Question

Three rods of equal length are joined to form an equilateral triangle $ABC$ and $D$ is the midpoint of $AB$. The coefficient of linear expansion is ${\alpha }_1$ for $AB$ and ${\alpha }_2$ for both $AC$ and $BC$. If the distance $DC$ remains constant for small changes in temperature, then what is the relation between ${\alpha }_1$ and ${\alpha }_2$?

• A. ${\alpha }_1={\alpha }_2$
• B. ${\alpha }_1=2{\alpha }_2$
• C. ${\alpha }_1=4{\alpha }_2$
• D. ${\alpha }_1=\frac{1}{2}{\alpha }_2$

Answer 1

The correct option is C ${\alpha }_1=4{\alpha }_2$


Let $l$ be the side length of the triangle.
$AD=BD=\frac{l}{2}$ [$D$ is midpoint of $AB$]
Before heating,
$D{C}^2=A{C}^2-A{D}^2$
$=l^2-\frac{l^2}{4}=\frac{3l^2}{4}$
After heating,
$D{C}^2=\left[l\right(1+{\alpha }_2\Delta t){]}^2-{\left[\frac{l(1+{\alpha }_1\Delta t)}{2}\right]}^2$
We can neglect the higher powers of ${\alpha }_1,{\alpha }_2$ as the coefficient of thermal expansion is generally a very small number.
As $DC$ is constant,
$\frac{3l^2}{4}=l^2(1+2{\alpha }_2\Delta t)-\frac{l^2}{4}(1+2{\alpha }_1\Delta t)$
$\Rightarrow \frac{3}{4}=(1+2{\alpha }_2\Delta t)-\frac{1}{4}(1+2{\alpha }_1\Delta t)$
$\Rightarrow 0=2{\alpha }_2\Delta t-\frac{{\alpha }_1\Delta t}{2}$
$\Rightarrow {\alpha }_2=\frac{{\alpha }_1}{4}$
Hence, option $\left(c\right)$ is the correct answer.