JEE Main 2026 — Rotational Motion Question with Solution

By the law of conservation of mechanical energy, the total initial kinetic energy of the rolling object is converted into potential energy at the maximum height.

Total initial kinetic energy for an object rolling without slipping is given by:
$K=K_{\text{translational}}+K_{\text{rotational}}$
$K=\frac{1}{2}m{v}_0^2+\frac{1}{2}I{\omega }^2$

Substituting $\omega =\frac{v_0}{R}$ and $I=m{k}^2$ (where $k$ is the radius of gyration):
$K=\frac{1}{2}m{v}_0^2+\frac{1}{2}(m{k}^2){\left(\frac{v_0}{R}\right)}^2=\frac{1}{2}m{v}_0^2\left(1+\frac{k^2}{R^2}\right)$

The potential energy at the maximum height $h$ is:
$U=mgh$

Given $h=\frac{7v_0^2}{10g}$, we have:
$U=mg\left(\frac{7v_0^2}{10g}\right)=\frac{7}{10}m{v}_0^2$

Equating kinetic and potential energy ($K=U$):
$\frac{1}{2}m{v}_0^2\left(1+\frac{k^2}{R^2}\right)=\frac{7}{10}m{v}_0^2$

$1+\frac{k^2}{R^2}=\frac{14}{10}=\frac{7}{5}$

$\frac{k^2}{R^2}=\frac{7}{5}-1=\frac{2}{5}$

The moment of inertia is $I=\frac{2}{5}m{R}^2$, which corresponds to a solid sphere.

Answer: solid sphere