A steel scale is to be prepared such that the millimeter intervals are to be accurate within `6 xx10^(-5)mm`. The maximum temperature variation form the temperature of calibration during the reading of the millimeter marks is `(alpha = 12 xx 10^(-6)//"^(@)C)`

Options

A. `4.0"^(@)C`
B. `4.5 "^(@)C`
C. `5.0"^(@)C`
D. `5.5 "^(@)C`

Answer

To solve the problem, we need to determine the maximum temperature variation (ΔT) that allows for the millimeter intervals on a steel scale to remain accurate within \(6 \times 10^{-5}\) mm. We will use the formula that relates the change in length (ΔL) to the change in temperature (ΔT) through the coefficient of linear expansion (α).

### Step-by-Step Solution:

1. **Identify the given values:**
- Desired accuracy (ΔL) = \(6 \times 10^{-5}\) mm
- Coefficient of linear expansion for steel (α) = \(12 \times 10^{-6} \, \text{°C}^{-1}\)
- Length of the interval (L) = 1 mm

2. **Convert the length from mm to meters for consistency in units:**
\[ L = 1 \, \text{mm} = 1 \times 10^{-3} \, \text{m} \]

3. **Use the formula for linear expansion:**
The formula relating change in length to change in temperature is given by:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]

4. **Rearrange the formula to solve for ΔT:**
\[ \Delta T = \frac{\Delta L}{L \cdot \alpha} \]

5. **Substitute the known values into the equation:**
\[ \Delta T = \frac{6 \times 10^{-5} \, \text{mm}}{1 \, \text{mm} \cdot 12 \times 10^{-6} \, \text{°C}^{-1}} \]

6. **Convert ΔL to the same unit as L:**
Since both ΔL and L are in mm, we can directly substitute:
\[ \Delta T = \frac{6 \times 10^{-5}}{1 \cdot 12 \times 10^{-6}} \]

7. **Calculate ΔT:**
\[ \Delta T = \frac{6 \times 10^{-5}}{12 \times 10^{-6}} = 5 \, \text{°C} \]

8. **Conclusion:**
The maximum temperature variation from the temperature of calibration should be \(5 \, \text{°C}\).

### Final Answer:
The maximum temperature variation (ΔT) is \(5 \, \text{°C}\).

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