What should be the sum of lengths of an aluminium and steel rod at `0^(@)C` is, so that all the temperature their difference in length is `0.25m`. (Take coefficient of linear expansion for aluminium and steel at `0^(@)C` as `22 xx 10^(-6)//.^(@)C` and `11 xx 10^(-5)//.^(@)C` respectively.)

Answer

To solve the problem, we need to find the sum of the lengths of an aluminum rod and a steel rod at 0°C, given that the difference in their lengths remains constant at 0.25 m for all temperatures. Let's denote:

- \( L_A \) = length of the aluminum rod
- \( L_S \) = length of the steel rod

### Step-by-step Solution:

1. **Understand the relationship between the lengths:**
We know that the difference in lengths is given by:
\[ L_S - L_A = 0.25 \, \text{m} \quad \text{(Equation 1)} \]

2. **Use the linear expansion formula:**
The change in length due to temperature change can be expressed as:
\[ \Delta L = L \alpha \Delta T \]
where \( \alpha \) is the coefficient of linear expansion and \( \Delta T \) is the change in temperature.

3. **Set up the equation for change in lengths:**
Since the temperature change is the same for both rods, we can write:
\[ \Delta L_A = L_A \alpha_A \Delta T \]
\[ \Delta L_S = L_S \alpha_S \Delta T \]

4. **Equate the ratios of lengths:**
From the above equations, we can set up the ratio:
\[ \frac{L_S}{L_A} = \frac{\alpha_A}{\alpha_S} \]
Given:
- \( \alpha_A = 22 \times 10^{-6} \, \text{°C}^{-1} \)
- \( \alpha_S = 11 \times 10^{-6} \, \text{°C}^{-1} \)

Thus,
\[ \frac{L_S}{L_A} = \frac{22 \times 10^{-6}}{11 \times 10^{-6}} = 2 \]
This implies that:
\[ L_S = 2 L_A \quad \text{(Equation 2)} \]

5. **Substitute Equation 2 into Equation 1:**
Replace \( L_S \) in Equation 1:
\[ 2 L_A - L_A = 0.25 \]
Simplifying this gives:
\[ L_A = 0.25 \, \text{m} \]

6. **Find \( L_S \):**
Using Equation 2:
\[ L_S = 2 L_A = 2 \times 0.25 = 0.5 \, \text{m} \]

7. **Calculate the sum of lengths:**
Now, we can find the sum of the lengths:
\[ L_A + L_S = 0.25 + 0.5 = 0.75 \, \text{m} \]

### Final Answer:
The sum of the lengths of the aluminum and steel rods at 0°C should be **0.75 m**.

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