Question

A uniform metal rod fixed at its ends of $2m{m^2}$ cross-section is cooled from $40^\circ C$ to $20^\circ C$. The coefficient of the linear expansion of the rod is $12 \times {10^{ - 6}}$ per degree Celsius and its Young’s modulus of elasticity is ${10^{11}}N/{m^2}$. The energy stored per unit volume of the rod is
$
  {\text{A}}{\text{. 2880 }}J/{m^3} \\
  {\text{B}}{\text{. 1500 }}J/{m^3} \\
  {\text{C}}{\text{. 5760 }}J/{m^3} \\
  {\text{D}}{\text{. 1440 }}J/{m^3} \\
$

Answer 1

Hint: The rod contracts when it is cooled. This exerts a stress and strain on the rod and their ratio is given by its Young’s modulus. The strain on the rod is equal to the ratio of the change in length of the rod due to contraction and the original length of the rod. The energy stored per unit volume equals half of the product of stress and strain on the rod.

Formula used:
The change in the length of the rod upon heating is given as
\[\Delta l = \alpha l\Delta T\] …(1)
Here $\Delta l$ is the change in the length of the rod, l is the original length of the rod, $\alpha $ is the coefficient of linear expansion of the material of the rod while $\Delta T$ signifies the difference between the final temperature and the initial temperature.
The strain is defined as
Strain $ = \dfrac{{\Delta l}}{l}$
While the Young’s modulus of a material is defined as
$Y = \dfrac{{stress}}{{strain}}$
The energy stored per unit volume of a material subjected to stress and strain is given as
$E = \dfrac{1}{2} \times stress \times strain$

Complete step-by-step answer:
We are given that a uniform metal rod fixed at its ends of $2m{m^2}$ cross-section is cooled from $40^\circ C$ to $20^\circ C$. Therefore, we have
$\Delta T = 20 - 40 = - 20^\circ C$
The coefficient of the linear expansion of the rod is $\alpha = 12 \times {10^{ - 6}}$ per degree Celsius and its Young’s modulus of elasticity is $Y = {10^{11}}N/{m^2}$.
The strain on the rod is given as $\dfrac{{\Delta l}}{l} = \alpha \Delta T$ (using equation 1)
We know that $Y = \dfrac{{stress}}{{strain}} \Rightarrow stress = Y \times strain = Y\alpha \Delta T$
The energy stored per unit volume of the rod is given as
$E = \dfrac{1}{2} \times stress \times strain$
Using the known expressions we get that
$E = \dfrac{1}{2} \times Y \times {\left( {\alpha \Delta T} \right)^2}$
Substituting the known values, we get
$
  E = \dfrac{1}{2} \times {10^{11}} \times {\left( {12 \times {{10}^{ - 6}} \times 20} \right)^2} \\
   = 2880J/{m^2} \\
$

So, the correct answer is “Option A”.

Note: It should be noted that in this question we do not need to calculate the value of stress explicitly. The Young’s modulus of elasticity gives the relation between stress and strain and we substitute the value of stress in terms of strain and the Young’s modulus of elasticity.