It is required to prepare a steel metre scale, such that the millimetre intervals are to be ac curate within `0.0005 mm` at a certain temperature. Determine the maximum temperature variation allowable during the rulling of millimetre marks. Given, `alpha` for steel `=1.322xx10^(-5) .^(@)C^(-1)`

Explanation

To determine the maximum temperature variation allowable during the ruling of millimetre marks on a steel metre scale, we can use the formula for linear expansion:

\[ \Delta L = L \cdot \alpha \cdot \Delta T \]

where:
- \(\Delta L\) is the change in length,
- \(L\) is the original length,
- \(\alpha\) is the coefficient of linear expansion,
- \(\Delta T\) is the change in temperature.

### Step-by-Step Solution:

1. **Identify the Given Values:**
- \(\Delta L = 0.0005 \, \text{mm} = 5 \times 10^{-4} \, \text{m}\)
- \(L = 1 \, \text{m}\)
- \(\alpha = 1.322 \times 10^{-5} \, \text{°C}^{-1}\)

2. **Rearrange the Formula:**
We need to find \(\Delta T\). Rearranging the formula gives:
\[ \Delta T = \frac{\Delta L}{L \cdot \alpha} \]

3. **Substitute the Values:**
Now, substitute the known values into the rearranged formula:
\[ \Delta T = \frac{5 \times 10^{-4} \, \text{m}}{1 \, \text{m} \cdot 1.322 \times 10^{-5} \, \text{°C}^{-1}} \]

4. **Calculate \(\Delta T\):**
Performing the calculation:
\[ \Delta T = \frac{5 \times 10^{-4}}{1.322 \times 10^{-5}} \approx 37.8 \, \text{°C} \]

5. **Conclusion:**
The maximum allowable temperature variation during the ruling of the millimetre marks is approximately \(37.8 \, \text{°C}\).