$A$ disc rolls without slipping with a constant velocity. What fraction of its total kinetic energy is in the form of rotational kinetic energy?

$A$ round body is given a translational velocity and released on a surface that has friction. Then:

$A$ sphere is rolling without slipping on a fixed horizontal plane surface. In the figure,$A$ is the point of contact,$B$ is the centre of the sphere and $C$ is its topmost point. Then,
$(A)$ $\vec{V}_C-\vec{V}_A=2(\vec{V}_B-\vec{V}_C)$
$(B)$ $\vec{V}_C-\vec{V}_B=\vec{V}_B-\vec{V}_A$
$(C)$ $|\vec{V}_C-\vec{V}_A|=2|\vec{V}_B-\vec{V}_C|$
$(D)$ $|\vec{V}_C-\vec{V}_A|=4|\vec{V}_B|$

An object is rolling on a horizontal surface without slipping. If its rotational kinetic energy and translational kinetic energy are equal,the object is a:

$A$ solid cylinder having radius $R$ and length $L$ is slipping on a rough horizontal plane. At time $t = 0$ the cylinder has a translational velocity $v_0 = 49 \text{ m/s}$, perpendicular to its axis and a rotational velocity $v_0/4R$ about the centre. The time taken by the cylinder to start rolling is . . . . . . seconds. (coefficient of kinetic friction $\mu_K = 0.25$ and $g = 9.8 \text{ m/s}^2$)

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