Q. 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, $i.e.,\ \alpha_{2}$ but that for $PQ$ is $\alpha_{1}.$ Then
A) $\alpha_{2} = 3\alpha_{1}$
B) $\alpha_{2} = 4\alpha_{1}$
C) $\alpha_{1} = 3\alpha_{2}$
D) $\alpha_{1} = 4\alpha_{2}$
Explanation
(d) $(OR)^{2} = (PR)^{2} - (PO)^{2} = l^{2} - \left( \frac{l}{2} \right)^{2}$
${= \left\lbrack l\left( 1 + \alpha_{2}t \right) \right\rbrack^{2} - \left\lbrack \frac{l}{2}\left( 1 + \alpha_{1}t \right) \right\rbrack^{2}
}{l^{2} - \frac{l^{2}}{4} = l^{2}\left( 1 + \alpha_{2}^{2}t^{2} + 2\alpha_{2}t \right) - \frac{l^{2}}{4}(1 + \alpha_{1}^{2}t^{2} + 2\alpha_{1}t)}$
Neglecting $\alpha_{2}^{2}t^{2}$ and $\alpha_{1}^{2}t^{2}$
$0 = l^{2}\left( 2\alpha_{2}t \right) - \frac{l^{2}}{4}\left( 2\alpha_{1}t \right) \Rightarrow 2\alpha_{2} = \frac{2\alpha_{1}}{4} \Rightarrow \alpha_{1} = 4\alpha_{2}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.