Question

Two rods of lengths $l_{1}$ and $l_{2}$ are made of materials whose coefficients of linear expansion are $\alpha_{1}$ and $\alpha_{2}$ respectively. If the difference between the two lengths is independent of temperature, then

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

Answer 1

Answer: (A)

$l_{1}'=l_{1}\left(1+\alpha_{1} \Delta T\right)$
$l_{2}'=l_{2}\left(1+\alpha_{2} \Delta T\right)$
Now, $\left(l_{1}'-l_{2}'\right)=\left(l_{1}-l_{2}\right)+\left(l_{1} \alpha_{1}-l_{2} \alpha_{2}\right) \Delta T$
As $\left(l_{1}'-l_{2}'\right)=\left(l_{1}-l_{2}\right) $
$\Rightarrow $ Coefficient of $\Delta T=0$
or $l_{1} \alpha_{1}=l_{2} \alpha_{2} $
$\Rightarrow \frac{l_{1}}{l_{2}}=\frac{\alpha_{2}}{\alpha_{1}}$666666