A solid cylinder, a circular disc, a solid sphere and a hollow cylinder of the same radius are placed on an inclined plane. Which of the following will have maximum acceleration at the bottom of the plane?

Let the inclination of the inclined plane be $\theta$. Acceleration of a rolling body is given as
$a=\frac{g \sin \theta}{\left(1+K^2 / R^2\right)}$ Here, $K$ is radius of gyration. For circular disc $I=\frac{1}{2} M R^2 \quad \therefore \frac{K^2}{R^2}=\frac{1}{2}$ $\therefore \quad a=\frac{2}{3} g \sin \theta$
For solid cylinder $I=\frac{1}{2} M R^2 \therefore \frac{K^2}{R^2}=\frac{1}{2}$ $\therefore \quad a=\frac{2}{3} g \sin \theta$
For solid sphere $I=\frac{2}{5} M R^2 \therefore \frac{K^2}{R^2}=\frac{2}{5}$ $\therefore \quad a=\frac{5}{7} g \sin \theta$
For hollow cylinder $I=M R^2 \quad \therefore \frac{K^2}{R^2}=1$ $\therefore a=\frac{1}{2} g \sin \theta$
Hence, from (i), (ii), (iii) and (iv) we conclude that the solid sphere will reach the bottom with maximum acceleration.