(c)
$N \cos \left(90^{\circ}-\theta\right)=\frac{m v^2}{r}$
$N \sin \left(90^{\circ}-\theta\right)=m g$
$N=\frac{m g}{\cos \theta} \ldots (i)$
and $N \sin \theta=\frac{m v^2}{r} \ldots (ii)$
dividing $(ii)$ and $(i)$
$m g \tan \theta=\frac{m v^2}{r}$
$g\left(\frac{h}{r}\right)=\frac{v^2}{r} \Rightarrow v=\sqrt{g h}$