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 $\frac{l}{3}$ distance from the top is $\left(\frac{\mathrm{W}}{\mathrm{A}}+\frac{2}{\gamma} \frac{\mathrm{w}}{\mathrm{A}}\right)$, where, A is the cross sectional area of the wire. The value of $\gamma$ is ____.

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Elasticity & SHM

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

\frac{l}{3}

distance from the top is

(\frac{W}{A} + \frac{2}{γ} \frac{w}{A})

, where, A is the cross sectional area of the wire. The value of

γ

is ____.

Answer: 3

The portion of the wire hanging below the point at

\frac{l}{3}

from the top has length

\frac{2 l}{3}

, so its own weight is

\frac{2}{3} w

. The tension carried by the cross-section there supports this part of the wire together with the load

W

, giving

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

. Dividing by the area

A

, the stress is

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

. Comparing with the given form

\frac{W}{A} + \frac{2}{γ} \frac{w}{A}

fixes

\frac{2}{γ} = \frac{2}{3}

, so

γ = 3

JEE Main · 2026 · 02 Apr 2026 · Shift 1 · Integer answer