A steel metre scale is to be ruled so that the millimetre intervals are accurate to within about `5 xx 10^(-5)`mm at a certain temperature. What is the maximum temperature variation allowable during the ruling ? Given `alpha` for steel `= 1.1 xx 10^(-5).^(@)C^(-1)`.

Options

A. `8^@C`
B. `9^@C`
C. `4.5^@C`
D. `10^@C`

Answer

To solve the problem, we need to determine the maximum allowable temperature variation (ΔT) during the ruling of a steel meter scale, given the accuracy requirement for the millimeter intervals and the coefficient of linear expansion for steel.

### Step-by-step Solution:

1. **Identify the Given Values:**
- Accuracy requirement (ΔL): \(5 \times 10^{-5}\) mm
- Coefficient of linear expansion for steel (α): \(1.1 \times 10^{-5} \, ^\circ C^{-1}\)
- Length of the scale (L): 1 mm (since we are considering the change for 1 mm interval)

2. **Use the Formula for Linear Expansion:**
The change in length (ΔL) due to a change in temperature (ΔT) is given by the formula:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]

3. **Substitute the Known Values:**
We know ΔL, L, and α, so we can substitute these values into the equation:
\[ 5 \times 10^{-5} = 1 \times 10^{-3} \cdot (1.1 \times 10^{-5}) \cdot \Delta T \]
Here, we converted 1 mm to meters for consistency in units.

4. **Rearranging the Equation:**
To find ΔT, we rearrange the equation:
\[ \Delta T = \frac{\Delta L}{L \cdot \alpha} \]

5. **Substituting Values into the Rearranged Equation:**
\[ \Delta T = \frac{5 \times 10^{-5}}{1 \times 10^{-3} \cdot 1.1 \times 10^{-5}} \]

6. **Calculating ΔT:**
\[ \Delta T = \frac{5 \times 10^{-5}}{1.1 \times 10^{-8}} = \frac{5}{1.1} \times 10^{3} \approx 4.545 \, ^\circ C \]

7. **Final Result:**
Rounding to two decimal places, the maximum allowable temperature variation is approximately:
\[ \Delta T \approx 4.5 \, ^\circ C \]

### Conclusion:
The maximum temperature variation allowable during the ruling of the steel meter scale is approximately **4.5 °C**.