12. A steel ball has a diameter of 60 mm and is 0,01 mm too large to pass through a hole in a brass plate when the ball and the plate are at a temperature of 303 K. Calculate the common temperature for the ball and plate at which the ball will just pass through the hole. The linear expansion of steel is 12 × 10-6/K and that of brass 19 × 10-6/K.
Let the original diameter of the hole in the brass plate be D. Since the steel ball is 0.01 mm too large to pass through the hole at 303 K, the diameter of the hole at 303 K is 60.01 mm.
Let the final diameter of the steel ball and the hole in the brass plate at the common temperature be d. Since both the steel ball and the brass plate expand, we can set up the equation:
(D + 12×10^-6×T) = (d + 19×10^-6×T)
where T is the common temperature.
Plugging in the values we know:
(D + 12×10^-6×303) = (d + 19×10^-6×303)
(D + 0.003636) = (d + 0.005757)
Since we know the original diameter of the hole is D=60mm, we can substitute this back into our equation:
(60 + 0.003636) = (d + 0.005757)
60.003636 = d + 0.005757
d = 60.003636 - 0.005757 = 60.001879
Therefore, the common temperature for the ball and plate at which the ball will just pass through the hole is 303 K.