Three rods of equal length l are joined to form an equilateral triangle PQR.O is the mid point of PQ. Distance OR remains same for small change in temperature. Coefficient of linear expansion for PR and RQ is same, i.e., α₂ but that for PQ is α₁. Then which of the following relation holds among the alphas?

As we know that,solid when expanded in one dimension (linear expansion) follows the following equation:

$l=l_o(1+\alpha \mathrm{\Delta }T)$

where ΔT is the change in temperature to cause the expansion.
Now applying pythagorus theorem in the given triangle based situation:

For figure,

$\Rightarrow (\mathrm{OR}{)}^2=(\mathrm{PR}{)}^2-(\mathrm{PO}{)}^2$

As given $PR=l,PO=\frac{PR}{2}$

$\Rightarrow OR=\sqrt{1^2-{\left(\frac{1}{2}\right)}^2}$

After applying the concept and formula of linear expansion,we get the extended value of OR :

$\begin{array}{l}\Rightarrow O{R}^{\mathrm{\prime }}\\ =\sqrt{{\left[l\left(1+{\alpha }_2\mathrm{t}\right)\right]}^2-{\left[\frac{l}{2}\left(1+{\alpha }_1\mathrm{t}\right)\right]}^2}\end{array}$

According to question, OR = OR'

Squaring both side, we get :

$\begin{array}{l}\Rightarrow l^2-\frac{l^2}{4}=l^2\left[1+{\mathrm{\alpha }}_2^2{\mathrm{t}}^2+2{\mathrm{\alpha }}_2\mathrm{t}\right]\\ -\frac{{\mathrm{l}}^2}{4}\left(1+{\mathrm{\alpha }}_2^2{\mathrm{t}}^2+2{\mathrm{\alpha }}_1\mathrm{t}\right)\end{array}$

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

$\begin{array}{l}\Rightarrow {\mathrm{l}}^2\left(2{\mathrm{\alpha }}_2\mathrm{t}\right)-\frac{{\mathrm{l}}^2}{4}\left(2{\mathrm{\alpha }}_1\mathrm{t}\right)=0\\ \Rightarrow {\mathrm{\alpha }}_1=4{\mathrm{\alpha }}_2\end{array}$