A ball and a thin plate are made from different materials and have the same init

Question

A ball and a thin plate are made from different materials and have the same initial temperature. The ball does not fit through a hole in the plate, because the diameter of the ball is slightly larger than the diameter of the hole. However, the ball will pass through the hole when the ball and the plate are both heated to a common higher temperature. In each of the arrangements in the drawing the diameter of the ball is 1.0 × 10-5 m larger than the diameter of the hole in the thin plate, which has a diameter of 0.11 m. The initial temperature of each arrangement is 22.0 °C. At what temperature will the ball fall through the hole in each arrangement?

Explanation / Answer

Assuming the arrangement to be a Gold ball and a Lead Plate, we write

The coefficients of linear expansion are

Gold= 14x10^-6 /oC
Lead= 29x10^-6 /oC

Lead (29×10^-6) minus Gold (14×10^-6) = 15×10^-6
1.5×10^-5 × (.11m) × (d)T = 10^-5m
(d)T = 1×10^-5m ÷ (1.5×10^-5×.1m) = 6.06°C
6.06°C + 22°C = 28.06°C

If there are any other arrangements, similiar method could be followed.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.