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(B) Let $m$ and $r$ be the mass and radius of both the hollow cylinder and the solid sphere.The moment of inertia of the hollow cylinder about its sta

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The solid sphere

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(B) Let $m$ and $r$ be the mass and radius of both the hollow cylinder and the solid sphere.The moment of inertia of the hollow cylinder about its standard axis is $I_{c} = m r^{2}$.The moment of inertia of the solid sphere about an axis passing through its centre is $I_{s} = \frac{2}{5} m r^{2}$.We use the relation $	au = I \alpha$,where $	au$ is the torque,$I$ is the moment of inertia,and $\alpha$ is the angular acceleration.Since the applied torque $	au$ is the same for both,we have $\alpha = \frac{	au}{I}$.Thus,$\alpha \propto \frac{1}{I}$.Since $I_{c} = m r^{2}$ and $I_{s} = 0.4 m r^{2}$,it is clear that $I_{s} Therefore,$\alpha_{s} > \alpha_{c}$.Using the relation $\omega = \omega_{0} + \alpha t$,for an initial angular velocity $\omega_{0} = 0$,we get $\omega = \alpha t$.Since $\alpha_{s} > \alpha_{c}$,it follows that $\omega_{s} > \omega_{c}$ for any given time $t$.Hence,the solid sphere will acquire a greater angular speed.

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