The length of two metallic rods at temperatures `theta` are `L_(A)` and `L_(B)` and their linear coefficient of expansion are `alpha_(A)` and `alpha_(B)` respectively. If the difference in their lengths is to remian constant at any temperature then

Options

A. `L_(A)//L_(B)=alpha_(A)//alpha_(B)`
B. `L_(A)//L_(B)=alpha_(B)//alpha_(A)`
C. `alpha_(A)=alpha_(B)`
D. `alpha_(A)alpha_(B)=1`

Explanation

To solve the problem, we need to analyze the condition under which the difference in lengths of the two metallic rods remains constant as the temperature changes. Let's denote the lengths of the rods at temperature θ as \( L_A \) and \( L_B \), and their linear coefficients of expansion as \( \alpha_A \) and \( \alpha_B \) respectively.

### Step-by-step Solution:

1. **Understanding Linear Expansion**:
The change in length of a rod due to temperature change can be expressed as:
\[ \Delta L = \alpha \cdot L_0 \cdot \Delta T \]
where \( \Delta L \) is the change in length, \( \alpha \) is the linear coefficient of expansion, \( L_0 \) is the original length, and \( \Delta T \) is the change in temperature.

2. **Change in Lengths of Rods**:
For rod A, the change in length when the temperature changes by \( \Delta T \) is:
\[ \Delta L_A = \alpha_A \cdot L_A \cdot \Delta T \]
For rod B, the change in length is:
\[ \Delta L_B = \alpha_B \cdot L_B \cdot \Delta T \]

3. **Setting Up the Condition**:
We want the difference in lengths \( L_A - L_B \) to remain constant. Therefore, the change in lengths must satisfy:
\[ \Delta L_A - \Delta L_B = 0 \]
This implies:
\[ \alpha_A \cdot L_A \cdot \Delta T = \alpha_B \cdot L_B \cdot \Delta T \]

4. **Cancelling \( \Delta T \)**:
Since \( \Delta T \) is common in both terms and is not zero, we can cancel it out:
\[ \alpha_A \cdot L_A = \alpha_B \cdot L_B \]

5. **Rearranging the Equation**:
Rearranging the equation gives us:
\[ \frac{L_A}{L_B} = \frac{\alpha_B}{\alpha_A} \]

### Conclusion:
Thus, the relationship that must hold for the difference in lengths to remain constant at any temperature is:
\[ \frac{L_A}{L_B} = \frac{\alpha_B}{\alpha_A} \]