Question

A uniform metal rod of $4{\text{mm}}^2$ cross-section is heated from $0^{\circ }\text{C}$ to ${40}^{\circ }\text{C}$. The coefficient of linear expansion of the rod is $2\times {10}^{-6}/{}^{\circ }\text{C},Y={10}^{11}{\text{N/m}}^2$. The energy stored per unit volume of the rod is

• A. $1440{\text{J/m}}^3$
• B. $320{\text{J/m}}^3$
• C. $2880{\text{J/m}}^3$
• D. $5760{\text{J/m}}^3$.

Answer 1

The correct option is B $320{\text{J/m}}^3$

We know that,
Energy stored during deformation $=\frac{1}{2}\times \text{stress}\times \text{strain}\times \text{volume}$
$\Rightarrow \frac{\text{Energy}}{\text{volume}}=\frac{1}{2}\times \text{stress}\times \text{strain}$

Modifying the above formula in terms of Young's modulus of elasticity $\left(Y\right)$, we get
$\frac{\text{Energy}}{\text{volume}}=\frac{1}{2}\times Y\times {\text{(strain)}}^2.....\left(1\right)$
But from definition, we know that
Longitudinal strain $\left(\frac{\Delta L}{L}\right)=\alpha \times \Delta T.....\left(2\right)$
Using $\left(2\right)$ in $\left(1\right)$, we get
$\frac{\text{Energy}}{\text{volume}}=\frac{1}{2}\times Y\times (\alpha \Delta T{)}^2.....(3)$

Given, coefficient of linear expansion $\left(\alpha \right)=2\times {10}^{-6}{/}^{\circ }\text{C}$
Young's modulus of elasticity of material $\left(Y\right)={10}^{11}{\text{N/m}}^2$
Area of cross-section of metal rod $\left(A\right)=4\times {10}^{-6}{\text{m}}^2$
Change in temperature $\left(\Delta T\right)={40}^{\circ }\text{C}$
From $\left(3\right)$,
Energy stored per unit volume $=\frac{1}{2}\times {10}^{11}\times (2\times {10}^{-6}\times 40{)}^2$
$=320{\text{J/m}}^3$
Thus, option (b) is the correct answer.