bottom by the solid sphere and the hollow sphere be $t_1$ and $t_2$, respectively, then
Answer: (C) $t_1 < t_2$
Rolling down an incline, $a = \dfrac{g\sin\theta}{1 + k^2/R^2}$.
- Solid sphere: $k^2/R^2 = \tfrac{2}{5}$, $a_1 = \tfrac{5}{7}g\sin\theta$
- Hollow sphere: $k^2/R^2 = \tfrac{2}{3}$, $a_2 = \tfrac{3}{5}g\sin\theta$
$a_1 > a_2$, and for the same length $t = \sqrt{2L/a}$, so $t_1 < t_2$.
Solution by Sreeraj P, M.Sc Physics