$A$ plane is inclined at an angle of $30^{\circ}$ with the horizontal. If a sphere rolls down this plane without slipping,what will be the linear acceleration of the sphere?

$A$ solid sphere is rolling without slipping on a semi-circular track of radius $R = 10 \ m$ as shown in the figure. The radius of the solid sphere is much smaller than the radius of the semi-circular track. At the lowest point,it has a velocity $v = 10 \ m/s$. To what maximum angle $\theta$ from the vertical will the sphere travel before it comes back down? Neglect the rolling friction between the sphere and the track. (Take $g = 10 \ m/s^2$)

$A$ 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 $g=10 \ m/s^2$)

$A$ solid cylinder $30\ cm$ in diameter at the top of an inclined plane $2.0\ m$ high is released and rolls down the incline without loss of energy due to friction. Its linear speed at the bottom is.......... $m/s$.

$A$ solid cylinder and a hollow cylinder,both of the same mass and same external diameter,are released from the same height at the same time on an inclined plane. Both roll down without slipping. Which one will reach the bottom first?

What is the velocity of a sphere starting from rest and rolling without slipping down an inclined plane of vertical height $h$?