We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains,say $10 \ cm$. We can use a bimetallic strip made of brass and iron each of different length whose length (both components) would change in such a way that the difference between their lengths remains constant. If $\alpha_{\text{iron}} = 1.2 \times 10^{-5} \ K^{-1}$ and $\alpha_{\text{brass}} = 1.8 \times 10^{-5} \ K^{-1}$,what should be the length of each strip?

(N/A) Let $l_{\text{iron}}$ and $l_{\text{brass}}$ be the lengths of the iron and brass strips respectively.
According to the problem,the difference in their lengths must remain constant at $10 \ cm$ for any temperature change $\Delta T$.
So,$l_{\text{iron}} - l_{\text{brass}} = 10 \ cm \ldots (1)$
When temperature changes by $\Delta T$,the new lengths are $l_{\text{iron}}' = l_{\text{iron}}(1 + \alpha_{\text{iron}} \Delta T)$ and $l_{\text{brass}}' = l_{\text{brass}}(1 + \alpha_{\text{brass}} \Delta T)$.
For the difference to be constant,$l_{\text{iron}}' - l_{\text{brass}}' = l_{\text{iron}} - l_{\text{brass}}$.
This implies $l_{\text{iron}} \alpha_{\text{iron}} \Delta T = l_{\text{brass}} \alpha_{\text{brass}} \Delta T$.
Therefore,$l_{\text{iron}} \alpha_{\text{iron}} = l_{\text{brass}} \alpha_{\text{brass}}$.
$\frac{l_{\text{iron}}}{l_{\text{brass}}} = \frac{\alpha_{\text{brass}}}{\alpha_{\text{iron}}} = \frac{1.8 \times 10^{-5}}{1.2 \times 10^{-5}} = \frac{3}{2}$.
So,$l_{\text{iron}} = 1.5 \ l_{\text{brass}}$.
Substituting this into equation $(1)$:
$1.5 \ l_{\text{brass}} - l_{\text{brass}} = 10 \ cm$.
$0.5 \ l_{\text{brass}} = 10 \ cm \implies l_{\text{brass}} = 20 \ cm$.
Then,$l_{\text{iron}} = 1.5 \times 20 \ cm = 30 \ cm$.
Thus,the lengths should be $30 \ cm$ for iron and $20 \ cm$ for brass.