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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
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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
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Verified Answer
The correct answer is:
$\alpha_1=4 \alpha_2$
For a change of temperature by $t$, change in length $D C$ is $\triangle D C$, then $\triangle D C^2=\triangle A C^2-\triangle A D^2$
$$
\begin{aligned}
& =l\left(1+\alpha_2 t\right)^2-\left(\frac{l}{2}\left(1+\alpha_2 t\right)\right)^2 \\
& =l^2\left(2 \alpha_2 t\right)-\frac{l^2}{4}\left(2 \alpha_1 t\right)
\end{aligned}
$$
(After neglecting terms $\alpha_2^2 t^2$ and $\alpha_1^2 t^2$, being very small)
As, $\quad \Delta D C=0 \Rightarrow l^2 2 \alpha_2 t-\frac{l^2}{4}\left(2 \alpha_1 t\right)=0$
$$
\Rightarrow \quad \alpha_1=4 \alpha_2
$$
$$
\begin{aligned}
& =l\left(1+\alpha_2 t\right)^2-\left(\frac{l}{2}\left(1+\alpha_2 t\right)\right)^2 \\
& =l^2\left(2 \alpha_2 t\right)-\frac{l^2}{4}\left(2 \alpha_1 t\right)
\end{aligned}
$$
(After neglecting terms $\alpha_2^2 t^2$ and $\alpha_1^2 t^2$, being very small)
As, $\quad \Delta D C=0 \Rightarrow l^2 2 \alpha_2 t-\frac{l^2}{4}\left(2 \alpha_1 t\right)=0$
$$
\Rightarrow \quad \alpha_1=4 \alpha_2
$$
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