$A$ 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

$A$ sphere of mass $m$ and radius $r$ rolls on a horizontal plane without slipping with the speed $u$. Now if it rolls up an incline, the maximum height it would attain will be

$A$ hollow cylinder and a solid cylinder,initially at rest at the top of an inclined plane,are rolling down without slipping. If the time taken by the hollow cylinder to reach the bottom of the inclined plane is $2 \ s$,the time taken by the solid cylinder to reach the bottom of the inclined plane is: (in $s$)

$A$ solid cylinder of mass $M$ and radius $R$ rolls down an inclined plane of height $h$. When it reaches the foot of the plane,its rotational kinetic energy is ($g=$ acceleration due to gravity).

$A$ uniform spherical object of mass $M$ and radius $R$ has a moment of inertia $I$. It rolls without slipping down an inclined plane of angle $\theta$. What is its acceleration?

$A$ uniform solid spherical ball is rolling down a smooth inclined plane from a height $h$. The velocity attained by the ball when it reaches the bottom of the inclined plane is $v$. If the ball is now thrown vertically upwards with the same velocity $v$,the maximum height to which the ball will rise is