$A$ solid sphere of mass $1\,kg$ rolls without slipping on a plane surface. Its kinetic energy is $7 \times 10^{-3}\,J$. The speed of the centre of mass of the sphere is $.........\,cm\,s^{-1}$.

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

$A$ uniform solid sphere of mass $m$ and radius $R$ is being pulled on a horizontal surface with a force $F$ parallel to the surface,applied at its topmost point. If the acceleration of the center of mass of the sphere is $a$ and it is rolling without slipping on the surface,the value of $F$ is:

If a solid sphere of mass $1\, kg$ and radius $0.1\, m$ rolls without slipping at a uniform velocity of $1\, m/s$ along a straight line on a horizontal floor,the kinetic energy is (in $, J$)

$A$ 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 $g = 9.8 \text{ m/s}^2$)

In the case of pure rolling,what will be the velocity of point $A$ of the ring of radius $R$?