A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about the horizontal axis. A massless string is wound round the cylinder with one end attached to it and other hanging freely. Tension in the string required to produce an angular acceleration of 2 rev ${\mathrm{s}}^{-2}$ is

                                                                                                 $\mathrm{Tr} = \mathrm{I\alpha }$

Applying torque equation about center of cylinder

$\mathrm{\tau } = \mathrm{I\alpha }$

$\alpha$ is the angular acceleration of cylinder and it is given

$\mathrm{\alpha } = 2 \mathrm{revolution}/{\mathrm{s}}^2 = 2\times 2\mathrm{\pi } = 4\mathrm{\pi } \mathrm{rad}/{\mathrm{s}}^2$

$\mathrm{T} = \frac{\mathrm{I\alpha }}{\mathrm{r}} = \frac{{\mathrm{mr}}^2}{2}\times \frac{\mathrm{\alpha }}{\mathrm{r}} = \frac{\mathrm{mr\alpha }}{2}$

 = $\frac{50\times 0.5\times 4\mathrm{\pi }}{2}\mathrm{N} = 157 \mathrm{N}$