First slide

Thermal expansion

Question

Two rods of length $L_{2}$ and coefficient of linear expansion $α_{2}$ are connected freely to a third rod of length $L_{1}$ of coefficient of linear expansion $α_{1}$ 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

Moderate

A

$\frac{L_{1}}{L_{2}} = \frac{α_{2}}{α_{1}}$
B

$\frac{L_{1}}{L_{2}} = \sqrt{\frac{α_{2}}{α_{1}}}$
C

$\frac{L_{1}}{L_{2}} = 2 \frac{α_{2}}{α_{1}}$
D

$\frac{L_{1}}{L_{2}} = 2 \sqrt{\frac{α_{2}}{α_{1}}}$

Solution

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
${(DC)}^{2} = L_{2}^{2} - {(\frac{L_{1}}{2})}^{2} (i)$
After expansion:
${(DC)}^{2} = {[L_{2} (1 + α_{2} t)]}^{2} - {[\frac{L_{1}}{2} (1 + α_{1} t)]}^{2}$
Equating Eqs. (i) and (ii), we get
$L_{2}^{2} - {(\frac{L_{1}}{2})}^{2} = {[L_{2} (1 + α_{2} t)]}^{2} - {[\frac{L_{1}}{2} (1 + α_{1} t)]}^{2} L_{2}^{2} - \frac{L_{1}^{2}}{4} = L_{2}^{2} + L_{2}^{2} \times 2 α_{2} \times t - \frac{L_{1}^{2}}{4} - \frac{L_{1}^{2}}{4} \times 2 α_{1} \times t$
(Neglecting higher terms)
$\Rightarrow \frac{L_{1}^{2}}{4} (2 α_{1} t) = L_{2}^{2} (2 α_{2} t) \Rightarrow \frac{L_{1}}{L_{2}} = 2 \sqrt{\frac{α_{2}}{α_{1}}}$

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