Given: Mass of the cylinder, $M = 50 \text{ kg}$ Radius of the cylinder, $R = 0.5 \text{ m}$ Angular acceleration, $\alpha = 2 \text{ rev/s}^2 = 2 \times 2\pi \text{ rad/s}^2 = 4\pi \text{ rad/s}^2$

The moment of inertia of a solid cylinder about its central axis is given by: $I = \frac{1}{2}MR^2$

The torque $\tau$ produced by the tension $T$ in the string acting at the rim of the cylinder is: $\tau = T \times R$

According to Newton's second law for rotational motion, we know that: $\tau = I\alpha$

Equating the two expressions for torque: $T \times R = \left(\frac{1}{2}MR^2\right)\alpha$ $T = \frac{1}{2}MR\alpha$

Substituting the given values into the equation: $T = \frac{1}{2} \times 50 \times 0.5 \times 4\pi$ $T = 12.5 \times 4\pi = 50\pi \text{ N}$

Taking $\pi \approx 3.14$: $T = 50 \times 3.14 = 157 \text{ N}$