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. $\frac{YA\alpha L{t}^2}{2}rad/s$
• B. $\frac{YA{\alpha }^{2}L{t}^2}{2}rad/s$
• C. $\frac{YA{\alpha }^2{L}^2{t}^2}{2}rad/s$
• D. $\frac{YA\alpha Lt}{2}rad/s$

Answer 1

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

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

Strain $=\frac{\Delta L}{L}=\alpha t$ .............(1)

Also from Hooke's law, $Y=\frac{Stress}{Strain}$$⟹$ $Stress=Y\times Strain$

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

Volume of the rod, $V=AL$

Thus total workdone, $W=$ $\frac{1}{2}\times Stress\times Strain\times Volume$

$W=$ $\frac{1}{2}Y\times (Strain{)}^2\times Volume$

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