Question

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

• A. $\frac{YA{\alpha }^{}{L}^{}{t}^2}{2}$
• B. $\frac{YA{\alpha }^2{L}^{}{t}^2}{2}$
• C. $\frac{YA{\alpha }^2{L}^2{t}^2}{2}$
• D. $\frac{YA{\alpha }^{}{L}^{}{t}^{}}{2}$$

Answer 1

The correct option is B $\frac{YA{\alpha }^2{L}^{}{t}^2}{2}$

Using $L^{\prime }=L(1+\alpha t)$, we get $\Delta L=L^{\prime }-L=L\alpha t$

Strain $=\frac{\Delta L}{L}=\alpha t$

Also from Hooke''s law ,$Y=\frac{stress}{strain}\Rightarrow stress=Y\times strain$

Work done per unit volume of rod, $W^{\prime }=\frac{1}{2}\times stress\times strain$

Volume of rod, $V=AL$

Thus,

Total work done $W=\frac{1}{2}\times stress\times strain\times volume$

$W=\frac{1}{2}Y\times (strain{)}^2\times volume$

$W=\frac{1}{2}Y\times (\alpha t{)}^2\times AL=\frac{YA{\alpha }^{2}L{t}^2}{2}$