Explanation

(d) When a body rolls down an inclined plane, it is accompanied by rotational and translational kinetic energies.
Rotational kinetic energy $= \frac{1}{2}I\omega^{2} = K_{R}$
Where $I$ is moment of inertia and $\omega$ the angular velocity.
Translational kinetic energy
$= \frac{1}{2}mv^{2} = K_{r} = \frac{1}{2}m(r\omega)^{2}$
where $m$ is mass, $v$ the velocity and $\omega$ the angular velocity.
Given,
Translational KE=rotational KE
$\frac{1}{2}mv^{2} = \frac{1}{2}I\omega^{2}$
Since, $v = r\omega$
$\therefore$ $\frac{1}{2}m\left( r^{2}\omega^{2} \right) = \frac{1}{2}I\omega^{2}$
$\Longrightarrow$ $I = mr^{2}$
We know that $mr^{2}$ is the moment of inertia of hollow cylinder about its axis is where $m$ is mass of hollow cylinderical body and $r$ the radius of cylinder.