Question

A metal rod of length L, cross-sectional area A, Young’s modulus Y and coefficient of linear expansion $\alpha$ is heated to $t^\circ C$. The work that can be performed by the rod when heated is:

A.)$\dfrac{YA\alpha L t^2}{2}$
B.)$\dfrac{YA\alpha^2 L t^2}{2}$
C.)$\dfrac{YA\alpha^2 L^2 t^2}{2}$
D.)$\dfrac{YA\alpha Lt}{2}$

Answer 1

Hint: Here the work done by the rod will be stored in the form of potential energy. So, by using the formula $Potential\; energy=Energy\; density\times volume$ of rod. The energy density of the rod will again be found out by using the formula, $Energy\; density=\dfrac{1}{2}\times Y \times \varepsilon^2$, where Y is the Young’s modulus and $\varepsilon$ is the strain generated in the rod upon expansion.

Complete step by step solution
We have been given that the rod is being heated, so the work done will be stored in the form of potential energy during the expansion due to heat.

We know that $Potential\; energy=Energy\; density\times volume$ of rod. So, as the work done will be equal to the potential energy, we need to find out the energy density of the rod and its volume from the data given to us.

We know that the Energy density is given by $\dfrac{1}{2}\times Y\times \varepsilon^2$, where Y is the Young’s modulus and $\varepsilon$ is the strain generated in the rod.

We can find the strain generated from the formula, $\varepsilon=\dfrac{\delta L}{L}=\alpha {\delta T}=\alpha t$, where $\alpha$ is the coefficient of linear expansion of rod, L is the length of rod and $\delta T$ is the change of temperature.

When we put these values in the formula of potential energy, we will get Energy density$=\dfrac{1}{2}\times Y\times (\alpha t)^2=\dfrac{Y\alpha^2 t^2}{2}$
Thus, work done$=\dfrac{Y\alpha^2 t^2}{2}\times V=\dfrac{Y\alpha^2 t^2}{2}\times AL=\dfrac{YA\alpha^2 Lt^2}{2}$

Hence option b is the correct answer.

Additional Information:
Potential energy or kinetic energy and work done are often the same thing. Potential or Kinetic energy is stored in a body by the virtue of its configuration or motion. The work done in most of the cases is generally the way of changing this energy from one body to another.
For example, if we consider a ball dropped from a certain height, the ball will have the potential energy stored in it and it will later get converted to kinetic energy when dropped due to its motion. Here, the work is being done by the gravity in changing its potential energy to kinetic energy. So, if in any case, complete transformation of energy takes place from one form to another, the work done will be the change in energy of the value of energy in its final state.

Note:
The key part to solve this question lies in the fact that the transformation of energy over time is the work done and one must be aware of how and when this concept should be applied. In this question, it has been assumed that linear expansion has taken place due to heat, that is every part of the rod has been affected equally by the heat. So, if in case the expansion is not the same for all the parts, the use of integration will come into picture.