$A$ solid cylinder of radius $R$ and mass $M$ rolls down an inclined plane without slipping and reaches the bottom with a speed $v$. The speed would be less than $v$ if we use:

$A$ 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 is rolling down an inclined plane without slipping. The ratio of its rotational kinetic energy to its total kinetic energy is:

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

$A$ solid disc of radius $a$ and mass $m$ rolls down without slipping on an inclined plane making an angle $\theta$ with the horizontal. The acceleration of the disc will be $\frac{2}{b} g \sin \theta$ where $b$ is $........$. (Round off to the Nearest Integer)
($g =$ acceleration due to gravity)
($\theta =$ angle as shown in figure)

$STATEMENT-1$: Two cylinders,one hollow (metal) and the other solid (wood) with the same mass and identical dimensions,are simultaneously allowed to roll without slipping down an inclined plane from the same height. The hollow cylinder will reach the bottom of the inclined plane first.
$STATEMENT-2$: By the principle of conservation of energy,the total kinetic energies of both the cylinders are identical when they reach the bottom of the incline.