When a block of mass M is suspended by a long wire of length L, the length of the wire becomes (L+l). The elastic potential energy stored in the extended wire is

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12MgL

Mgl

MgL

12Mgl

answer is D.

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Detailed Solution

Stress =FA=MgA, Strain =ΔLL=L+l-LL=lL Energy stored in the wire is, U=12× Stress × Strain × Volume =12×MgA×lL×A×L=12Mgl

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