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
Concept. Rolling up without slipping (static friction does no work): all the KE, translational and rotational, becomes PE: $h=\dfrac{v_0^2}{2g}\left(1+\dfrac{k^2}{R^2}\right)$.
Step 1. Compare. $\dfrac{v_0^2}{2g}\left(1+\dfrac{k^2}{R^2}\right)=\dfrac{7v_0^2}{10g}\Rightarrow1+\dfrac{k^2}{R^2}=\dfrac75$.
Step 2. Identify. $\dfrac{k^2}{R^2}=\dfrac25$: a solid sphere.
Answer. Option D.
Solution by Sreeraj P, M.Sc Physics
