$A$ solid sphere rolls down without slipping on a $30^{\circ}$ inclined plane. If $g = 10\,m/s^2$,the acceleration of the rolling sphere is

$A$ solid sphere having mass $m$ and radius $r$ rolls down an inclined plane. Then its kinetic energy is

$A$ ring is rolling on an inclined plane. The ratio of the linear and rotational kinetic energies will be

$A$ solid cylinder and a solid sphere having same mass and same radius roll down on the same inclined plane. The ratio of the acceleration of the cylinder '$a_{c}$' to that of sphere '$a_{s}$' is

$A$ ring and a solid sphere roll down the same inclined plane without slipping. They start from rest. The radii of both bodies are identical and the ratio of their kinetic energies is $\frac{7}{x}$ where $x$ is . . . . . . .

$A$ uniform disk of mass $m$ and radius $R$ rolls without slipping down an inclined plane of length $l$ and inclination $\theta$. Initially,the disk was at rest at the top of the inclined plane. Its angular momentum about the point of contact with the inclined plane when it reaches the bottom will be equal to: