A steel ball has a diameter of 60mm and is 0,01mm too large to pass through a hole in a brass plate when the ball and the plate are at 303k CALCULATE common temperature for the ball and plate at which the ball will just pass through the hole the linear expansion of stell is 12×10

Answer 1

-6 K^-1 and for brass it is 19×10^-6 K^-1.

Let the final diameter of the ball after being heated to the common temperature be d, and the final diameter of the hole in the brass plate be D.

Assuming that the temperature will be in between their initial temperatures and final temperature, we have

d = 60mm + 12×10^-6 × 60mm × (T - 303K)
D = d - 0.01mm = D + 19×10^-6 × D × (T - 303K)

At the common temperature, d = D. Substituting d = D into the above equations, we get

D = 60mm + 12×10^-6 × 60mm × (T - 303K)
D = D - 0.01mm + 19×10^-6 × D × (T - 303K)

By solving these equations, we can find the common temperature at which the ball will just pass through the hole in the brass plate.