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Q.

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 isd2g2AL3NY . Then N=_______ 

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answer is 6.

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Detailed Solution

dUdV=12×stress×strain=(stress)22Y=(dxy)22Y 

∵ stress=dxgAA and dV=Adx

∫dU=12Y∫d2x2g2Adx=d2g2AL36Y   ⇒  N=6 

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