Question:

A uniform wire of length \(l\) of weight \(w\) is suspended from the roof with a weight \(W\) at the other end. The stress in the wire at \( \frac{l}{3} \) distance from the top is} \[ \left(\frac{W}{A} + \frac{2}{\gamma}\frac{w}{A}\right) \] where \(A\) is the cross sectional area of the wire. The value of \(\gamma\) is ____.}

Updated On: Apr 10, 2026
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Correct Answer: 3

Solution and Explanation

Concept: Stress at a point in a hanging wire equals the force acting on the portion of wire below that point divided by cross-sectional area. \[ \text{Stress} = \frac{\text{Force}}{A} \]
Step 1:Weight of the wire} Total weight = \(w\) Weight per unit length: \[ \frac{w}{l} \]
Step 2:Weight of portion below \(l/3\)} Length below that point: \[ l - \frac{l}{3} = \frac{2l}{3} \] Weight of this portion: \[ \frac{w}{l} \times \frac{2l}{3} \] \[ = \frac{2w}{3} \]
Step 3:Total force at that point} \[ F = W + \frac{2w}{3} \]
Step 4:Stress} \[ \text{Stress} = \frac{W + \frac{2w}{3}}{A} \] \[ = \frac{W}{A} + \frac{2}{3}\frac{w}{A} \] Comparing with given form: \[ \frac{W}{A} + \frac{2}{\gamma}\frac{w}{A} \] \[ \frac{2}{\gamma} = \frac{2}{3} \] \[ \gamma = 3 \] \[ \boxed{3} \]
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