A steel scale measures the length of a copper rod as `l_(0)` when both are at `20^(@)C`, which is the calibration temperature for the scale . The scale reading when both are at `40^(@)C`, is :-

Options

A. `(1 + 20alpha_(C))l_(0)`
B. `(1+20alpha_(S))l_(0)`
C. `((1+20alpha_(S))/(1+20alpha_(C)))l_(0)`
D. `((1+20alpha_(C))/(1+20alpha_(S)))l_(0)`

Explanation

To solve the problem, we need to understand how the lengths of the steel scale and the copper rod change with temperature due to thermal expansion.

### Step-by-Step Solution:

1. **Understanding Thermal Expansion**:
- The length of an object changes with temperature according to the formula:
\[ L = L_0 (1 + \alpha \Delta T) \]
where:
- \( L \) is the new length,
- \( L_0 \) is the original length,
- \( \alpha \) is the coefficient of linear expansion,
- \( \Delta T \) is the change in temperature.

2. **Identify Given Values**:
- Initial temperature \( T_0 = 20^\circ C \)
- New temperature \( T = 40^\circ C \)
- Change in temperature \( \Delta T = 40 - 20 = 20^\circ C \)
- Let \( \alpha_C \) be the coefficient of linear expansion for copper.
- Let \( \alpha_S \) be the coefficient of linear expansion for steel.
- The initial reading of the scale (at \( 20^\circ C \)) is \( L_0 \).

3. **Calculate the Length of the Steel Scale at \( 40^\circ C \)**:
- The new length of the steel scale at \( 40^\circ C \) is given by:
\[ L_S = L_0 (1 + 20 \alpha_S) \]

4. **Calculate the Length of the Copper Rod at \( 40^\circ C \)**:
- The new length of the copper rod at \( 40^\circ C \) is given by:
\[ L_C = L_0 (1 + 20 \alpha_C) \]

5. **Aligning the Readings**:
- When both the steel scale and the copper rod are at \( 40^\circ C \), the reading on the steel scale (which has expanded) will be equal to the length of the copper rod (which has also expanded).
- Therefore, we set the two lengths equal:
\[ L_0 (1 + 20 \alpha_C) = L_S (1 + 20 \alpha_S) \]

6. **Expressing the Scale Reading**:
- Rearranging the equation gives us the scale reading \( L_S \):
\[ L_S = \frac{L_0 (1 + 20 \alpha_C)}{(1 + 20 \alpha_S)} \]

7. **Final Expression**:
- Thus, the scale reading when both are at \( 40^\circ C \) is:
\[ L_S = L_0 \cdot \frac{1 + 20 \alpha_C}{1 + 20 \alpha_S} \]

### Conclusion:
The correct option for the scale reading when both the steel scale and the copper rod are at \( 40^\circ C \) is:
\[ L_S = \left(1 + 20 \alpha_C \right) \div \left(1 + 20 \alpha_S \right) \cdot L_0 \]
This corresponds to option D in the provided choices.