b) Lets suppose that the angle made by the string and the vertical line is x.

Then sin x = 5/2x50 = 1/20

so cos x = 0.9987

Since the tension in the string will balance the force along the string, The net force along the string will be zero.

The force perpendicular to the string will be

F= mgsinx - F_{e}cosx = 0.1x9.8x0.05 - 0.9987x3.6x10^{-4} = 0.0486 N