Question

A metal rod of length '$L$', cross-sectional area '$A$', Young's modulus '$Y$' and coefficient of linear expansion ${}^{\prime }{\alpha }^{\prime }$ is heated to '$t$'. 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. $\frac{YA\alpha Lt}{2}$

Answer 1

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

The potential energy will be equal to the work done here, so

Potential energy $=$ Energy density $\times$ Volume of rod

Energy density $=\frac{1}{2}\times \left(YoungModulus\right)\times {\left(strain\right)}^2$

Strain $=\frac{△l}{l}=\alpha △T$

Strain $=\frac{△l}{l}=\alpha t$

Putting values in the formula

Energy density $=\frac{1}{2}\times Y\times {\alpha }^2{t}^2=\frac{Y{\alpha }^2{t}^2}{2}$

Now, Work done $=\frac{Y{\alpha }^2{t}^2}{2}\times AL=\frac{ALY{\alpha }^2{t}^2}{2}$