$A$ solid sphere of mass $m$ and radius $r$ rolls down an inclined plane. What is the ratio of its rotational kinetic energy to its translational kinetic energy?

$A$ 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

Two identical uniform discs roll without slipping on two different surfaces $AB$ and $CD$ (see figure) starting at $A$ and $C$ with linear speeds $v_1$ and $v_2$,respectively,and always remain in contact with the surfaces. If they reach $B$ and $D$ with the same linear speed and $v_1 = 3 \ m/s$,then $v_2$ in $m/s$ is $(g = 10 \ m/s^2)$.

Solid cylinders of radii $r_1, r_2$ and $r_3$ roll down an inclined plane from the same place simultaneously. If $r_1 > r_2 > r_3$,which one would reach the bottom first?

$A$ solid sphere rolls purely on an inclined plane of inclination $\theta$. Which of the following statements are correct?
$(1)$ The friction force acting on the sphere is $f = \mu mg \cos \theta$.
$(2)$ Friction is a dissipative force.
$(3)$ Friction increases angular velocity and decreases linear velocity.
$(4)$ If $\theta$ decreases,friction decreases.

$A$ hollow cylinder is rolling on an inclined plane,inclined at an angle of $30^{\circ}$ to the horizontal. Its speed after travelling a distance of $10 \ m$ will be .......... $m/s$.