JEE Main 2026 — Mechanical Properties of Solids Question with Solution

Let $T$ be the tension in the wire at a distance $\frac{l}{3}$ from the top.

The length of the wire below this point is $l-\frac{l}{3}=\frac{2l}{3}$.

The weight of this lower part of the wire is $\frac{w}{l}\times \frac{2l}{3}=\frac{2w}{3}$.

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:

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

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

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

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

$\gamma =3$

Answer: $3$