A solid cylinder of mass 50 kg and radius 0.5 m is free to rotate about a horizontal axis. A massless string is wound round the cylinder with one end attached to it and the other hanging freely. Tension in the string required to produce an angular acceleration of 2 revolutions/s² is:

The moment of inertia of a solid cylinder about its axis is given by:

I = (1/2)MR²

where M is the mass and R is the radius of the cylinder.

Given:
M = 50 kg
R = 0.5 m
Angular acceleration (α) = 2 revolutions/s² = 2 * 2π rad/s² = 4π rad/s²

The torque (τ) acting on the cylinder is given by:

τ = Iα = (1/2)MR²α

The tension (T) in the string provides the torque. The torque is also given by:

τ = TR

Therefore, we can equate the two expressions for torque:

TR = (1/2)MR²α

T = (1/2)MRα

Substituting the given values:

T = (1/2) * 50 kg * 0.5 m * 4π rad/s²

T = 50π N

T ≈ 78.54 N

Therefore, the tension in the string required to produce an angular acceleration of 2 revolutions/s² is approximately 78.5 N.