$A$ solid cylinder of mass $M$ and radius $R$ rolls down a smooth inclined plane about its own axis and reaches the bottom with velocity $v$. What is the height of the inclined plane? (Assume $g$ is the acceleration due to gravity.)

$A$ thin uniform,circular ring is rolling down an inclined plane of inclination $30^{\circ}$ without slipping. Its linear acceleration along the inclined plane will 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]

$A$ ring and a disc are initially at rest, side by side, at the top of an inclined plane which makes an angle $60^{\circ}$ with the horizontal. They start to roll without slipping at the same instant of time along the shortest path. If the time difference between their reaching the ground is $(2-\sqrt{3}) / \sqrt{10} \,s$, then the height of the top of the inclined plane, in metres, is. . . . . . Take $g=10 \,m \,s^{-2}$.

$A$ solid sphere rolls without slipping,first horizontally and then up to a point $X$ at height $h$ on an inclined plane before rolling down,as shown in the figure below. The initial horizontal speed of the sphere is

An object slides down a smooth incline and reaches the bottom with velocity $v$. If the same mass is in the form of a ring and it rolls down an inclined plane of the same height and angle of inclination,then its velocity at the bottom of the inclined plane will be ............