Q 11-06-049JEE MainJEE Main 2026 (5 Apr, Shift 2)Medium
An object of uniform density rolls up the curved path with the initial velocity $v_0$ as shown in the figure. If the maximum height attained by an object is $\dfrac{7v_0^2}{10g}$ ($g =$ acceleration due to gravity), the object is a ______.
Answer: (D) solid sphere
Rolling without slipping, all the kinetic energy becomes potential energy at the top:
$$\frac{1}{2}mv_0^2\left(1 + \frac{k^2}{R^2}\right) = mgh \;\Rightarrow\; h = \frac{v_0^2}{2g}\left(1 + \frac{k^2}{R^2}\right)$$
$h = \dfrac{7v_0^2}{10g}$ needs $1 + \dfrac{k^2}{R^2} = \dfrac{7}{5}$, i.e. $\dfrac{k^2}{R^2} = \dfrac{2}{5}$: a solid sphere.
Solution by Sreeraj P, M.Sc Physics