$A$ 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

$A$ solid sphere rolls without slipping on an inclined plane at an angle $\theta$. The ratio of total kinetic energy to its rotational kinetic energy is

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

$A$ 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?

$A$ solid homogeneous sphere rolls without slipping down an inclined plane of height $h$. The velocity of the sphere at the bottom is:

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