A hollow sphere rolls down a 30^o incline of length 6m without slipping. The speed of centre of mass at the bottom of plane is

Given $\mathrm{\theta }={30}^{\mathrm{o}}, \mathrm{l}=6\mathrm{m}$

For hollow sphere
 $\frac{{\mathrm{k}}^2}{{\mathrm{R}}^2}=\frac{2}{3}$

Velocity at the bottom be v.

According to conservation of mechanical energy :

$\mathrm{mglsin\theta } = \frac{1}{2}{\mathrm{mv}}^2\left(1+\frac{{\mathrm{k}}^2}{{\mathrm{R}}^2}\right)$

$\mathrm{v}=\sqrt{\frac{2\mathrm{gl} \mathrm{sin\theta }}{1+\frac{{\mathrm{k}}^2}{{\mathrm{R}}^2}}}=\sqrt{\frac{2 \mathrm{x} 10 \mathrm{x} 6 \mathrm{x}{\displaystyle \frac{1}{2}}}{1+{\displaystyle \frac{2}{3}}}}=6 \mathrm{m}/\mathrm{s}$