A solid sphered rolls down without slipping on an inclined plane of angle of inclination ${\sin }^{-1}\left(\frac{21}{50}\right).$ If the acceleration due to gravity is $10{\mathrm{ms}}^{-2},$ the acceleration of the rolling sphere is _____

Given,

the inclination of the inclined plane,

$\mathrm{\theta }={\sin }^{-1}\left(\frac{21}{50}\right)\Rightarrow \mathrm{sin\theta }=0.42, \mathrm{g}=10\mathrm{m}/{\mathrm{s}}^2$

The formula for the acceleration of a solid sphere rolling on an inclined plane without sliding is

$\mathrm{a}=\frac{\mathrm{gsin\theta }}{1+{\displaystyle \frac{\mathrm{I}}{{\mathrm{MR}}^2}}}$

But,

$\mathrm{l}=\frac{2}{5}{\mathrm{mr}}^2$

$\mathrm{a}=\frac{\mathrm{gsin\theta }}{1+{\displaystyle \frac{\frac{2}{5}{\mathrm{MR}}^2}{{\mathrm{MR}}^2}}}=\frac{5}{7}\mathrm{gsin\theta }$

$\mathrm{a}=\frac{5}{7}\times 10\times 0.42=3\mathrm{m}/{\mathrm{s}}^2$

Hence the correct answer is $3m/s^2.$