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A uniform metal rope of length L and cross sectional area A is suspended from ceiling vertically. D is density and Y is Young’s modulus of material of the wire respectively. Due to its own weight rope sags such the strain energy stored is $\frac{d^{2} g^{2} A L^{3}}{N Y}$ . Then $N =_______$

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answer is 6.

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Detailed Solution

$\frac{d U}{d V} = \frac{1}{2} \times s t r e s s \times s t r a i n = \frac{{(s t r e s s)}^{2}}{2 Y} = \frac{{(d x y)}^{2}}{2 Y}$

$∵ stress = \frac{dxgA}{A} and dV = Adx$

$\int d U = \frac{1}{2 Y} \int d^{2} x^{2} g^{2} A d x = \frac{d^{2} g^{2} A L^{3}}{6 Y} \Rightarrow N = 6$

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