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An isosceles triangle is formed with a thin rod of length $l_{1}$ and coefficient of linear expansion $α_{1}$ , as the base and two thin rods each of length $l_{2}$ and coefficient of linear expansion $α_{2}$ as the two sides. The distance between the apex and the midpoint of the base remain unchanged as the temperature is varied. The ratio of lengths $\frac{l_{1}}{l_{2}}$ is

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$\sqrt{\frac{α_{2}}{α_{1}}}$

$2 \sqrt{\frac{α_{1}}{α_{2}}}$

$2 \sqrt{\frac{α_{2}}{α_{1}}}$

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

answer is C.

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Detailed Solution

The distance between the apex and the midpoint of the base, using Pythagoras theorem $l = \sqrt{{(l_{2})}^{2} - {(\frac{l_{1}}{2})}^{2}}$

Or $l^{2} = {(l_{2})}^{2} - {(\frac{l_{1}}{2})}^{2}$

Differentiating (i) w.r.t temperature

$0 = 2 l_{2} \times \frac{d l_{2}}{d T} - 2 (\frac{l_{1}}{2}) . \frac{1}{2} \times \frac{d l_{1}}{d T}$

$\frac{l_{1}}{2} \times l_{1} α_{1} = 2 l_{2} \times l_{2} α_{2}$

${(\frac{l_{1}}{l_{2}})}^{2} = 4 \frac{α_{2}}{α_{1}} \Rightarrow \frac{l_{1}}{l_{2}} = 2 \sqrt{\frac{α_{2}}{α_{1}}}$

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