A uniform wire of length \(l\) of weight \(w\) is suspended from the roof with a weight of \(W\) at the other end. The stress in the wire at \(\dfrac{l}{3}\) distance from the top is \(\left(\dfrac{W}{A} + \dfrac{2}{\gamma}\dfrac{w}{A}\right)\), where, \(A\) is the cross sectional area of the wire. The value of \(\gamma\) is _______.

Let \(T\) be the tension in the wire at a distance \(\dfrac{l}{3}\) from the top.

The length of the wire below this point is \(l - \dfrac{l}{3} = \dfrac{2l}{3}\).

The weight of this lower part of the wire is \(\dfrac{w}{l} \times \dfrac{2l}{3} = \dfrac{2w}{3}\).

The total downward force (tension \(T\)) at this point is the sum of the weight \(W\) attached at the bottom and the weight of the lower part of the wire:

\(T = W + \dfrac{2w}{3}\)

The stress at this point is given by the tension divided by the cross-sectional area \(A\):

\(\text{Stress} = \dfrac{T}{A} = \dfrac{W + \dfrac{2w}{3}}{A} = \dfrac{W}{A} + \dfrac{2}{3}\dfrac{w}{A}\)

Comparing this with the given expression for stress, \(\dfrac{W}{A} + \dfrac{2}{\gamma}\dfrac{w}{A}\), we get:

\(\gamma = 3\)

Answer: \(3\)