$A$ 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:

$A$ disc is rolling (without slipping) on a horizontal surface. $C$ is its centre and $Q$ and $P$ are two points on the same horizontal line passing through $C$,such that $Q$ is at a distance $r$ from $C$ and $P$ is at a distance $r$ from $C$ on the opposite side. Let $V_P, V_Q$ and $V_C$ be the magnitudes of velocities of points $P, Q$ and $C$ respectively,then:

$A$ solid sphere is rolling without slipping on a horizontal plane. The ratio of the linear kinetic energy of the centre of mass of the sphere and rotational kinetic energy is $:$

$A$ ring and a disc roll on a horizontal surface without slipping with the same linear velocity. If both have the same mass and the total kinetic energy of the ring is $6 \ J$,then the total kinetic energy of the disc is: (in $/2 \ J$)

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

In a bicycle,the radius of the rear wheel is twice the radius of the front wheel. If $r_F$ and $r_r$ are the radii,and $v_F$ and $v_r$ are the speeds of the topmost points of the wheels respectively,then: