A 25.01 -mm-diameter brass ball sits at room temperature on a 25.00 - mm- diameter hole made in an aluminum plate. The ball and plate are heated uniformly in a furnace, so both are at the same temperature at all times. At what temperature will the ball fall through the plate?

We are given the following information: - Initial diameter of the brass ball (D1_ball) = 25.01 mm - Initial diameter of the hole in the aluminum plate (D1_hole) = 25.00 mm - Linear expansion coefficients for brass (α_brass) = 19 x 10^{-6} °C^{-1} - Linear expansion coefficients for aluminum (α_aluminum) = 23 x 10^{-6} °C^{-1}

We need to find the temperature (T) at which the brass ball falls through the aluminum plate hole. To do this, we will first find the relationship between the change in diameter of the ball and the hole due to temperature. From the formula for linear expansion, we have: ΔD = α * D1 * ΔT For the brass ball: ΔD_ball = α_brass * D1_ball * ΔT For the hole in the aluminum plate: ΔD_hole = α_aluminum * D1_hole * ΔT The ball will fall through the hole when the diameter of the hole becomes larger than the diameter of the ball, i.e., when: D1_ball + ΔD_ball = D1_hole + ΔD_hole Now substitute the expressions from the linear expansion formula: D1_ball + α_brass * D1_ball * ΔT = D1_hole + α_aluminum * D1_hole * ΔT Now we can solve for ΔT. Rearrange the equation to make ΔT the subject: ΔT = (D1_hole - D1_ball) / (α_brass * D1_ball - α_aluminum * D1_hole) Substitute the given values and calculate ΔT: ΔT = (25.00 - 25.01) / (19 x 10^{-6} * 25.01 - 23 x 10^{-6} * 25.00) = -0.01 / (-4 x 10^{-6}) ΔT = 2500 °C The change in temperature is 2500 °C. Now, since we assumed room temperature, we will take room temperature to be 20 °C. Thus, the final temperature when the ball falls through the hole is: T = 20 + ΔT T = 20 + 2500 T = 2520 °C

The ball will fall through the hole when both the ball and the aluminum plate are heated uniformly to 2520 °C.