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Q.

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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a

12MgL

b

Mgl

c

MgL

d

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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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