Torques of equal magnitude are applied to a hollow cylinder and a solid sphere,

Let m and r be the respective masses of the hollow cylinder and the solid sphere.

The moment of inertia of the hollow cylinder about its standard axis, $I_{1}=m r^{2}$

The moment of inertia of the solid sphere about an axis passing through its centre, $I_{11}=\frac{2}{5} m r^{2}$

We have the relation:

$\tau=I \alpha$

Where,

$\alpha=$ Angular acceleration

$\mathrm{T}=$ Torque

$I=$ Moment of inertia

For the hollow cylinder, $\tau_{1}=I_{1} \alpha_{1}$

For the solid sphere, $\tau_{\|}=I_{\|} \alpha_{\|}$

As an equal torque is applied to both the bodies, $\tau_{1}=\tau_{2}$

$\therefore \frac{\alpha_{\mathrm{II}}}{\alpha_{\mathrm{I}}}=\frac{I_{1}}{I_{\mathrm{II}}}=\frac{m r^{2}}{\frac{2}{5} m r^{2}}=\frac{2}{5}$

$\alpha_{\mathrm{n}}>\alpha_{1}$      $\ldots(i)$

Now, using the relation:

$\omega=\omega_{0}+\alpha t$

Where,

$\omega_{0}=$ Initial angular velocity

$t=$ Time of rotation

$\omega=$ Final angular velocity

For equal $\omega_{0}$ and $t$, we have:

$\omega \propto \alpha \ldots(i i)$

From equations (i) and (ii), we can write:

$\omega_{\|}>\omega_{1}$

Hence, the angular velocity of the solid sphere will be greater than that of the hollow cylinder.