Two rods of length $L_2$ and coefficient of linear expansion ${\alpha }_2$are connected freely to a third rod of length $L_1$ of coefficient of linear expansion ${\alpha }_1$to form an isosceles triangle. The arrangement is supported on the knife edge at the midpoint of $L_1$ which is horizontal. The apex of the isosceles triangle is to remain at a constant distance from the knife edge if:

The apex of the isosceles triangle to remain at a constant distance from the knife edge $DC$should remains constant before and after heating.

Before expansion: In triangle $ADC$

$(DC{)}^2=L_2^2-{\left(\frac{L_1}{2}\right)}^2$………….(i)

After expansion

$(DC{)}^2={\left[L_2\left(1+{\alpha }_{2}t\right)\right]}^2-{\left[\frac{L_1}{2}\left(1+{\alpha }_{1}t\right)\right]}^2$…………………….(ii)

Equating (i) and (ii) we get $L_2^2-{\left(\frac{L_1}{2}\right)}^2$

$={\left[L_2\left(1+{\alpha }_{2}t\right)\right]}^2-{\left[\frac{L_1}{2}\left(1+{\alpha }_{1}t\right)\right]}^2$

$\Rightarrow L_2^2-\frac{L_1^2}{4}=L_2^2+L_2^2\times 2{\alpha }_2\times t-\frac{L_1^2}{4}-\frac{L_1^2}{4}\times 2{\alpha }_1\times t$

Neglecting higher terms

$\Rightarrow \frac{L_1^2}{4}\left(2{\alpha }_{1}t\right)=L_2^2\left(2{\alpha }_{2}t\right)\Rightarrow \frac{L_1}{L_2}=2\sqrt{\frac{{\alpha }_2}{{\alpha }_1}}$