A hollow sphere rolls down a 30⁰ incline of length 6m without slipping. The speed of centre of mass at the bottom of plane is

From the question of applying conservation of energy,

${\mathrm{E}}_{\mathrm{i}}=\mathrm{mgh}=\mathrm{m}\times 10\left(6\sin {30}^{°}\right)=30\mathrm{m}$

${\mathrm{E}}_{\mathrm{f}}=\frac{1}{2}{\mathrm{mv}}^2+\frac{1}{2}{\mathrm{l\omega }}^2=\frac{1}{2}{\mathrm{mR}}^2{\mathrm{\omega }}^2+\frac{1}{2}\left(\frac{2}{3}{\mathrm{mR}}^2\right){\mathrm{\omega }}^2$

${\mathrm{E}}_{\mathrm{f}}=\frac{5}{6}{\mathrm{mR}}^2{\mathrm{\omega }}^2$

$\mathrm{Since}, {\mathrm{E}}_{\mathrm{i}}={\mathrm{E}}_{\mathrm{f}}$

${\mathrm{R}}^2{\mathrm{\omega }}^2=36$

${\mathrm{v}}_{\mathrm{cm}}=6$

Hence the correct answer is 6m/s.