A solid cylinder of mass 50 k g and radius 0.5 m is free to rotate about horizontal axis. A massless string…

In this question we have to calculate tension

Given M = 50 kg 

R = 0.5 m

angular acceleration = 2 revolution/$s^2$

Now, Balancing the Torque

$\left(T\right)\times \left(R\right)=\left(\frac{M{R}^2}{2}\right)\left(\alpha \right)$

$T=\left(\frac{MR}{2}\right)\left(\alpha \right)=\left(\frac{50\times 0.5}{2}\right)\left(2\times 2\pi \right)=157 \mathrm{N}$

So, the tension will be 157 N