A solid oylinder and a solid sphere having same mass and same radius roll down on the same inclined plane. The fatio of the acceleration of the cylinder ' \(\mathrm{a}_{\mathrm{c}}\) ' to that of sphere ' \(\mathrm{a}_{\mathrm{s}}\) ' is

For sphere: M.I. \(\mathrm{I}_{\mathrm{S}}=\frac{2}{5} \dot{\mathrm{MR}}^2\)
For cylinder: M.I. \(\mathrm{I}_{\mathrm{C}}=\frac{1}{2} \mathrm{MR}^2\)
Acceleration: \(\mathrm{a}=\frac{\mathrm{g} \sin \theta}{1+\frac{\mathrm{I}}{\mathrm{MR}^2}}\)
\(\begin{aligned}
& \therefore \quad \frac{\mathrm{a}_{\mathrm{c}}}{\mathrm{a}_{\mathrm{s}}}=\frac{1+\frac{\mathrm{I}_{\mathrm{s}}}{\mathrm{MR}^2}}{1+\frac{\mathrm{I}_{\mathrm{C}}}{\mathrm{MR}^2}}=\frac{\mathrm{MR}^2+\mathrm{I}_{\mathrm{s}}}{\mathrm{MR}^2+\mathrm{I}_{\mathrm{c}}}=\frac{\mathrm{MR}^2+\frac{2}{5} \mathrm{MR}^2}{\mathrm{MR}^2+\frac{1}{2} \mathrm{MR}^2} \\
& \therefore \quad \frac{\mathrm{a}_{\mathrm{c}}}{\mathrm{a}_{\mathrm{s}}}=\frac{7}{5} \times \frac{2}{3}=\frac{14}{15}
\end{aligned}\)