Question

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

• A.
• B.
• C.
• D.

Answer 1

The correct option is D


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 $(DC{)}^2=L_2^2-{\left(\frac{L_1}{2}\right)}^2\dots \dots \left(i\right)$
After expansion : $(DC{)}^2=\left[L_2\right(1+{\alpha }_{2}t){]}^2-{\left[\frac{L_1}{2}(1+{\alpha }_{1}t)\right]}^2\dots \dots \left(ii\right)$
Equating (i) and (ii) we get $L_2^2-{\left(\frac{L_1}{2}\right)}^2=\left[L_2\right(1+{\alpha }_{2}t){]}^2-{\left[\frac{L_1}{2}(1+{\alpha }_{1}t)\right]}^2$
$\Rightarrow$ $L_2^2-\frac{L_1^2}{4}=L_2^2+L_2^2\times 2{\alpha }_2\times t-\frac{L_1^2}{4}-\frac{L_1^2}{4}\times 2{\alpha }_1\times t$ [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$ $\frac{L_1}{L_2}=2\sqrt{\frac{{\alpha }_2}{{\alpha }_1}}$