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One end of a uniform wire of length L and of weight W is attached rigidly to a point in the ceiling and a weight W₁ is suspended from its lower end. If S is the area of cross-section of the wire; the stress in the wire at a height 3 L/4 from its lower end is

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W₁/S

$\frac{(W_{1} + \frac{W}{4})}{S}$

$\frac{(W_{1} + \frac{3 W}{4})}{S}$

$\frac{(W_{1} + W)}{S}$

answer is C.

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Detailed Solution

Since the wire is uniform, i.e. its mass per unit length is constant over its entire length L, the total downward weight = the weight of the suspended mass + weight of length $\frac{3 L}{4}$ of the wire or F = $W_{1} + \frac{3 W}{4}$ .

$∴ S t r e s s = \frac{f o r c e}{a r e a} = \frac{F}{S} = \frac{W_{1} + \frac{3}{4} W}{S}$

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