A unit scale is to be prepared whose length does not change with temperature and remains $$20 \text{ cm}$$, using a bimetallic strip made of brass and iron each of different length. The length of both components would change in such a way that difference between their lengths remains constant. If length of brass is $$40 \text{ cm}$$ and length of iron will be ______ cm.
$$\alpha_{\text{iron}} = 1.2 \times 10^{-5} \text{ K}^{-1}$$ and $$\alpha_{\text{brass}} = 1.8 \times 10^{-5} \text{ K}^{-1}$$.

For the bimetallic strip to maintain a constant length difference (20 cm), both metals must expand by the same amount when the temperature changes.

The change in length for each material is:

$$\Delta L = L \alpha \Delta T$$

For the difference to remain constant, the expansions must be equal:

$$\Delta L_{\text{brass}} = \Delta L_{\text{iron}}$$

$$L_{\text{brass}} \times \alpha_{\text{brass}} \times \Delta T = L_{\text{iron}} \times \alpha_{\text{iron}} \times \Delta T$$

Cancelling $$\Delta T$$ from both sides:

$$L_{\text{brass}} \times \alpha_{\text{brass}} = L_{\text{iron}} \times \alpha_{\text{iron}}$$

$$40 \times 1.8 \times 10^{-5} = L_{\text{iron}} \times 1.2 \times 10^{-5}$$

$$L_{\text{iron}} = \dfrac{40 \times 1.8}{1.2} = \dfrac{72}{1.2} = 60 \text{ cm}$$

We can verify: The difference $$L_{\text{iron}} - L_{\text{brass}} = 60 - 40 = 20 \text{ cm}$$, which is the required scale length.

Therefore, the length of iron is $$\boxed{60}$$ cm.