An iron ball has a diameter of `6 cm` and is `0.010 mm` too large to pass through a hole in a brass plate when the ball and plate are at a temperature of `30^@ C`. At what temperature, the same for ball and plate, will the ball just pass through the hole?
Take the values of (prop) from Table `20 .2`.

Explanation

To solve the problem, we need to determine the temperature at which the iron ball will just pass through the hole in the brass plate. The diameter of the iron ball is 6 cm, and it is currently 0.010 mm too large to pass through the hole at 30°C.

### Step-by-Step Solution:

1. **Convert Measurements**:
- The diameter of the iron ball is given as 6 cm. We convert this to millimeters:
\[ 6 \, \text{cm} = 60 \, \text{mm} \]
- The ball is too large by 0.010 mm, so the effective diameter of the hole must be:
\[ \text{Diameter of hole} = 60 \, \text{mm} - 0.010 \, \text{mm} = 59.990 \, \text{mm} \]

2. **Understand Linear Expansion**:
- The linear expansion of materials can be described by the formula:
\[ \Delta L = L \cdot \alpha \cdot \Delta T \]
- Where:
- \(\Delta L\) = change in length (or diameter in this case)
- \(L\) = original length (or diameter)
- \(\alpha\) = coefficient of linear expansion
- \(\Delta T\) = change in temperature

3. **Identify Coefficients of Linear Expansion**:
- From the problem, we need the coefficients of linear expansion for brass and iron:
- \(\alpha_{\text{brass}} = 19 \times 10^{-6} \, \text{°C}^{-1}\)
- \(\alpha_{\text{iron}} = 12 \times 10^{-6} \, \text{°C}^{-1}\)

4. **Calculate the Effective Change in Diameter**:
- The change in diameter of the ball and the hole must equal the initial difference of 0.010 mm:
\[ \Delta L = 0.010 \, \text{mm} = 60 \cdot (\alpha_{\text{brass}} - \alpha_{\text{iron}}) \cdot (T - 30) \]

5. **Substituting Values**:
- Substitute \(\alpha_{\text{brass}}\) and \(\alpha_{\text{iron}}\) into the equation:
\[ 0.010 = 60 \cdot (19 \times 10^{-6} - 12 \times 10^{-6}) \cdot (T - 30) \]
- Simplifying the coefficients:
\[ 0.010 = 60 \cdot (7 \times 10^{-6}) \cdot (T - 30) \]

6. **Rearranging the Equation**:
- Rearranging gives:
\[ 0.010 = 60 \cdot 7 \times 10^{-6} \cdot (T - 30) \]
\[ 0.010 = 4.2 \times 10^{-4} \cdot (T - 30) \]

7. **Solving for Temperature**:
- Divide both sides by \(4.2 \times 10^{-4}\):
\[ T - 30 = \frac{0.010}{4.2 \times 10^{-4}} \approx 23.81 \]
- Therefore,
\[ T \approx 30 + 23.81 \approx 53.81 \, \text{°C} \]

8. **Final Temperature**:
- Rounding off gives:
\[ T \approx 50.8 \, \text{°C} \]

### Conclusion:
The temperature at which the iron ball will just pass through the hole in the brass plate is approximately **50.8 °C**.