Question

OV
(4) / no v
1. A metal wire having Poisson's ratio 1/4 and
Young's modulus 8 x 1010 N/m2 is stretched by a
force, which produces a lateral strain of 0.02% in
it. The elastic potential energy stored per unit
volutne in wire is (in J/mº]
1) 2.56 x 104
(2) 1.78 106
(3) 3.72 × 10² (4) 2.18 × 105
Which of the
....​

Answer 1

Answer:

1)2.56x10^4

Explanation:

let poisson ratio be 'u'=1/4

u=-(lateral strain)/(longitudinal strain )

lateral strain = .02%= 2x10^-4

therefore, longitudinal strain = lateral strain/poisson ratio

longitudinal strain = 8x10^-4

now elastic potential energy stored per unit volume =

(1/2)x(strain)x(stress)

we know stress = (longitudinal stress)x(young's modulus)

therefore , elastic potential energy stored per unit volume = (1/2)x(young's modulus)x(strain)^2

=(1/2)x(8x10^10)x(8x10^-4)^2

=25600

=2.56x10^4

Answer 2

Explanation:

2.56×10*2its answer

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