Let M and R be mass and radius of the hollow cylinder and the solid sphere, then, Moment of inertia of the hollow cylinder about its axis of symmetry,$I_{1}=M R^{2}$Moment of inertia of the solid sphere about its axis through its centre $I_{2}= \frac{2}{5} M R^{2}$Torque $\tau = I \alpha$where $\alpha$ is angular acceleration.$I_{1} \alpha_{1}=I_{2} \alpha_{2}$$\frac{\alpha_{2}}{\alpha_{1}} = \frac{I_{1}}{I_{2}} = \frac{M R^{2}}{\frac{2}{5} M R^{2}} = \frac{5}{2}$$\alpha_{2}=2.5 \alpha_{1}$ or $\alpha_{2} > \alpha_{1}$from $\omega = \omega_{0} + \alpha t,$ we find that for given $\omega_{0}$ and $t, \omega_{2} > \omega_{1},$ angular speed of solid sphere will be greater than the angular speed of hollow cylinder.