Question

What will be energy stored in a strained wire?

• A. $ \frac{1}{2}\times$ load $\times$ extension
• B. $ \frac{1}{2}\times$ stress $\times$ strain
• C. $ \frac{1}{2}\times$ load $\times$ strain
• D. $ \frac{1}{2}\times$ load $\times$ stress

Answer 1

Answer: (B)

When a wire is stretched, work is done against the interatomic forces. This work is stored in the form of elastic potential energy. Let $Y$ is Young's modulus, $F$ is force, $A$ is area, $L$ is length of wire and $l$ is increase in length then
$Y=\frac{\text { stress }}{\text { strain }}=\frac{F / A}{l / L}=\frac{F L}{A l}$
$\Rightarrow F=\frac{Y A}{L} L$
When wire is further increased by infinitesimal length all, then work done is
$d W =F \times d l=\frac{Y A}{L} l d l $
$W =\int\limits_{0}^{l} \frac{Y A}{L} l d l=\frac{Y A}{L}\left[\frac{l^{2}}{2}\right]_{0}^{l}=\frac{1}{2} Y A \frac{l^{2}}{L} $
$=\frac{1}{2}\left(y \frac{l}{L}\right)\left(\frac{l}{L}\right)(A)$
$=\frac{1}{2} \times $ stress $ \times$ strain $ \times $ volume
So, elastic potential energy per unit volume is
$U=\frac{1}{2}$ strees $ \times$ strain