Two metre scales, one of steel and the other of aluminium, have...

Let the true length of $1\text{ cm}$ on both scales at 20^{}^{{}^{\circ}}\text{C} be $l_0$.

At 40^{}^{{}^{\circ}}\text{C}, the physical length of $1\text{ cm}$ on the aluminium scale (aluminium-centimetre) becomes:

$l_{Al} = l_0(1 + \alpha_{Al}\Delta T)$

Similarly, the physical length of $1\text{ cm}$ on the steel scale (steel-centimetre) becomes:

$l_{St} = l_0(1 + \alpha_{St}\Delta T)$

The ratio of aluminium-centimetre to steel-centimetre at 40^{}^{{}^{\circ}}\text{C} is given by:

$\dfrac{l_{Al}}{l_{St}} = \dfrac{1 + \alpha_{Al}\Delta T}{1 + \alpha_{St}\Delta T}$

Since $\alpha_{St}\Delta T \ll 1$, using the binomial approximation $(1 + x)^{-1} \approx 1 - x$:

$\dfrac{l_{Al}}{l_{St}} \approx (1 + \alpha_{Al}\Delta T)(1 - \alpha_{St}\Delta T) \approx 1 + (\alpha_{Al} - \alpha_{St})\Delta T$

Given \alpha_{Al} = 2.3 \times 10^{-5}^{}^{{}^{\circ}}\textC^{-1}, \alpha_{St} = 1.1 \times 10^{-5}^{}^{{}^{\circ}}\textC^{-1}, and \Delta T = 40^{}^{{}^{\circ}}\text{C} - 20^{}^{{}^{\circ}}\text{C} = 20^{}^{{}^{\circ}}\text{C}:

$\dfrac{l_{Al}}{l_{St}} = 1 + (2.3 \times 10^{-5} - 1.1 \times 10^{-5}) \times 20$

$\dfrac{l_{Al}}{l_{St}} = 1 + (1.2 \times 10^{-5}) \times 20$

$\dfrac{l_{Al}}{l_{St}} = 1 + 24 \times 10^{-5} = 1.00024$