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Two rods of length $l_{2}$ each 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. If the arrangement (as shown) is supported by a knife edge at D such that BD=DC. Then on uniform heating which of the relations are possible?
(Use appropriate approximation)

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AB=AC always

AB= $\sqrt{B D^{2} + A D^{2}}$ always

BD=DC always

AD is constant if $α_{1} l_{1}^{2} = 4 l_{2}^{2} α_{2}$

answer is A, B, C, D.

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Detailed Solution

For AD remain constant $\sqrt{A B^{2} - B D^{2}} = \sqrt{l_{2}^{2} - {(\frac{l_{1}}{2})}^{2}}$ = Constant over the temperature change

$∴ \sqrt{l_{2}^{2} {(1 + α_{2} t)}^{2} - \frac{1}{4} l_{1}^{2} {(1 + α_{1} t)}^{2}} = \sqrt{l_{2}^{2} - {(\frac{l_{1}}{2})}^{2}}$

Solve to get $α_{1} l_{1}^{2} = 4 l_{2}^{2} α_{2}$ , So choice (d) is possible.

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