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Question

# When a uniform metallic wire is stretched the lateral strain produced in it is β. If ν and Y are the Poisson's ratio and Young's modulus for the wire, then elastic potential energy density of wire is

A
Yβ22
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B
Yβ22ν2
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C
Yνβ22
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D
Yν22β
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Solution

## The correct option is B Yβ22ν2β= Strain (lateral) ν= Poisson's ratio Y= Young's modulus Elastic potential energy density U=12×Y×(longitudinal strain )2 ..... (1) Also Poisson's ratio=Lateral strainLongitudinal strain ⇒βLongitudinal strain=ν ⇒Longitudinal strain=βν Substituting the value in equation (1), Elastic P.E density =12×Y×(βν)2=Yβ22ν2

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