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 rolls down on an inclined plane of inclination $\theta$. What is the acceleration as the sphere reaches the bottom?

$A$ sphere is rolling down an incline without slipping. If the height of the incline is $14 \ m$,find its linear velocity at the bottom.

$A$ 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 frictionless surface,as shown in the figure,with a translational velocity $v \, m/s$. If it is to climb the inclined surface to a height $h$,then $v$ should be:

An inclined plane makes an angle $30^{\circ}$ with the horizontal. $A$ solid sphere rolling down an inclined plane from rest without slipping has a linear acceleration (where $g$ is the acceleration due to gravity and $\sin 30^{\circ} = 0.5$).