A stretched wire of a material whose young's modulus \(Y=2 \times 10^{11} \mathrm{Nm}^{-2}\) has poisson's ratio 0.25 . Its lateral strain \(\varepsilon_l=10^{-3}\). The elastic energy density of the wire is

Given, Young's modulus, \(Y=2 \times 10^{11} \mathrm{Nm}^{-2}\)
Poisson ratio, \(\sigma=0.25\)
Lateral strain, \(\varepsilon_1=10^{-3}\)
Elastic potential energy density is given by, \(\mathrm{PE}\) \(=\frac{1}{2} \times Y \times(\text { strain })^2\)
Poisson ratio
\(\begin{aligned} & =\frac{\text { Lateral strain }}{\text { longitudinal strain }}=\frac{\varepsilon_1}{\text { longitudinal strain }} \\ & \Rightarrow \sigma=\frac{\varepsilon_1}{\text { longitudinal strain }} \\ & \Rightarrow \text { longitudinal strain }=\frac{\varepsilon_1}{\sigma}=\frac{10^{-3}}{0.25}=c \times 10^{-3}\end{aligned}\)
Elastic potential energy density \(=\frac{1}{2} \times Y \times\left(\frac{E_l}{\sigma}\right)^2\)
\(\begin{aligned} & =\frac{1}{2} \times 2 \times 10^{11} \times\left(4 \times 10^{-3}\right)^2 \\ & =10^{11} \times 16 \times 10^{-6}=16 \times 10^5 \mathrm{Jm}^{-3}\end{aligned}\)