$A$ solid sphere and a disc of same mass and radius start rolling down a rough inclined plane from the same height. The ratio of the time taken in the two cases is:

$A$ rigid body of mass $M$ and radius $R$ rolls without slipping on an inclined plane of inclination $\theta$, under gravity. Match the type of body in Column-$I$ with the magnitude of the force of friction in Column-$II$.

Column-$I$	Column-$II$
$(A)$ Ring	$(I)$ $\frac{Mg \sin \theta}{3.5}$
$(B)$ Solid sphere	$(II)$ $\frac{Mg \sin \theta}{2}$
$(C)$ Solid cylinder	$(III)$ $\frac{Mg \sin \theta}{3}$
$(D)$ Hollow cylinder	$(IV)$ $\frac{Mg \sin \theta}{2.5}$

$A$ body slides down a smooth inclined plane of inclination $\theta$ and reaches the bottom with velocity $V$. If the same body is a ring which rolls down the same inclined plane, then the linear velocity at the bottom of the plane is:

When a uniform solid sphere and a disc of the same mass and of the same radius roll down a rough inclined plane from rest to the same distance,then the ratio of the time taken by them is

$A$ sphere of mass $2 \, kg$ and radius $0.5 \, m$ is rolling with an initial speed of $1 \, m/s$ up an inclined plane which makes an angle of $30^{\circ}$ with the horizontal plane,without slipping. How long will the sphere take to return to the starting point $A$? (in seconds)

$A$ solid cylinder and a solid sphere having the same mass and radius roll down on the same smooth inclined plane. The ratio of the acceleration of the cylinder $(a_c)$ to that of the sphere $(a_s)$ is