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 × 10^{–6} °C^{–1} and 23 × 10^{–6} °C^{–1} respectively.

#### Solution

Let the original length of iron rod be L_{Fe} and L^{'}^{}^{}_{Fe }be its length when temperature is increased by ΔT.

Let the original length of aluminium rod be L_{Al} and L^{'}^{}^{}_{Al }be its length when temperature is increased by ΔT.

Coefficient of linear expansion of iron,

\[\alpha_{Fe}\] = 12 × 10^{–6} °C

\[-\] 1

Coefficient of linear expansion of aluminium, α_{Al} = 23 × 10^{–6} °C^{}

\[-\] 1

Since the difference in length is independent of temperature, the difference is always constant.

\[L '_{Fe} = L_{Fe} \left( 1 + \alpha_{Fe} \times ∆ T \right)\]

\[and\] \[ L '_{Al} = L_{Al} \left( 1 + \alpha_{Al} \times ∆ T \right)\]

\[ \Rightarrow \begin{array}\[L '_{Fe} - L '_{Al} = L_{Fe} - L_{Al} + L_{Fe} \times \alpha_{Fe} ∆ T - L_{Al} \times \alpha_{Al} \times ∆ T & - (1)\end{array}\]

\[Given: \]

\[L '_{Fe} - L '_{Al} = L_{Fe} - L_{Al} \]

\[Hence, L_{Fe} \alpha_{Fe} = L_{Al} \alpha_{Al} [using (1)]\]

\[ \Rightarrow \frac{L_{Fe}}{L_{Al}} = \frac{23}{12}\]

The ratio of the lengths of the iron to the aluminium rod is 23:12.