$A$ solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is $:$

$A$ thin uniform circular disc rolls with a constant velocity without slipping on a horizontal surface. Its total kinetic energy is

In the figure,$E$ and $v_{cm}$ represent the total energy and speed of the centre of mass of an object of mass $1 \ kg$ in pure rolling. The object is

$A$ disc is rolling (without slipping) on a horizontal surface. $C$ is its centre and $Q$ and $P$ are two points on the same horizontal line passing through $C$,such that $Q$ is at a distance $r$ from $C$ and $P$ is at a distance $r$ from $C$ on the opposite side. Let $V_P, V_Q$ and $V_C$ be the magnitudes of velocities of points $P, Q$ and $C$ respectively,then:

$A$ circular disc is rolling on a horizontal plane. Its total kinetic energy is $300 \, J$. The translational kinetic energy of the disc is:

If a solid sphere is rolling without slipping on a horizontal plane,then the ratio of its rotational and total kinetic energies is