A sphere rolls down on an inclined plane of the inclination $\mathrm{\theta }$. What is the acceleration as the sphere reaches bottom ?

Acceleration of a body rolling down an inclined plane is given by , $\mathrm{a}=\frac{\mathrm{g} \mathrm{sin\theta }}{1+{\displaystyle \frac{{\mathrm{K}}^2}{{\mathrm{R}}^2}}}$

In case of a solid sphere, we have $\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2}=\frac{\left[\mathrm{I} /\mathrm{M}\right]}{{\mathrm{R}}^2}=\frac{\mathrm{I}}{{\mathrm{MR}}^2}=\frac{{\displaystyle \left(\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$5$}\right.\right)} {\mathrm{MR}}^2}{{\mathrm{MR}}^2}=\frac{2}{5}$

Substituting $\frac{{\mathrm{K}}^2}{{\mathrm{R}}^2}=\frac{2}{5}$ in equation (1) we get $\mathrm{a}=\frac{5}{7}\mathrm{g} \mathrm{sin\theta }$