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 AFΔL2AL 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