Correct option is (1) \(\frac{mg}2\)
Let a be the acceleration of block and α be the angular acceleration of solid cylinder.
Let T be tension in string.
Forces acting on the system is shown below.
Net forces on block is mg - T = ma ...(i)
Torque about centre tangent is TR = Iα, where, using parallel axis theorem, \(I = \frac{2mR^2}2 + mR^2\) is the moment of inertia.
We know for rolling without slipping on a stationary surface, a = αR.
So, \(a = \frac{mg \times R}{mR^2 + \frac{2mR^2}{2}} \times R = \frac{mgR}{2mR} = \frac g2\)
Thus, when released, the tension in the string is \(T = mg -a= \frac {mg}2\)
