A stretched wire of a material whose young's modulus Y=2 10¹¹ Nm ⁻² has poisson's ratio 0.25 . Its lateral strain _l=10⁻³ . The elastic energy density of the wire is

Step-by-step solution

Given, Young's modulus, Y=2 10¹¹ Nm ⁻² Poisson ratio, =0.25 Lateral strain, ₁=10⁻³ Elastic potential energy density is given by, PE = 1 2 Y ( strain )^2 Poisson ratio aligned & = Lateral strain longitudinal strain = ₁ longitudinal strain & = ₁ longitudinal strain & longitudinal strain = ₁ = 10⁻³ 0.25 =c 10⁻³ aligned Elastic potential energy density = 1 2 Y ( E_l )^2 aligned & = 1 2 2 10¹¹ (4 10⁻³ )^2 & =10¹¹ 16 10⁻⁶=16 10^5 Jm ⁻³ aligned