Question

When a uniform metallic wire is stretched the lateral strain produced in it $\beta .If\sigma$ and Y are the pisson 's' ration Young's modulus for wire,then elastic potential energy density of wire is

• A. $\frac{Y{\beta }^2}{2}$
• B. $\frac{Y{\beta }^2}{2{\sigma }^2}$
• C. $\frac{Y\sigma {\beta }^2}{2}$
• D. $\frac{Y{\sigma }^2}{2\beta }$

Answer 1

Answer: (A)

$\frac{Y{\beta }^2}{2}$

$\begin{array}{c}\text{Potential energy density}=\frac{1}{2}\times \text{stress}\times \text{strain}\\ \text{we know that}Y=\frac{\text{stress}}{\text{strain}}\\ \text{So,}(P\cdot E/v)=\frac{1}{2}\times Y(\text{strain}{)}^2=\frac{Y{\beta }^2}{2}\end{array}$