Question

A steel rod of length $l$, area of cross section $A$, Young’s modulus $E$ and linear coefficient of expansion $a$ is heated through $t^\circ {\text{C}}$. The work that can be performed by rod when heated is
A. $\left( {EAat} \right) \times \left( {lat} \right)$
B. $\dfrac{1}{2}\left( {EAat} \right) \times \left( {lat} \right)$
C. $\dfrac{1}{2}\left( {EAat} \right) \times \dfrac{1}{2}\left( {lat} \right)$
D. $2\left( {EAat} \right)\left( {lat} \right)$

Answer 1

Use the formula for work done per unit volume in terms of the stress and strain. Use the formula for the coefficient of linear thermal expansion. Use the formula for Young’s modulus for the material of an object and the formula for strain in an object. Using these equations, derive the formula for stress and strain in terms of coefficient of linear thermal expansion and substitute it in the formula for work per unit volume.

Formulae used:
The formula for work per unit volume is
$\dfrac{W}{V} = \dfrac{1}{2} \times {\text{stress}} \times {\text{strain}}$ …… (1)
Here, $W$ is the work and $V$ is the volume.
The formula for coefficient of linear thermal expansion $\alpha$ is
$\alpha = \dfrac{{\Delta L}}{{L\Delta T}}$ …… (2)
Here, $\Delta L$ is the change in length, is original length and $\Delta T$ is change in temperature of the rod.
The formula for Young’s modulus $Y$ for material of a substance is
$Y = \dfrac{{{\text{Stress}}}}{{{\text{Strain}}}}$ …… (3)
The strain is given by
${\text{Strain}} = \dfrac{{\Delta L}}{L}$ …… (4)
Here, $L$ is the original length of the rod and $\Delta L$ is the change in length of the rod.

Complete step by step answer:
We have given that the length of the steel rod is $l$, the cross sectional area of the steel rod is $A$, Young’s modulus of the steel is $E$ and the coefficient of linear thermal expansion for the steel rod is $a$. We have also given that the steel rod is heated by a temperature of $t^\circ {\text{C}}$. We have asked to calculate the work done by the steel rod when it is heated. We can use equation (1) to calculate the work performed by the rod.
Rearrange equation (1) for the work performed by the rod.
$W = \dfrac{1}{2} \times {\text{stress}} \times {\text{strain}} \times V$ …… (5)

Let us first calculate the value of stress.
Substitute $E$ for $Y$ and $\dfrac{{\Delta l}}{l}$ for ${\text{Strain}}$ in equation (3).
$E = \dfrac{{{\text{Stress}}}}{{\dfrac{{\Delta l}}{l}}}$
$\Rightarrow {\text{Stress}} = \dfrac{{E\Delta l}}{l}$
According to equation (2), substitute $at$ for $\dfrac{{\Delta l}}{l}$ in the above equation.
$\Rightarrow {\text{Stress}} = Eat$
Rewrite equation (4) for the strain as
${\text{Strain}} = \dfrac{{\Delta l}}{l}$

According to equation (2), substitute $at$ for $\dfrac{{\Delta l}}{l}$ in the above equation.
${\text{Strain}} = at$
The volume of the steel rod can be written as
$V = Al$
Substitute $Eat$ for ${\text{stress}}$, $at$ for ${\text{strain}}$ and $Al$ for $V$ in equation (5).
$W = \dfrac{1}{2} \times \left( {Eat} \right) \times \left( {at} \right) \times \left( {Al} \right)$
$\therefore W = \dfrac{1}{2} \times \left( {EAat} \right) \times \left( {lat} \right)$
Therefore, the work performed by the rod during expansion is $\dfrac{1}{2} \times \left( {EAat} \right) \times \left( {lat} \right)$.

Hence, the correct option is B.

Note: The students should be careful while deriving the relation for stress and strain acting on the rod in terms of the coefficient of linear thermal expansion of the rod. If the formula for stress and strain are not derived correctly then the final derivation for the work performed by the rod during its expansion will be incorrect.