A solid sphere, disc and solid cylinder all of the same mass and made of the same material are allowed to roll down (from rest) on the inclined plane, then

Solution :
Let us consider that solid sphere, disc and solid cylinder are rolling on an inclined plane. $$M, I$$  and $$R$$ be mass, moment of inertia and radius of the rolling section in each case.
(i) Solid sphere The moment of inertia of a solid sphere about its diameter is given by $$I = \frac{2}{5}M{R^2}$$
$${\text{or}}\,\,K = \frac{I}{{M{R^2}}} = \frac{2}{5}$$
As from the concept, acceleration
$$\eqalign{ & a = \frac{{g\sin \theta }}{{1 + K}} \cr & {\text{So,}}\,\,a = \frac{{g\sin \theta }}{{1 + \frac{2}{5}}} = \frac{5}{7}g\sin \theta \cr} $$
(ii) Disc The moment of inertia of disc about an axis perpendicular to the plane of disc and passing through its centre is given by $$I = \frac{1}{2}M{R^2}$$
$${\text{or}}\,\,\frac{I}{{M{R^2}}} = \frac{1}{2}$$
$$\therefore a = \frac{{g\sin \theta }}{{1 + \frac{1}{2}}} = \frac{2}{3}g\sin \theta $$
(iii) Solid cylinder The moment of inertia of a cylinder about the axis passing through its centre and perpendicular to its plane is given by $$I = \frac{1}{2}M{R^2}$$
$$\eqalign{ & {\text{or}}\,\,\frac{I}{{M{R^2}}} = \frac{1}{2} \cr & \therefore a = \frac{{g\sin \theta }}{{1 + \frac{1}{2}}} = \frac{2}{3}g\sin \theta \cr} $$
So, acceleration of solid sphere is more. It implies that solid sphere reaches the bottom first.