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: $3$