Question

A metal wire having Poisson's ratio $\frac{1}{4}$ Young's modulud $8\times {10}^{10}N/m^2$ is stretched force, which produces a lateral strain of 0.02 it. The elastic potential energy stored per volume in wire is$\left[inJ/m^3\right]$

• A. $2.56\times {10}^4$
• B. $1.78\times {10}^2$
• C. $3.72\times {10}^2$
• D. $3.18\times {10}^5$

Answer 1

The correct option is A $2.56\times {10}^4$

$\begin{array}{c}\text{Given poisson ratio}=1/4\text{, young modulus}=8\times {10}^{10}N/m^2\\ \Rightarrow 1/4=\frac{\text{lateral strain}}{\text{longi}\text{tudinal Strain}}\end{array}$

$\begin{array}{c}\text{given lateral strain}=0.02\%=2\times {10}^{-4}\\ \Rightarrow \text{longi}\text{tudinal strain}=8\times {10}^{-4}\end{array}$

$\begin{array}{cc}\text{PE density}& =\frac{1}{2}\times Y\times (\mathrm{Strain}{)}^2\\ & =\frac{1}{2}\times 4\times {10}^{10}\times {(8\times {10}^{-4})}^2\\ & =4\times {10}^{10}\times 64\times {10}^{-8}\\ & =256\times {10}^2\\ & =2.56\times {10}^{4}J/m^3\end{array}$