A

Let the length of the aluminium rod be L_a and the length of the steel rod be L_s at 0 °C.
The change in length for the aluminium rod when temperature changes by T degrees can be given as:
\( \Delta L_a = L_a \cdot \alpha_a \cdot T \)
where \( \alpha_a = 22 \times 10^{-6} / ^\circ C \) (coefficient of linear expansion for aluminium).

Similarly, for the steel rod:
\( \Delta L_s = L_s \cdot \alpha_s \cdot T \)
where \( \alpha_s = 11 \times 10^{-6} / ^\circ C \) (coefficient of linear expansion for steel).

The difference in length between the two rods at any temperature T is:
\( \Delta L = \Delta L_a - \Delta L_s = (L_a \cdot \alpha_a - L_s \cdot \alpha_s) \cdot T \)
We want this difference in length to be constant and equal to 0.25 m at all temperatures, which implies:
\( L_a \cdot \alpha_a - L_s \cdot \alpha_s = \frac{0.25}{T} \)
For this equation to hold true at all temperatures T, we need the coefficients of T to cancel out, which gives:
\( L_a \cdot \alpha_a = L_s \cdot \alpha_s \)
Hence:
\( \frac{L_a}{L_s} = \frac{\alpha_s}{\alpha_a} = \frac{11 \times 10^{-6}}{22 \times 10^{-6}} = \frac{1}{2} \)
Therefore, we can say \( L_a = \frac{1}{2}L_s \). Let L_s = 2L_a.
The total length is L_a + L_s = L_a + 2L_a = 3L_a.
Since L_a is in terms of L_s, we can express:
Total length = 3L_a = 3 \cdot \frac{L_s}{2} = \frac{3}{2}L_s
Hence, we can choose any value for L_a and L_s satisfying this ratio to maintain that the difference remains 0.25m.
Therefore, the sum of the lengths of aluminium and steel rods can be expressed in multiples of L_a and L_s as a constant based on this calculation.