Question

As shown in the figure, an equilateral triangle $A B C$ is formed by joining three rods of equal lengths and $D$ is the midpoint of $A B$. Coefficient of linear expansion of the material of $A B$ is $\alpha_{1}$ and that of $A C$ and $B C$ is $\alpha_{2}$. If the length $D C$ 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)

$\alpha_1 = 4\alpha_2$

Answer 2

For a change of temperature by $t$, change in length $D C$ is $\Delta D C$, 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, $\Delta D C=0$
$\Rightarrow l^{2} 2 \alpha_{2} t-\frac{l^{2}}{4}\left(2 \alpha_{1} t\right)=0$
$\Rightarrow \alpha_{1}=4 \alpha_{2}$