Question

If in a wire of Young’s modulus $Y,$ longitudinal strain $X$ is produced then the potential energy stored in its unit volume will be:
\[\begin{align}
  & \text{A}\text{. }0.5Y{{X}^{2}} \\
 & \text{B}\text{. }0.5{{Y}^{2}}X \\
 & \text{C}\text{. }2Y{{X}^{2}} \\
 & \text{D}\text{. }Y{{X}^{2}} \\
\end{align}\]

Answer 1

Hint: In order to stretch a wire, some work must be done against the restoring force. This work done is stored in the wire in the form of potential energy of wire. The energy stored per unit volume of the stretched wire is called its energy density. We will calculate the stress, strain in the wire and the young’s modulus of wire. Potential energy stored in the wire can be calculated using the stress and strain in the wire.

Formula used:
Potential energy stored in an elastic body, $PE=\dfrac{1}{2}\times \left( \text{stress} \right)\times \left( \text{strain} \right)$
$\text{Strain}=\dfrac{\text{change in length}}{\text{original length}}$
Young’s modulus $Y=\dfrac{\text{stress}}{\text{strain}}$

Complete step by step answer:
Stress is defined as the force applied on an object per unit area, or in mathematical terms, it is equal to force applied divided by the area upon which the force acts.
Strain can be defined as change in length of object per unit length of object. This strain is the linear strain.
So strain is: $\dfrac{\text{change in length}}{\text{original length}}$
When a stress is present in an object, it leads to strain, or vice versa.
The potential energy of an elastic body can be defined as the energy stored in an elastic body, when a load is applied on it. It is a result of stress and strain produced in the body.
Potential energy stored in a wire under a load producing both stress and strain in the wire, can be formulated as:
$PE=\dfrac{1}{2}\times \left( \text{stress} \right)\times \left( \text{strain} \right)$
Now, Young’s modulus of an object can be defined as the measure of an elastic object to withstand the changes in its physical structure under any applied load.
Mathematically, Young’s modulus is the ratio of stress and strain.
So, $Y=\dfrac{\text{stress}}{\text{strain}}$
Or, $\text{stress}=Y\times \left( \text{strain} \right)$
Now, for the given wire of Young’s modulus $Y,$ longitudinal strain $X$ is produced in it,
So the stress in the wire will be:
$\text{stress}=Y\times X=YX$
Thus the potential energy stored in the wire will be,
$PE=\dfrac{1}{2}\times \left( \text{stress} \right)\times \left( \text{strain} \right)$
Substituting $\text{stress}=Y\times X=YX$
And $\text{strain}=X$
We have,
$PE=\dfrac{1}{2}\times YX\times X$
$PE=\dfrac{1}{2}Y{{X}^{2}}$
Or, $PE=0.5Y{{X}^{2}}$
The potential energy stored in the wire is $0.5Y{{X}^{2}}$
Hence, the correct option is A.

 Note:
Students should note that the stress is the force acting a wire while it is being stretched. The change in the length of the wire due to applied force, stress, is called strain. Also, Strain is not the same as the physical pressure applied on a body. Pressure is defined as force per unit area and its unit is Pascal or newton per meter square while strain is defined as the change in the length of wire per unit original length and strain has no unit.