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Question: Two rods of length \(L_{2}\) and coefficient of linear expansion \(\alpha_{2}\) are connected freely...

Two rods of length L2L_{2} and coefficient of linear expansion α2\alpha_{2} are connected freely to a third rod of length L1L_{1} of coefficient of linear expansion α1\alpha_{1} to form an isosceles triangle. The arrangement is supported on the knife edge at the midpoint of L1L_{1} which is horizontal. The apex of the isosceles triangle is to remain at a constant distance from the knife edge if

A

L1L2=α2α1\frac{L_{1}}{L_{2}} = \frac{\alpha_{2}}{\alpha_{1}}

B

L1L2=α2α1\frac{L_{1}}{L_{2}} = \sqrt{\frac{\alpha_{2}}{\alpha_{1}}}

C

L1L2=2α2α1\frac{L_{1}}{L_{2}} = 2\frac{\alpha_{2}}{\alpha_{1}}

D

L1L2=2α2α1\frac{L_{1}}{L_{2}} = 2\sqrt{\frac{\alpha_{2}}{\alpha_{1}}}

Answer

L1L2=2α2α1\frac{L_{1}}{L_{2}} = 2\sqrt{\frac{\alpha_{2}}{\alpha_{1}}}

Explanation

Solution

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=L22−(L12)2(DC)^{2} = L_{2}^{2} - \left( \frac{L_{1}}{2} \right)^{2}

.....(i)

After expansion : (DC)2=[L2(1+α2t)]2−[L12(1+α1t)]2(DC)^{2} = \lbrack L_{2}(1 + \alpha_{2}t)\rbrack^{2} - \left\lbrack \frac{L_{1}}{2}(1 + \alpha_{1}t) \right\rbrack^{2} .(ii)

Equating (i) and (ii) we get

L22−(L12)2=[L2(1+α2t)]2−[L12(1+α1t)]2L_{2}^{2} - \left( \frac{L_{1}}{2} \right)^{2} = \lbrack L_{2}(1 + \alpha_{2}t)\rbrack^{2} - \left\lbrack \frac{L_{1}}{2}(1 + \alpha_{1}t) \right\rbrack^{2}⇒ L22−L124=L22+L22×2α2×t−L124−L124×2α1×tL_{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]

⇒ L124(2α1t)=L22(2α2t)\frac{L_{1}^{2}}{4}(2\alpha_{1}t) = L_{2}^{2}(2\alpha_{2}t) ⇒ L1L2=2α2α1\frac{L_{1}}{L_{2}} = 2\sqrt{\frac{\alpha_{2}}{\alpha_{1}}}