Question

An equilateral triangle $ABC$ is formed by joining three rods of equal lengths 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 distance $DC$ remains constant for small changes in temperature is

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

Answer 1

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

Taking length of the equilateral triangle to be I,

$D{C}^2=A{C}^2-A{D}^2$

$=1^2-\frac{1^2}{2^2}$

$=[{\left(1\right(1+\alpha 2t\left)\right]}^2-[\frac{1}{2}(1+{\alpha 1}^2{t}^2+2{\alpha }_{1}t)$

$=1^2-\frac{1^2}{4}=1^2(1+{\alpha }^{2}2t^2+2\alpha 1t)$

Neglecting ${\alpha }^{2}2t^2$and ${\alpha }^{2}1t^2,$

$0=I^2\left(2\alpha 2t\right)-\frac{I^2}{4}\left(2\alpha 1t\right)$

$\Rightarrow 2\alpha 2=2\frac{\alpha 1}{4}$

$\Rightarrow \alpha 1=4\alpha 2$

Hence, the correct option is (B) is $\alpha 1=4\alpha 2$