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Question

If force F is applied for stretching a wire of length L by ΔL, then elastic potential energy stored per unit volume is
(A is cross sectional area of wire)

A
FΔL2AL
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B
FA2L
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C
FL2A
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D
FL2
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Solution

The correct option is A FΔL2AL
Elastic potential energy stored per unit volume is given by,
ΔU=12×stress×strain ...(i)
where,
Stress=FA, Strain=ΔLL
F= Force applied to the cross-sectional area A of the wire
L=Original Length of the wire
ΔL=Elongation in length of wire due to force F
Hence from Eq.(i)
ΔU=12×(FA)×(ΔLL)
ΔU=FΔL2AL

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