$A$ solid ball of mass $m$ and radius $r$ rolls without slipping along the track shown in the figure. The radius of the circular part of the track is $R$. The ball starts rolling down the track from rest from a height of $8R$ from the ground level. When the ball reaches the point $P$,then its velocity will be

$A$ solid sphere rolling without friction on a horizontal surface with a constant speed of $2 \,m/s$, rolls up on an inclined ramp which is inclined at $30^{\circ}$. The maximum distance travelled by the sphere on the inclined ramp is (acceleration due to gravity $g=10 \,m/s^2, \sin 30^{\circ}=1/2$) (in $\,m$)

$A$ ring,a solid sphere,and a disc are rolling down from the top of an inclined plane of the same height. What is the sequence in which they reach the surface?

$A$ sphere,a disc,a ring,and a hollow sphere of the same radius are rolled down from the same height on an inclined plane simultaneously. The order in which these objects reach the bottom is:

$A$ cylinder is rolling down on an inclined plane of inclination $60^{\circ}$. Its acceleration during rolling down will be $\frac{x}{\sqrt{3}} \ m/s^2$,where $x=$ . . . . . . . (use $g=10 \ m/s^2$)

$A$ 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 $g=10 \,m \,s^{-2}$.