A uniform solid cylinder and a uniform solid sphere, having the...

Let the mass of both objects be $M$ and their radius be $R$.

The moment of inertia of a uniform solid cylinder about its central axis is $I_{cyl} = \dfrac{1}{2}MR^2$.

The moment of inertia of a uniform solid sphere about its central axis is $I_{sph} = \dfrac{2}{5}MR^2$.

From Newton's second law for rotation, the angular acceleration is given by $\alpha = \dfrac{\tau}{I}$.

Since both objects start from rest, their final angular velocity after time $t$ is $\omega = \alpha t = \dfrac{\tau t}{I}$.

Given that the applied torque $\tau$ and time $t$ are the same for both, the angular velocity is inversely proportional to the moment of inertia ($\omega \propto \dfrac{1}{I}$).

Therefore, the ratio of their angular velocities is:
$\dfrac{\omega_{cyl}}{\omega_{sph}} = \dfrac{I_{sph}}{I_{cyl}} = \dfrac{\dfrac{2}{5}MR^2}{\dfrac{1}{2}MR^2} = \dfrac{2}{5} \times \dfrac{2}{1} = \dfrac{4}{5}$

The ratio is $4 : 5$.

Answer: $4 : 5$