Solution:

Let the length of the other rod be \( L \).

Since the difference in their lengths remains the same at all temperatures, the expansion of each rod must be identical.

For the first rod:
\[
\Delta L_1 = 40 \times \alpha_1 \times \Delta T
\]

For the second rod:
\[
\Delta L_2 = L \times \alpha_2 \times \Delta T
\]

Since \(\Delta L_1 = \Delta L_2\):
\[
40 \times \alpha_1 = L \times \alpha_2
\]

Substitute \(\alpha_1 = 6 \times 10^{-6}/^\circ \text{C}\) and \(\alpha_2 = 4 \times 10^{-6}/^\circ \text{C}\):
\[
40 \times 6 \times 10^{-6} = L \times 4 \times 10^{-6}
\]

Dividing both sides by \(4 \times 10^{-6}\):
\[
L = \frac{40 \times 6}{4} = 60 \, \text{cm}
\]

So, the length of the other rod is \( 60 \, \text{cm} \).