Question

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

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

Answer 1

The correct option is B $\frac{Y{\beta }^2}{2{\nu }^2}$

$\beta =$ Strain (lateral)
$\nu =$ Poisson's ratio
$Y=$ Young's modulus
Elastic potential energy density
$U=\frac{1}{2}\times Y\times {\text{(longitudinal strain )}}^2$ ..... (1)
Also $\text{Poisson's ratio}=\frac{\text{Lateral strain}}{\text{Longitudinal strain}}$
$\Rightarrow \frac{\beta }{\text{Longitudinal strain}}=\nu$
$\Rightarrow \text{Longitudinal strain}=\frac{\beta }{\nu }$
Substituting the value in equation (1),
Elastic P.E density $=\frac{1}{2}\times Y\times {\left(\frac{\beta }{\nu }\right)}^2=\frac{Y{\beta }^2}{2{\nu }^2}$