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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
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$16 \times 10^5 \mathrm{Jm}^{-3}$
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 $=\frac{1}{2} \times Y \times\left(\frac{E_l}{\sigma}\right)^2$
$\begin{aligned} & =\frac{1}{2} \times 2 \times 10^{11} \times\left(4 \times 10^{-3}\right)^2 \\ & =10^{11} \times 16 \times 10^{-6}=16 \times 10^5 \mathrm{Jm}^{-3}\end{aligned}$
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 $=\frac{1}{2} \times Y \times\left(\frac{E_l}{\sigma}\right)^2$
$\begin{aligned} & =\frac{1}{2} \times 2 \times 10^{11} \times\left(4 \times 10^{-3}\right)^2 \\ & =10^{11} \times 16 \times 10^{-6}=16 \times 10^5 \mathrm{Jm}^{-3}\end{aligned}$
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