$A$ uniform solid sphere of mass $m$ and radius $r$ rolls without slipping down an inclined plane,inclined at an angle $45^{\circ}$ to the horizontal. Find the magnitude of the minimum frictional coefficient at which slipping is absent.

Three bodies, a ring, a solid disc, and a solid sphere, roll down the same inclined plane without slipping. The radii of the bodies are identical, and they start from rest. If $V_S, V_R$, and $V_D$ are the speeds of the sphere, ring, and disc, respectively, when they reach the bottom, then the correct option is:

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

$A$ disc of mass $3 \, kg$ rolls down an inclined plane of height $5 \, m$. The translational kinetic energy of the disc on reaching the bottom of the inclined plane is .......... $J$. (Take $g = 10 \, m/s^2$)

$A$ sphere of radius $a$ and mass $m$ rolls along a horizontal plane with constant speed $v_{0}$. It encounters an inclined plane at angle $\theta$ and climbs upward. Assuming that it rolls without slipping,how far up the sphere will travel?

$A$ 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