Question

The elastic energy stored per unit volume in a stretched wire is
(A) $\dfrac{1}{2}\dfrac{(stress)}{Y}$
(B) $\dfrac{1}{2}\dfrac{{{(stress)}^{2}}}{Y}$
(C) $\dfrac{1}{2}\dfrac{{{(stress)}^{2}}}{{{Y}^{2}}}$
(D) $\dfrac{1}{2}\dfrac{(stress)}{{{Y}^{2}}}$

Answer 1

When we stretch a wire, the interatomic forces that exist within it will oppose the change that occurs in it , that is, a restoring force occurs in the string and it is applied by the interatomic molecules. The work that is done against the restoring forces is stored as its elastic potential energy.

Complete step by step solution:
When a force F is applied on a wire by causing the length to elongate from $l$ to $\Delta l$. During the process of elongation, the restoring force applied by the wire on the system will increase from 0 to F.

Hence, the average increase in the internal force applied by the wire would be $\dfrac{0+F}{2}=\dfrac{F}{2}$
Hence the work done would be $W=\dfrac{F}{2}\Delta l$ (product of average force and increased length)
Since the work done is stored as the elastic potential energy U,
$\begin{aligned} & U=\dfrac{F\Delta l}{2}=\dfrac{F\Delta l(Al)}{2(Al)}=\dfrac{1}{2}(\dfrac{F}{A})\times \dfrac{\Delta l}{l}\times Al \\\ &\Rightarrow U=\dfrac{1}{2}\times stress\times strain\times volume \\\ \end{aligned}$
Potential energy per unit volume is
$U=\dfrac{1}{2}\times stress\times strain$
Young’s modulus Y is the ratio between stress and strain. Expressing strain in terms of Y gives
$\therefore U=\dfrac{1}{2}\times \dfrac{{{(stress)}^{2}}}{Y}$

Thus, option B is the correct answer.

Note: Students must note the point that stress is a dimensionless figure whereas strain has the dimension of pressure.Also, young’s modulus can be defined as the ratio of the longitudinal stress to the longitudinal strain, within the elastic limits. Now, elastic limit is the maximum stress within which a body is able to regain its former shape and size after the removal of deforming force. All these small definitions are necessary for problem solving.