$A$ solid cylinder rolls without slipping down an inclined plane of height $h$. The velocity of the cylinder when it reaches the bottom is

$A$ solid sphere of mass $m$ and radius $r$ is allowed to roll without slipping from the highest point of an inclined plane of length $L$ that makes an angle $30^{\circ}$ with the horizontal. The speed of the sphere at the bottom of the plane is $v_1$. If the angle of inclination is increased to $45^{\circ}$ while keeping $L$ constant,the new speed of the sphere at the bottom of the plane is $v_2$. The ratio of $v_1^2 : v_2^2$ is

$A$ tennis ball (treated as a hollow spherical shell) starting from $O$ rolls down a hill. At point $A$,the ball becomes airborne,leaving at an angle of $30^\circ$ with the horizontal. The ball strikes the ground at $B$. What is the value of the distance $AB$ (in $m$)? (Moment of inertia of a spherical shell of mass $m$ and radius $R$ about its diameter is $I = \frac{2}{3}mR^2$).

$A$ 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 at rest rolls down an inclined plane of vertical height $h$ without sliding. Its speed on reaching the bottom of the plane is ($g=$ acceleration due to gravity).

$A$ 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