Question

Find the ratio of the lengths of an iron rod and an aluminium rod for which the difference in the lengths is independent of temperature. Coefficients of linear expansion of iron and aluminium are $12\times {10}^{-6}{}^{\circ }{C}^{-1}$ and $23\times {10}^{-6}{}^{\circ }{C}^{-1}$ respectively.

Answer 1

$\begin{array}{cc}\text{Iron Rod}& \text{Aluminium Rod}\\ L_{Fe}& L_{Al}\\ {\alpha }_{Fe}=12\times {10}^{-6}{/}^{\circ }C& {\alpha }_{Al}=23\times {10}^{-6}{/}^{\circ }C\end{array}$

Since the difference in length is independent of temperature the diffrernce always remains constant

$L_{Fe}^{\prime }=L_{Fe}(1+{\alpha }_{Fe}\times \Delta T)\cdots \left(1\right)$

$\underset{–}{L_{Al}^{\prime }=L_{Al}(1+{\alpha }_{Al}\times \Delta T)}\cdots \left(2\right)$

$L_{Fe}^{\prime }-L_{Al}^{\prime }=L_{Fe}-L_{Al}+L_{Fe}\times {\alpha }_{Fe}\Delta T-L_{Al}\times {\alpha }_{Al}\times \Delta T$

Given

$L_{Fe}^{\prime }-L_{Al}^{\prime }=L_{Fe}-L_{Al}$

Hence $L_{Fe}{\alpha }_{Fe}=L_{Al}{\alpha }_{Al}$

$\frac{L_{Fe}}{L_{Al}}=\frac{23}{12}$