Question

If 'S' is stress and 'Y' is Young's modulus of a wire material, then energy stored in the wire per unit volume is?
A. $\dfrac{{{S^2}}}{{2Y}}$
B. $\dfrac{{2Y}}{{{S^2}}}$
C. $\dfrac{S}{{2Y}}$
D. $2{S^2}Y$

Answer 1

Hint: When we apply force to stretch or to compress a particular material there will be change in length of that material. That change can happen along the direction of applied force or perpendicular to the direction of applied force or length might change in three dimensions due to that application of force.

Complete answer:
When a force is applied on the given cross section area then it will produce the stress which is given by the ratio of force applied to the cross-section area. Due to the application of force the ratio of change in length to the original length is called strain. The ratio of stress and strain is called as young’s modulus which is denoted by ${\text{Y}}$
Now when we apply force in one direction then length will change along that direction and will also change along perpendicular direction. So strain will be created in both directions. The perpendicular strain is called lateral strain and the parallel strain to the direction of application of force is called longitudinal strain. The ratio of lateral to the longitudinal strain is called the poisson's ratio and denoted by $\eta $.
Here we will consider only about longitudinal strain.
The ratio of stress and strain is called young’s modulus.
Due to longitudinal stress the potential energy stored per unit volume in the wire will be
$\eqalign{
  & U = \dfrac{1}{2}\left( {{\text{stress}}} \right)\left( {{\text{strain}}} \right) \cr
  & \Rightarrow {\text{stress/strain = Y}} \cr
  & \Rightarrow U = \dfrac{1}{2}\left( {{\text{stress}}} \right)\left( {\dfrac{{{\text{stress}}}}{Y}} \right) \cr
  & \Rightarrow U = \dfrac{{{{\left( {{\text{stress}}} \right)}^2}}}{{2Y}} \cr
  & \therefore U = \dfrac{{{S^2}}}{{2Y}} \cr} $

Hence option A will be the answer.

Note:
Due to application of force there will be lateral strain too. But in this derivation we are not considering it. We can assume this as a normal spring which is elongated. When the spring is elongated energy will be stored in it. We don’t consider lateral elongation of spring over there. This is exactly like that.