A steel metre scale and an aluminium metre scale are calibrated to...

Let the true length of $1 \text{ cm}$ at 20^{}^{{}^{\circ}} C be $L_0$.

At 100^{}^{{}^{\circ}} C, the change in temperature is \Delta T = 100^{}^{{}^{\circ}} C - 20^{}^{{}^{\circ}} C = 80^{}^{{}^{\circ}} C.

The length of $1 \text{ cm}$ on the aluminium scale at 100^{}^{{}^{\circ}} C is $L_{Al} = L_0 (1 + \alpha_{Al} \Delta T)$.

The length of $1 \text{ cm}$ on the steel scale at 100^{}^{{}^{\circ}} C is $L_{Steel} = L_0 (1 + \alpha_{Steel} \Delta T)$.

The ratio of aluminium-centimetre to steel-centimetre at 100^{}^{{}^{\circ}} C is:

$\dfrac{L_{Al}}{L_{Steel}} = \dfrac{1 + \alpha_{Al} \Delta T}{1 + \alpha_{Steel} \Delta T}$

Using the binomial approximation $(1+x)(1+y)^{-1} \approx 1 + x - y$ for small $x$ and $y$:

$\dfrac{L_{Al}}{L_{Steel}} \approx 1 + (\alpha_{Al} - \alpha_{Steel}) \Delta T$

Substituting the given values:

$\dfrac{L_{Al}}{L_{Steel}} = 1 + (2.3 \times 10^{-5} - 1.1 \times 10^{-5}) \times 80$

$\dfrac{L_{Al}}{L_{Steel}} = 1 + (1.2 \times 10^{-5}) \times 80$

$\dfrac{L_{Al}}{L_{Steel}} = 1 + 96 \times 10^{-5} = 1 + 0.00096 = 1.00096$