$A$ solid homogeneous sphere of mass $M$ and radius $r$ is moving on a rough horizontal surface,partly rolling and partly sliding. During this kind of motion of this sphere,

$A$ 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:
$(i) \vec{V}_C - \vec{V}_A = 2(\vec{V}_B - \vec{V}_C)$
$(ii) \vec{V}_C - \vec{V}_B = \vec{V}_B - \vec{V}_A$
$(iii) |\vec{V}_C - \vec{V}_A| = 2|\vec{V}_B - \vec{V}_C|$
$(iv) |\vec{V}_C - \vec{V}_A| = 4|\vec{V}_B|$

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

$A$ solid sphere of mass $5 \,kg$ rolls on a plane surface. Find its kinetic energy at an instant when its centre moves with speed $4 \,m/s$. (in $\,J$)

Two spheres are rolling with the same velocity (for their $C.M.$). Their ratio of kinetic energy is $2:1$ and the ratio of their radii is $2:1$. Their mass ratio will be:

$A$ spool consists of a cylinder of radius $R_1$ wrapped with thread,and its end caps have radius $R_2$ as shown in the end view illustrated in the figure. The mass of the spool,including the thread,is $m$,and its moment of inertia about an axis through its center is $I$. The spool is placed on a rough horizontal surface so that it rolls without slipping. When a force $\vec{T}$ acting to the right is applied to the free end of the thread,the friction force exerted by the surface on the spool is given by: