Question

As shown in the figure, an equilateral triangle $ABC$ is formed by joining three rods of equal lengths and $D$ is the midpoint of $AB$. Coefficient of linear expansion of the material of $AB$ is ${\alpha }_1$ and that of $AC$ and $BC$ is ${\alpha }_2$. If the length $DC$ remains constant for small changes in temperature, then

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

Answer 1

Answer: (B)

For a change of temperature by $t$, change in length $DC$ is $\Delta DC$, then $\Delta D{C}^2=\Delta A{C}^2-\Delta A{D}^2$
$=l{\left(1+{\alpha }_{2}t\right)}^2-{\left(\frac{l}{2}\left(1+{\alpha }_{2}t\right)\right)}^2$
$=l^2\left(2{\alpha }_{2}t\right)-\frac{l^2}{4}\left(2{\alpha }_{1}t\right)$
(After neglecting terms ${\alpha }_2^2{t}^2$ and ${\alpha }_1^2{t}^2$, being very small)
As, $\Delta DC=0$
$\Rightarrow l^{2}2{\alpha }_{2}t-\frac{l^2}{4}\left(2{\alpha }_{1}t\right)=0$
$\Rightarrow {\alpha }_1=4{\alpha }_2$