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Three rods of equal length are joined to form an equilateral triangle ABC . D is the midpoint of AB . The coefficient of linear expansion is $α 1$ for AB and $α 2$ for AC and BC . If the distance DC remains constant for small changes in temperature.

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α 1 = α 2

α 1 = 2 α 2

α 1 = 4 α 2

2 α 1 = α 2

answer is C.

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Detailed Solution

Before heating,

$\begin{aligned} {DC}^{2} = {AC}^{2} - {AD}^{2} \\ = L^{2} - \frac{L^{2}}{4} = \frac{3 L^{2}}{4} \end{aligned}$

After heating,

${DC}^{2} = {[L (1 + a_{2} Δt)]}^{2} - {[\frac{L (1 + a_{1} Δt)}{2}]}^{2}$

As DC is constant,

$\frac{3 L^{2}}{4} = {[L (1 + a_{2} Δt)]}^{2} - {[\frac{L (1 + a_{1} Δt)}{2}]}^{2}$
$\Rightarrow α_{1} = 4 α_{2}$

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