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 consider the linear expansion of both the copper rod and the steel scale when the temperature changes from 20°C to 40°C.

### Step-by-Step Solution:

1. **Identify the Variables:**
- Let \( L_0 \) be the initial length of the copper rod at 20°C.
- Let \( \alpha_c \) be the coefficient of linear expansion for copper.
- Let \( \alpha_s \) be the coefficient of linear expansion for steel.
- The change in temperature, \( \Delta T = 40°C - 20°C = 20°C \).

2. **Calculate the Length of the Copper Rod at 40°C:**
The formula for linear expansion is given by:
\[ L = L_0 (1 + \alpha \Delta T) \]
For the copper rod, substituting the values:
\[ L = L_0 (1 + \alpha_c \cdot 20) \]

3. **Calculate the Length of the Steel Scale at 40°C:**
Similarly, for the steel scale, we can denote its initial length as \( L_1 \) (which is equal to \( L_0 \) at calibration):
\[ L' = L_1 (1 + \alpha_s \cdot 20) \]
Since the scale is calibrated at 20°C, we can assume \( L_1 = L_0 \) for this calculation.

4. **Set the Lengths Equal:**
At 40°C, the reading on the steel scale will be equal to the expanded length of the copper rod. Therefore, we equate the two lengths:
\[ L_0 (1 + \alpha_c \cdot 20) = L_1 (1 + \alpha_s \cdot 20) \]
Since \( L_1 = L_0 \), we can simplify this to:
\[ L_0 (1 + \alpha_c \cdot 20) = L_0 (1 + \alpha_s \cdot 20) \]

5. **Solve for the Scale Reading:**
Dividing both sides by \( L_0 \) (assuming \( L_0 \neq 0 \)):
\[ 1 + \alpha_c \cdot 20 = 1 + \alpha_s \cdot 20 \]
Rearranging gives:
\[ \alpha_c \cdot 20 = \alpha_s \cdot 20 \]
Thus, the scale reading when both are at 40°C can be expressed as:
\[ L' = L_0 \left( \frac{1 + \alpha_c \cdot 20}{1 + \alpha_s \cdot 20} \right) \]

### Final Expression:
The final reading of the scale when both the copper rod and the steel scale are at 40°C is:
\[ L' = L_0 \left( \frac{1 + 20 \alpha_c}{1 + 20 \alpha_s} \right) \]