A body of radius R and mass m is rolling horizontally without slipping with speed  $v$. It then rolls up a hill to a maximum height $h=\frac{3v^2}{4g}$ . The body might be a 

Let $l$  be the moment of inertia of the body. Then total 
$KE=\frac{1}{2}m{v}^2+\frac{1}{2}l{\omega }^2$
or  $KE=\frac{1}{2}m{v}^2+\frac{1}{2}l\left(\frac{V^2}{R^2}\right)$       $(\omega =\frac{V}{R})$     
According to energy conservation, loss in KE = gain in PE
or   $\frac{1}{2}(m+\frac{l}{R^2})V^2=mgh=mg\left(\frac{2V^2}{4g}\right)$
Solving this, we get  $l=\frac{1}{2}m{R}^2$
i.e the solid body is a disc.