Question

An inclined plane makes an angle of ${30}^{\circ }$ with the horizontal. A solid sphere rolling down this inclined plane from rest without slipping has a linear acceleration equal to $\frac{5g}{n}$, then value of $n$ is

Answer 1

For pure rolling,
$a=\frac{g\sin \theta }{1+\frac{I}{M{R}^2}}=\frac{g\sin {30}^{\circ }}{1+\frac{2}{5}}=\frac{5}{14}g$