$A$ cylinder is rolling down an inclined plane of inclination $30^\circ$ without slipping. Its linear acceleration along the inclined plane will be:

An inclined plane makes an angle of $30^\circ$ with the horizontal. $A$ solid sphere starts rolling down from rest without slipping. Its linear acceleration will be:

$A$ sphere of radius $a$ and mass $m$ rolls along a horizontal plane with constant speed $v_{0}$. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping,how far up the sphere will travel?

$A$ solid sphere of mass $2 \,kg$ rolls on a smooth horizontal surface at $10 \,m/s$. It then rolls up a smooth inclined plane of inclination $30^{\circ}$ with the horizontal. The height attained by the sphere before it stops is [take $g=10 \,m/s^2$].

$A$ solid sphere of mass $m$ rolls down an inclined plane without slipping,starting from rest at the top of an inclined plane. The linear speed of the sphere at the bottom of the inclined plane is $v$. The kinetic energy of the sphere at the bottom is

$A$ uniform solid sphere of mass $m$ and radius $r$ rolls without slipping down an inclined plane,inclined at an angle $45^{\circ}$ to the horizontal. Find the magnitude of the minimum frictional coefficient at which slipping is absent.