$A$ 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 $\,m/s$)

$A$ solid sphere is rolling on a surface as shown in the figure,with a translational velocity $v \, ms^{-1}$. If it is to climb the inclined surface continuing to roll without slipping,then the minimum velocity for this to happen is:

When a sphere with moment of inertia $I$ rolls down an inclined plane,what percentage of its total energy is rotational kinetic energy?

$A$ solid cylinder rolls without slipping down an inclined plane of height $h$. The velocity of the cylinder when it reaches the bottom is

$A$ disc of radius $R$ is rolling down an inclined plane whose angle of inclination is $\theta$. Its acceleration would be

The following bodies,
$(1)$ a ring
$(2)$ a disc
$(3)$ a solid cylinder
$(4)$ a solid sphere,
of same mass $m$ and radius $R$ are allowed to roll down without slipping simultaneously from the top of an inclined plane. The body which will reach first at the bottom of the inclined plane is ...........
[Mark the body as per their respective numbering given in the question]