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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 <img alt="W_1" src="https://entrancecorner.oncodecogs.com/gif.latex?W_1" /

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 W_1 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 \mathrm{3 L / 4} from its lower end is

Option: 1

\mathrm{W_1 / S}


Option: 2

\mathrm{\left(W_1+\frac{W}{4}\right) / S}


Option: 3

\mathrm{\left(W_1+\frac{3 W}{4}\right) / S}


Option: 4

\mathrm{\left(W_1+W\right) / S}


Answers (1)

best_answer

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 \mathrm{\frac{3 L}{4}} of the wire or \mathrm{F= W_1+\frac{3 W}{4}.}

\mathrm{ \therefore \quad \text { Stress }=\frac{\text { force }}{\text { area }}=\frac{F}{S}=\frac{W_1+\frac{3}{4} W}{S} }

Hence the correct choice is (c).

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Nehul

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