Question

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 $\left[inJ/m^3\right]$

• A. $2.56\times 10^4$
• B. $1.78\times 10^6$
• C. $3.72\times 10^2$
• D. $2.18\times 10^5$

Answer 1

Answer: (A)

$\frac{\text{ Lateral strain }}{\text{ Longitudinal strain }}=$ Poisson's ratio
$\frac{0.02/100}{\Delta /1}=\frac{1}{4}$
$\frac{\Delta l}{I}=\frac{0.08}{100}$
$Y=($ Young's modulus $)$
$=8\times 10^{10}$ (given)
Poission's ratio $=\frac{1}{4}$ (given)
Lateral strain $=0.02\%$ (given)
$\Delta U$ (Elastic potential energy per unit volume $=\frac{1}{2}\times Y\times ($ Longitudinal strain))
Substituting values
$\Delta U=\frac{1}{2}\times 8\times 10^{10}\times {\left(\frac{0.08}{100}\right)}^2$
$\Delta U=2.56\times 10^{4}J/m^3$