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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 Yβ22ν2
β= 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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