A solid sphere, solid cylinder and a disc of same radius are allowed to roll down from the top of an incline plane from the same height. Then

let $l$ is the length of the incline surface and $\theta$ is the angle of inclination.

time taken to reach the bottom of the incline is given as,

$\mathrm{t}=\sqrt{\frac{2l\left(1+\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2}\right)}{\mathrm{gsin}\theta }}$

for solid sphere: $\frac{K^2}{R^2}=\frac{2}{5}$

for solid cylinder: $\frac{K^2}{R^2}=\frac{1}{2}$

for a disc: $\frac{K^2}{R^2}=\frac{1}{2}$

$\frac{K^2}{R^2}$ for solid sphere is less than that of solid cylinder and disc.

so, Sphere will reach the bottom first