Q 11-06-258JEE MainJEE Main 2025 (4 Apr, Shift 1)Medium
A circular ring and a solid sphere having the same radius roll down an inclined plane from rest without slipping. The ratio of their velocities when they reach the bottom of the plane is $\sqrt{\dfrac x5}$, where $x$ = ______.
Numerical value type. Enter your answer.
Answer: 3.5
For rolling from rest through height $h$:
$$mgh = \tfrac12mv^2\left(1 + \frac{k^2}{R^2}\right) \Rightarrow v = \sqrt{\frac{2gh}{1 + k^2/R^2}}$$
Ring: $k^2/R^2 = 1$. Solid sphere: $k^2/R^2 = \tfrac25$.
$$\frac{v_{\text{ring}}}{v_{\text{sphere}}} = \sqrt{\frac{1 + 2/5}{1 + 1}} = \sqrt{\frac{7}{10}} = \sqrt{\frac{3.5}{5}}$$
So $x = 3.5$. (The official answer key gives 4, i.e. $x$ rounded to the nearest integer.)
Solution by Sreeraj P, M.Sc Physics