$A$ uniform solid cylindrical roller of mass $m$ is being pulled on a horizontal surface with force $F$ parallel to the surface and applied at its centre. If the acceleration of the cylinder is $a$ and it is rolling without slipping,then the value of $F$ is:

$A$ 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,
$(A)$ $\vec{V}_C-\vec{V}_A=2(\vec{V}_B-\vec{V}_C)$
$(B)$ $\vec{V}_C-\vec{V}_B=\vec{V}_B-\vec{V}_A$
$(C)$ $|\vec{V}_C-\vec{V}_A|=2|\vec{V}_B-\vec{V}_C|$
$(D)$ $|\vec{V}_C-\vec{V}_A|=4|\vec{V}_B|$

Read each statement below carefully,and state,with reasons,if it is true or false;
$(a)$ During rolling,the force of friction acts in the same direction as the direction of motion of the $CM$ of the body.
$(b)$ The instantaneous speed of the point of contact during rolling is zero.
$(c)$ The instantaneous acceleration of the point of contact during rolling is zero.
$(d)$ For perfect rolling motion,work done against friction is zero.
$(e)$ $A$ wheel moving down a perfectly frictionless inclined plane will undergo slipping (not rolling) motion.

Consider a point $P$ at the contact point of a wheel on the ground,which rolls on the ground without slipping. Find the displacement of point $P$ when the wheel completes half of a rotation (given the radius of the wheel is $1 \ m$).

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

$A$ circular disc rolls on a horizontal floor without slipping and the centre of the disc moves with a uniform velocity $v$. Which of the following values can the velocity of a point on the rim of the disc have?