JEE Main 2026 — 2 April, Morning Shift — Question 18

The tension at a distance of $\frac{l}{3}$ from the top of the hanging wire consists of the load $W$ at the bottom plus the weight of the portion of the wire hanging below that point.

The length of the wire below this point is $\frac{2l}{3}$. Given that the total weight of the wire is $w$ and its length is $l$, the linear weight density is $\frac{w}{l}$.

Weight of the wire below the point = $\left(\frac{2l}{3}\right) \left(\frac{w}{l}\right) = \frac{2w}{3}$

Total tension at this point ($T$) = $W + \frac{2w}{3}$

The stress at this cross-section of area $A$ is:

$$\text{Stress} = \frac{T}{A} = \frac{W + \frac{2w}{3}}{A} = \frac{W}{A} + \frac{2w}{3A}$$

Comparing this expression with the given form $\frac{W}{A} + \frac{2w}{\gamma A}$, we equate the denominators of the second term:

$$\gamma = 3$$