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Thermal Expansion

Two rods each...

Question

Two rods each of length $L_{2}$ and coefficient of linear expansion $α_{2}$ each are connected freely to a third rod of length $L_{1}$ and coefficient of expansion $α_{1}$ to form an isosceles triangle. The arrangement is supported on a knife-edge at the midpoint of $L_{1}$ which is horizontal. What relation must exist between $L_{1}$ and $L_{2}$ so that the apex of the isosceles triangle is to remain at a constant height from the knife edge as the temperature changes.

4 L_{2}^{2} α_{2} = L_{1}^{2} α_{1}

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4 L_{2}^{2} α_{2} = 5 L_{1}^{2} α_{1}

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2 L_{2}^{2} α_{2} = 3 L_{1}^{2} α_{1}

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L_{2}^{2} α_{2} = 3 L_{1}^{2} α_{1}

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Solution

The correct option is A $4 L_{2}^{2} α_{2} = L_{1}^{2} α_{1}$
The rods AC and BC are of equal length, the length being $L_{2} . A B = L_{1}$ . The rod AB is supported at its midpoints D so that the apex C is exactly above D and let CD = h.
For all temperature changes CD should remain a constant. Let a temperature increase of $Δ T$ take place and AB extends to $A_{1} B_{1}$ .

$A_{1} B_{1} = L_{1} + Δ L_{1} N o w A_{1} C = B_{1} C i s t h e n e w l e n g t h o f r o d s A C a n d B C A_{1} C = L_{2} + Δ L_{2} L e t u s c o n s i d e r t h e t r i a n g l e A C D, C D^{2} = h^{2} = L_{2}^{2} - {(\frac{L_{1}}{2})}^{2} \dots \dots (1) I n t h e t r i a n g l e A_{1} C D C D^{2} = h^{2} = (L_{2} + Δ L_{2})^{2} - {(\frac{L_{1} + Δ L_{1}}{2})}^{2} = L_{2}^{2} + 2 L_{2} Δ L_{2} - [{(\frac{L_{1}}{2})}^{2} + 2 \frac{L_{1}}{2} . \frac{Δ L_{1}}{2}] \dots \dots (2) O m i t t i n g t h e t e r m s (Δ L_{1})^{2} a n d (Δ L_{2})^{2} b e c a u s e o f (Δ L_{1}) a n d (Δ L_{2}) b e i n g v e r y s m a l l . F r o m e q n . (1) a n d (2) L_{2}^{2} - {(\frac{L_{1}^{2}}{2})}^{2} = L_{2}^{2} + 2 L_{2} Δ L_{2}) - {(\frac{L_{1}^{2}}{2})}^{2} - \frac{L_{1} Δ L_{1}}{2} o r 4 L_{2} Δ L_{2} = L_{1} Δ L_{1} A s Δ L_{1} = L_{1} α_{1} Δ T a n d Δ L_{2} = L_{2} α_{2} Δ T H e n c e, 4 L_{2}^{2} α_{2} = L_{1}^{2} α_{1}$

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