JEE Main 2026 — Apr 2, Shift 1 — Physics solution

JEE Main2026Apr 2, Shift 1Physics

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 ______.

Your answer

Solution

Consider the cross-section located at distance $ℓ /3$ from the top. The tension there must support everything hanging below it, namely the weight $W$ at the bottom plus the lower $2 ℓ /3$ of the wire (which weighs $\frac{2 w}{3}$ ).

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

Stress $= \frac{T}{A} = \frac{W}{A} + \frac{2 w}{3 A} = \frac{W}{A} + \frac{2}{6} \cdot \frac{w}{A}$

Comparing with the given form, $γ = 6$ .

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