Q 11-08-028JEE MainJEE Main 2026 (2 Apr, Shift 1)Easy
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 ______.
Numerical value type. Enter your answer.
Answer: 3
The tension at a point supports everything below it: the load $W$ and the part of the wire below the point.
At $l/3$ from the top, the length below is $\tfrac{2l}{3}$, whose weight is $\tfrac23w$.
$$\text{Stress}=\frac{W+\frac23w}{A}=\frac{W}{A}+\frac23\frac{w}{A}$$
Comparing with $\dfrac{W}{A}+\dfrac{2}{\gamma}\dfrac{w}{A}$ gives $\gamma=3$.
Solution by Sreeraj P, M.Sc Physics