• # question_answer An inclined plane makes an angle of ${{30}^{o}}$ with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to: A)  $\frac{g}{3}$                       B)  $\frac{5g}{7}$C)  $\frac{2g}{3}$                     D)  $\frac{5g}{14}$

As the popular formula for acceleration of the rolling bodies on an inclined plane, is $a=\frac{g\,\sin \theta }{\left( 1+\frac{1}{m{{R}^{2}}} \right)}=\frac{g\sin {{30}^{o}}}{\left( 1+\frac{2}{5} \right)}=\frac{5}{14}g$