$A$ sphere of mass $50 \, g$ and diameter $20 \, cm$ is rolling without slipping with a velocity of $5 \, cm/s$. Its total kinetic energy is:

Consider a point $P$ on the circumference of a disc rolling along a horizontal surface. If $R$ is the radius of the disc,the distance through which $P$ moves in one full rotation of the disc is:

$A$ solid homogeneous sphere is rolling with sliding on a horizontal rough surface. For this type of motion of the sphere:

$A$ sphere of mass $M$ and radius $r$ slips on a rough horizontal plane. At some instant,it has translational velocity $V_0$ and rotational velocity about the centre $\frac{V_0}{2r}$. The translational velocity when the sphere starts pure rolling motion is

$A$ small roller of diameter $20 \ cm$ has an axle of diameter $10 \ cm$. It is on a horizontal floor and a meter scale is positioned horizontally on its axle with one edge of the scale on top of the axle. The scale is now pushed slowly on the axle so that it moves without slipping on the axle,and the roller starts rolling without slipping. After the roller has moved $50 \ cm$,the position of the scale will look like (figures are schematic and not drawn to scale)-

The ratio of kinetic energies of two spheres rolling with equal centre of mass velocities is $2 : 1$. If their radii are in the ratio $2 : 1$,then the ratio of their masses will be