An equilateral triangle ABC is formed by joining three rods of equal length, and D is the mid-point of AB. The coefficient of linear expansion for AB is $\alpha_1$ and for AC and BC is $\alpha_2$. The relation between $\alpha_1$ and $\alpha_2$, if the distance DC remains constant for small changes in temperature, is
Answer: (B) $\alpha_1 = 4\alpha_2$
$DC^2 = AC^2 - AD^2$, where $AD = \dfrac{AB}{2}$. With side $L$, heating by $\Delta T$ changes $AC$ by $L\alpha_2\Delta T$ and $AD$ by $\dfrac{L}{2}\alpha_1\Delta T$.
For $DC$ to stay constant, $d(AC^2) = d(AD^2)$:
$$2L(L\alpha_2\Delta T) = 2\cdot\frac{L}{2}\left(\frac{L}{2}\alpha_1\Delta T\right) \;\Rightarrow\; \alpha_2 = \frac{\alpha_1}{4} \;\Rightarrow\; \alpha_1 = 4\alpha_2$$
Solution by Sreeraj P, M.Sc Physics