Question

The energy stored per unit volume in copper wire, which produces longitudinal strain of $0.1$% is: $(Y=1.1\times {10}^{11}N/m^2)$

• A. $11\times {10}^{3}J/m^3$
• B. $5.5\times {10}^{3}J/m^3$
• C. $5.5\times {10}^{4}J/m^3$
• D. $11\times {10}^{4}J/m^3$

Answer 1

The correct option is C $5.5\times {10}^{4}J/m^3$

When a wire is stretched, some work is done against the internal restoring forces, acting between particles of the wire. This work done appears as elastic potential energy in the wire. Elastic potential energy $\left(U\right)$ is given by
$U=\frac{1}{2}F\times \Delta l$
$=\frac{1}{2}\times \frac{F}{a}\times \frac{\Delta l}{l}\times al$ ......(i)
where $l$ is length of wire, $a$ is area of cross section of wire, $F$ is stretching force and $\Delta l$ is increase in length.
Equation (i), may be written as
$U=\frac{1}{2}\times$ stress $\times$ strain $\times$ volume of the wire
$\therefore$ Elastic potential energy per unit volume of the wire
$u=\frac{U}{al}=\frac{1}{2}\times$ stress $\times$ strain
$=\frac{1}{2}\times$ (Young's modulus $\times$ strain) $\times$ strain
$=\frac{1}{2}\times \left(Y\right)\times {\left(strain\right)}^2$
Hence, $u=\frac{1}{2}\times 1.1\times {10}^{11}\times {\left(\frac{0.1}{100}\right)}^2$
$=5.5\times {10}^{4}J/m^3$