A metal wire having Poisson's ratio $1 / 4$ and Young's modulus $8 \times 10^{10} \,N / m ^2$ is stretched by a force, which produces a lateral strain of $0.02 \%$ in it. The elastic potential energy stored per unit volume in wire is $[$in $J / m ^3]$
Medium
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$\frac{\text { Lateral strain }}{\text { Longitudinal strain }}=$ Poisson's ratio
$\frac{0.02 / 100}{\Delta l / l}=\frac{1}{4}$ $\left\{\begin{array}{l}Y=\text { (Young's modulus) } \\ \quad=8 \times 10^{10} \text { (given) } \\ \text { Poission's ratio }=\frac{1}{4} \text { (given) } \\ \text { Lateral strain }=0.02 \% \text { (given) }\end{array}\right.$
$\frac{\Delta l}{l}=\frac{0.08}{100}$
$\Delta U ($Elastic potential energy per unit volume $=\frac{1}{2} \times Y \times($ Longitudinal strain $\left.)\right)$
Substituting values
$\Delta U=\frac{1}{2} \times 8 \times 10^{10} \times\left(\frac{0.08}{100}\right)^2$
$\Delta U=2.56 \times 10^4 \,J / m ^3$
$\frac{0.02 / 100}{\Delta l / l}=\frac{1}{4}$ $\left\{\begin{array}{l}Y=\text { (Young's modulus) } \\ \quad=8 \times 10^{10} \text { (given) } \\ \text { Poission's ratio }=\frac{1}{4} \text { (given) } \\ \text { Lateral strain }=0.02 \% \text { (given) }\end{array}\right.$
$\frac{\Delta l}{l}=\frac{0.08}{100}$
$\Delta U ($Elastic potential energy per unit volume $=\frac{1}{2} \times Y \times($ Longitudinal strain $\left.)\right)$
Substituting values
$\Delta U=\frac{1}{2} \times 8 \times 10^{10} \times\left(\frac{0.08}{100}\right)^2$
$\Delta U=2.56 \times 10^4 \,J / m ^3$
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