Question

A metal wire having poisson ratio $1/4$ and young's modulus $4\times {10}^{10}{\text{N/m}}^2$ is stretched by a force, which produces a lateral strain of $0.01\%$ in it. The elastic potential energy stored per unit volume in wire is (in ${\text{J/m}}^3$)

• A. $2500$
• B. $3200$
• C. $6400$
• D. $1250$

Answer 1

The correct option is B $3200$

We know,

$\frac{\text{Lateral strain}}{\text{Longitudinal strain}}=\text{Poisson ratio}$

$\frac{0.01/100}{\text{Longitudinal strain}}=\frac{1}{4}$

$\therefore$ Longitudinal strain $=4\times {10}^{-4}$
Given, Young's modulus $\left(Y\right)=4\times {10}^{10}{\text{N/m}}^2$

Elastic potential energy per unit volume,
$\Delta U=\frac{1}{2}\times Y\times {\text{(Longitudinal strain)}}^2$
$\Delta U=\frac{1}{2}\times Y\times {\text{(Longitudinal strain)}}^2$
$=\frac{1}{2}\times 4\times {10}^{10}\times (4\times {10}^{-4}{)}^2$
$\Delta U=3200{\text{J/m}}^3$