Three rods of equal lengths are joined to form an equilateral triangle $A B C . D$ is the mid-point of $A B$…
- $\alpha_1=2 \alpha_2$
- $\alpha_1=4 \alpha_2$
- $\alpha_1=8 \alpha_2$
- $\alpha_1=\alpha_2$
Solution

$C D^2=(A C)^2-(A D)^2=l^2-\left(\frac{l}{2}\right)^2$ After small change in temperature $C D^2=(A C)^2-(A D)^2$ $=\left[l\left(1+\alpha_2 t\right)\right]^2-\left[\frac{l}{2}\left(1+\alpha_1 t\right)\right]^2$ $\therefore \quad l^2-\frac{l^2}{4}=l^2\left[1+\alpha_2^2 t^2+2 \alpha_2 t\right]$ $-\frac{l^2}{4}\left[1+\alpha_1^2 t^2+2 \alpha_1 t\right]$ Neglecting $\alpha_2^2 t^2$ and $\alpha_1^2 t^2$, (because very small quantity) $0=l^2\left(2 \alpha_2 t\right)-\frac{l^2}{4}\left(2 \alpha_1 t\right)$ or $\quad \alpha_1=4 \alpha_2$
Asked in: AP EAMCET 2010
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