An object is rolling on an inclined plane. Its linear and rotational kinetic energies are equal. The object 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

$A$ solid cylinder and a solid sphere having the same mass and radius roll down on the same smooth inclined plane. The ratio of the acceleration of the cylinder $(a_c)$ to that of the sphere $(a_s)$ is

$A$ solid cylinder of mass $M$ and radius $R$ rolls without slipping down an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is:

If a solid sphere is rolling,the ratio of its rotational kinetic energy to the total kinetic energy is given by

$A$ solid cylinder of mass $M$ and radius $R$ rolls down an inclined plane of height $h$. The angular velocity of the cylinder when it reaches the bottom of the plane will be