$A$ disc is rolling without slipping on a straight surface. The ratio of its translational kinetic energy to its total kinetic energy is

Three solid spheres are made to move on a rough horizontal surface. Sphere $P$ is given a spin and released. Sphere $Q$ is given a forward linear velocity. Sphere $R$ is given linear and rotational motions as shown in the figure. Directions of the friction force on spheres $P, Q, R$ are respectively

$A$ horizontal force $F$ is applied at the center of a sphere as shown in the figure. The coefficient of friction between the sphere and the ground is $\mu$. What is the maximum value of $F$ if the sphere does not slip?

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

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

When a body is rolling without slipping on a rough horizontal surface,the work done by friction is ........