Q. The coefficient of linear expansion of brass and steel are $\alpha_{1}$ and $\alpha_{2}$. If we take a brass rod of length $l_{1}$ and steel rod of length $l_{2}$ at $0℃$, their difference in length $(l_{2} - l_{1})$ will remain the same at a temperature if
A) $\alpha_{1}l_{2} = \alpha_{2}l_{1}$
B) $\alpha_{1}l_{2}^{2} = \alpha_{2}l_{1}^{2}$
C) $\alpha_{1}^{2}l_{1} = \alpha_{2}^{2}l_{2}$
D) $\alpha_{1}l_{1} = \alpha_{2}l_{2}$
Explanation
(d) $L_{2} = l_{2}(1 + \alpha_{2}\mathrm{\Delta}\theta)$ and $L_{1} = l_{1}(1 + \alpha_{1}\mathrm{\Delta}\theta)$
$\Rightarrow \left( L_{2} - L_{1} \right) = \left( l_{2} - l_{1} \right) + \mathrm{\Delta}\theta(l_{2}\alpha_{2} - l_{1}\alpha_{1})$
Now $\left( L_{2} - L_{1} \right) = \left( l_{2} - l_{1} \right)$ so, $l_{2}\alpha_{2} - l_{1}\alpha_{1} = 0$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.