Question

Energy per unit volume in a stretched string is
A. half of load × strain
B. load × strain
C. stress × strain
D. half of stress × strain

Answer 1

Hint: When a wire is stretched , the interatomic forces of rod will oppose the change i.e a restoring force will be applied by the interatomic molecules. So the work has to be done against this restoring forces. This work done will be stored in wire as elastic potential energy.

Formulas used:
Work Done, $W=\text{ Average force }\times \text{ increase in length=}\dfrac{F}{2}\times \Delta l$
 Energy per unit volume in a stretched string is
$u=\dfrac{1}{2}\text{Stress }\times \text{strain}$

Complete step by step answer:
Suppose a force $F$ is applied on a wire of length $l$ due to which the length of wire increases by $\Delta l$. Before this force is applied the internal restoring force in the wire is zero. When the length increases by $\Delta l$, the internal restoring force will be increased from $0\text{ to }F\left( =\text{ applied force} \right)$.
So the average internal force for an increase in length $\Delta l$ of wire is $\dfrac{0+F}{2}=\dfrac{F}{2}$
So work done on the wire is
$W=\text{ Average force }\times \text{ increase in length=}\dfrac{F}{2}\times \Delta l$
This work done is stored is stored as elastic potential energy $U$ in the wire , so
$\begin{align}
  & U=\dfrac{F}{2}\times \Delta l \\
 & \Rightarrow U=\dfrac{1}{2}\times \text{ Stretching force }\times \text{ increase in length} \\
\end{align}$
Let $A$ be the area of crosssection of the wire, then
$\begin{align}
  & U=\dfrac{1}{2}\dfrac{F}{A}\times \dfrac{\Delta l}{l}\times Al \\
 & \Rightarrow U=\dfrac{1}{2}\text{Stress }\times \text{strain }\times \text{Volume of the wire} \\
\end{align}$
So elastic potential energy per unit volume of the wire is given by
$u=\dfrac{U}{\text{Volume}}$
So,$u=\dfrac{1}{2}\text{Stress }\times \text{strain}$
So the correct answer is option D. half of stress × strain

Additional Information:
Energy per unit volume in a stretched wire or string is given by
$u=\dfrac{1}{2}\text{Stress }\times \text{strain}$
But $\text{stress = Young }\!\!'\!\!\text{ s modulus }\times \text{ strain}$
So elastic potential energy per unit volume is given by
$u=\dfrac{1}{2}\text{Young }\!\!'\!\!\text{ s modulus }\times \text{strai}{{\text{n}}^{2}}$
Stress: The internal restoring force setup per unit area of cross-section of a deformed body is called stress. This restoring force is opposite to the applied deforming force.so
$\text{Stress=}\dfrac{\text{Applied Force}}{\text{Area}}$
Strain: The ratio of change in dimension produced per original dimension is called strain. i.e.
$\text{Strain=}\dfrac{\text{Change in dimension}}{\text{Original dimension}}$
Elastic limit: the maximum stress within which a body regains its original size and shape after removal of deforming force is called elastic limit.
Young;s modulus: Within elastic limit the ratio of longitudinal stress to longitudinal strain is called Young’s modulus
$Y=\dfrac{\text{Longitudinal Stress}}{\text{Longitudinal Strain}}=\dfrac{\left( \dfrac{F}{A} \right)}{\left( \dfrac{\Delta l}{l} \right)}=\left( \dfrac{F}{A} \right)\left( \dfrac{l}{\Delta l} \right)$

Note:
Note that the Strain has no dimension But the stress has the dimension of pressure. Elastic limit is the limit before any material is elastic. Beyond the elastic limit the material loses its elastic property and becomes plastic. At that point the body will not return to its original shape or original configuration.