A solid sphere of mass M rolls without slipping on an inclined plane of inclination e. What should be the
minimum coefficient of friction, so that the sphere rolls down without slipping? [NEET 2011]

This is the case of rolling on rough inclined plane. Force of friction in this case will be backward.

Hence,    $a=\frac{Mg\sin \theta -f}{M}$                 …(i)

and    $\alpha =\frac{fR}{I}$                               …(ii)

For pure rolling to take place,  $a=R\alpha$             …(iii)

Solving Eqs. (i), (ii) and (iii), we get

$f=\frac{Mg\sin \theta }{1+\frac{M{R}^2}{I}}$                              …(iv)

Further, the force of friction calculated in Eq. (iv) for pure rolling to take place should be less than or equal to the maximum friction,

$\text{ i.e. }\frac{Mg\sin \theta }{1+\frac{M{R}^2}{I}}\le \mu Mg\cos \theta$            $\left(\because I=\frac{2}{5}M{R}^2\right)$

$\Rightarrow {\mu }_{min}\le \frac{\tan \theta }{1+\frac{M{R}^2}{I}}=\frac{\tan \theta }{1+\frac{5}{2}}=\frac{2}{7}\tan \theta$