Question

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 $0^{\circ} c ,$ their difference in length $\left(l_{2}-l_{1}\right)$ will remain the same at a temperature if.

• A. $\alpha_{1} l_{1}=\alpha_{2} l_{2}$
• B. $\alpha_{1} l_{1}^{2}=\alpha_{2} l_{2}^{2}$
• C. $\alpha_{1}^{2} l_{1}=\alpha_{2}^{2} l_{2}$
• D. $\alpha_{1} l_{2}=\alpha_{2} l_{1}$

Answer 1

Answer: (A)

$L_{2}=\ell_{2}\left(1+\alpha_{2} \Delta \theta\right) \ldots$(1)
$L_{1}=\ell_{1}\left(1+\alpha_{1} \Delta \theta\right) \ldots$ (2)
Subtract eq (2) by $(1),$ we get
$\left(L_{2}-L_{1}\right)=\left(\ell_{2}-\ell_{1}\right)+\Delta \theta\left(\ell_{2} \alpha_{2}-\ell_{1} \alpha_{1}\right)$
$\Delta \theta\left(\ell_{2} \alpha_{2}-\ell_{1} \alpha_{1}\right)=0$
$\alpha_{2} \ell_{2}=\alpha_{1} \ell_{1}$