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Three metal rods of the same length and area of cross-section form an equilateral triangle as shown in Fig. D is the mid-point of side BC. If AD is independent for small change in temperature, then ( $α_{1}$ is the coefficient of linear expansion for rod BC and $α_{2}$ for rods AB and AC) :

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$α_{1} = 4 α_{2}$

$α_{1} = 2 α_{2}$

$α_{1} = \frac{α_{2}}{2}$

$α_{2} = α_{1}$

answer is D.

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Detailed Solution

${(A D)}^{2} = {(A B)}^{2} - {(B D)}^{2}$ ${(A D)}^{2} = l^{2} - \frac{l^{2}}{4}$ $= {[l (1 + α_{2} t)]}^{2} - {[\frac{l}{2} (1 + α_{1} t)]}^{2}$

$l^{2} - \frac{l^{2}}{4} = {[l (1 + α_{2} t)]}^{2} - {[\frac{l}{2} (1 + α_{1} t)]}^{2}$ $l^{2} - \frac{l^{2}}{4} = l^{2} (1 + α_{2}^{2} t^{2} + 2 α_{2} t) - \frac{l^{2}}{4} (1 + α_{1}^{2} t^{2} + 2 α_{1} t)$

Neglecting higher order terms, on simplification we get $α_{1} = 4 α_{2}$

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