Question

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 $BC$ 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

Answer: (B)

${\alpha }_1=4{\alpha }_2$

Taking length of the equilateral triangle to be $l$,

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

$l^2-\frac{l^2}{4}=l^2(1+{\alpha }_2^2{t}^2+2{\alpha }_{2}t)-\frac{l^2}{4}(1+{\alpha }_1^2{t}^2+2{\alpha }_{1}t)$

Neglecting ${\alpha }_2^2{t}^2$ and ${\alpha }_1^2{t}^2,$

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

$⟹2{\alpha }_2=2\frac{{\alpha }_1}{4}$

$⟹{\alpha }_1=4{\alpha }_2$