To solve the problem, we need to find the temperature at which the steel ball will just pass through the hole in the brass plate.
Let's denote:
- \( d_s \) = diameter of the steel ball = 60 mm,
- \( d_h \) = diameter of the hole = \( d_s - 0.01 \) mm = 60 mm - 0.01 mm = 59.99 mm,
- \( \alpha_s \) = linear expansion coefficient of steel = \( 12 \times 10^{-6} , \text{K}^{-1} \),
- \( \alpha_b \) = linear expansion coefficient of brass = \( 19 \times 10^{-6} , \text{K}^{-1} \),
- \( T_0 \) = initial temperature = 303 K.
Step 1: Calculate the linear expansion of the steel ball and the brass hole.
The diameter of the steel ball at temperature \( T \) when it expands can be given by:
\[ d_s(T) = d_s \cdot \left(1 + \alpha_s (T - T_0)\right) \]
The diameter of the hole in the brass plate at temperature \( T \) when it expands can be given by:
\[ d_h(T) = d_h \cdot \left(1 + \alpha_b (T - T_0)\right) \]
Step 2: Set up the equation for when the ball just fits through the hole.
At the temperature \( T \) when the ball just fits through the hole, we have:
\[ d_s(T) = d_h(T) \]
Substituting the expressions for \( d_s(T) \) and \( d_h(T) \):
\[ d_s \left(1 + \alpha_s (T - T_0)\right) = d_h \left(1 + \alpha_b (T - T_0)\right) \]
Step 3: Substitute values into the equation.
Using \( d_s = 60 , \text{mm} \) and \( d_h = 59.99 , \text{mm} \):
\[ 60 \left(1 + 12 \times 10^{-6} (T - 303)\right) = 59.99 \left(1 + 19 \times 10^{-6} (T - 303)\right) \]
Step 4: Expand and simplify the equation.
Expanding both sides:
\[ 60 + 60 \cdot 12 \times 10^{-6} (T - 303) = 59.99 + 59.99 \cdot 19 \times 10^{-6} (T - 303) \]
Now we simplify:
- For the left side:
\[ 60 + 720 \times 10^{-6} (T - 303) \]
- For the right side:
\[ 59.99 + 1.13981 \times 10^{-3} (T - 303) \]
Now we have:
\[ 60 + 720 \times 10^{-6} (T - 303) = 59.99 + 1.13981 \times 10^{-3} (T - 303) \]
Step 5: Isolate \( T \).
Rearranging gives us:
\[ 60 - 59.99 = 1.13981 \times 10^{-3} (T - 303) - 720 \times 10^{-6} (T - 303) \]
Which simplifies to:
\[ 0.01 = (1.13981 \times 10^{-3} - 0.000720) (T - 303) \]
Calculating:
\[ 1.13981 \times 10^{-3} - 720 \times 10^{-6} = 0.41981 \times 10^{-3} \]
So we have:
\[ 0.01 = (0.41981 \times 10^{-3})(T - 303) \]
Dividing both sides by \( 0.41981 \times 10^{-3} \):
\[ T - 303 = \frac{0.01}{0.41981 \times 10^{-3}} \approx 23.84 \]
Then:
\[ T \approx 303 + 23.84 = 326.84 , \text{K} \]
Thus, when rounding to two decimal places, the common temperature at which the steel ball will just pass through the hole in the brass plate is:
\[ \boxed{326.84 , \text{K}} \]