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 $\left[in\, J / m ^{3}\right]$

• 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} $
$\Delta U=2.56 \times 10^{4} J / m ^{3}$