Question

What amount of work is done in increasing the length of a wire through unity?
A. \[\dfrac{YL}{2A}\]
B. \[\dfrac{Y{{L}^{2}}}{2A}\]
C. \[\dfrac{YA}{2L}\]
D. \[\dfrac{YL}{A}\]

Answer 1

Hint: The amount of work done in increasing the wire would be equal to the amount of elastic potential energy developed in the wire while stretching it. This energy is given by:
\[\text{U=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ Stress }\!\!\times\!\!\text{ Strain }\!\!\times\!\!\text{ Volume}\]

Complete step by step answer:
When the wire is stretched through unity some work is done against the restoring force that is developed in the wire. This work is stored in the form of elastic potential energy in the wire which is:
\[\text{U=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ Stress }\!\!\times\!\!\text{ Strain }\!\!\times\!\!\text{ Volume}\]
When a deforming force is applied on a body restoring forces set up that try to bring it back to its original configuration. This restoring force is called stress and it is given by restoring force per unit area.
To simplify things further:
We know that Young’s Modulus is defined as the ratio of normal stress to longitudinal strain within elastic limits:
\[\text{Y=}\dfrac{\text{Stress}}{\text{Strain}}\]
Substituting in above equation we get
\[U=\dfrac{1}{2}\times Y\times {{\left( \text{Stress} \right)}^{2}}\times \text{Volume}\]
Where strain is defined as the ratio of change in configuration to original configuration:
\[\text{Strain=}\dfrac{\Delta L}{L}\]
In this case strain is:
\[\text{Strain=}\dfrac{1}{L}\]
Volume of the wire is the product of cross-sectional area and length of the wire.
\[\text{V=AL}\]
Substituting these values in the above equation we get:
\[\text{U=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ Y }\!\!\times\!\!\text{ (}\dfrac{1}{L}{{\text{)}}^{2}}\text{ }\!\!\times\!\!\text{ AL}\]
On further simplification we get,
\[\text{U=}\dfrac{YA}{\text{2L}}\]
Hence, the correct answer is option C. \[\dfrac{YA}{2L}\]

Note: Another way of doing this question would be by using the formula for work done by the elastic wire:
\[\text{W=}\dfrac{\text{1}}{\text{2}}\text{ }\!\!\times\!\!\text{ F }\!\!\times\!\!\text{ Extension}\]
Here force is taken in terms of Young’s modulus and change in length is unity,
\[F=\dfrac{YA}{L}\]
On substituting and solving we will obtain the same result as above.