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
$Y=\frac{\text{ stress }}{\text{ strain }}=\frac{F/A}{l/L}=\frac{FL}{Al}$
$\Rightarrow F=\frac{YA}{L}L$
When wire is further increased by infinitesimal length all, then work done is
$dW=F\times dl=\frac{YA}{L}ldl$
$W=\underset{0}{\overset{l}{\int }}\frac{YA}{L}ldl=\frac{YA}{L}{\left[\frac{l^2}{2}\right]}_0^l=\frac{1}{2}YA\frac{l^2}{L}$
$=\frac{1}{2}\left(y\frac{l}{L}\right)\left(\frac{l}{L}\right)(A)$
$=\frac{1}{2}\times$ stress $\times$ strain $\times$ volume
So, elastic potential energy per unit volume is
$U=\frac{1}{2}$ strees $\times$ strain