$A$ solid cylinder is released from rest from the top of an inclined plane of inclination $30^{\circ}$ and length $60 \,cm$. If the cylinder rolls without slipping, then the speed when it reaches the bottom is (in $\,m/s$)

$A$ 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	$(IV)$ $\frac{Mg \sin \theta}{2.5}$

Three bodies, a ring, a solid disc, and a solid sphere, roll down the same inclined plane without slipping. The radii of the bodies are identical, and they start from rest. If $V_S, V_R$, and $V_D$ are the speeds of the sphere, ring, and disc, respectively, when they reach the bottom, then the correct option is:

$A$ solid cylinder of diameter $30 \ cm$ is rolled down an inclined plane from a height of $2 \ m$. If there is no energy loss due to friction,the angular velocity at the base of the plane is ....... $rad/s$. (Take $g = 10 \ m/s^2$)

$A$ ring,a solid sphere,a disc,and a solid cylinder of the same radii roll down an inclined plane. Which one would reach the bottom last?

The speed of a homogeneous solid sphere after rolling down an inclined plane of vertical height $h$,from rest without sliding,is