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) $\alpha_1 = 4\alpha_2$
Concept. $DC$ is the altitude: $DC^2 = AC^2 - AD^2$ with $AD = \dfrac{AB}{2}$. It stays constant if $d(AC^2) = d(AD^2)$.
Step 1. Changes in length. With side $L$: $\Delta(AC) = L\alpha_2\Delta T$ and $\Delta(AD) = \dfrac{L}{2}\alpha_1\Delta T$.
Step 2. Condition.
$$2L \cdot L\alpha_2\Delta T = 2 \cdot \frac{L}{2} \cdot \frac{L}{2}\alpha_1\Delta T \;\Rightarrow\; \alpha_2 = \frac{\alpha_1}{4}$$
Step 3. Relation. $\alpha_1 = 4\alpha_2$.
Answer. $\alpha_1 = 4\alpha_2$ (Option B).
Solution by Sreeraj P, M.Sc Physics