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 rod of length $L_1,$ which is horizontal. The apex of the isosceles triangle remains at a constant distance from the knife edge, if $\frac{L_1}{L_2}=n{\left(\frac{{\alpha }_2}{{\alpha }_1}\right)}^{\frac{1}{2}}$. Then value of $n$ is

Answer 1

The apex of the isosceles triangle to remain at a constant distance from the knife edge. Thus, DC should remain 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:
$(DC{)}^2=[L_2(1+{\alpha }_{2}t){]}^2-{\left[\frac{L_1}{2}\right(1+{\alpha }_{1}t\left)\right]}^2\left(ii\right)$
Equating Eqs. (i) and (ii), we get
$L_2^2-{\left(\frac{L_1}{2}\right)}^2=\left[L_2\right(1+{\alpha }_{2}t){]}^2-{\left[\frac{L_1}{2}\right(1+{\alpha }_{1}t\left)\right]}^2$
$L_2^2-\frac{L_1^2}{4}=L_2^2+L_2^2\times 2{\alpha }_2\times t-\frac{L_1^2}{4}-\frac{L_1^2}{4}\times 2{\alpha }_1\times t$
(As, ${\alpha }_{1}t<<1$ and ${\alpha }_{2}t<<1$, neglecting higher terms )
$\Rightarrow L_2^2\left(2{\alpha }_2\right)t=\frac{L_1^2}{4}\left(2{\alpha }_{1}t\right)\Rightarrow \frac{L_1}{L_2}=2\sqrt{\frac{{\alpha }_2}{{\alpha }_1}}$

Why this question? Key Concept: Application of binomial theorem $(1+x{)}^n=1+nx\text{when}|x|<<1$ in physics. Caution: Neglecting higher order terms is must for simplification.