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 26.0 °C. At what temperature will the ball fall through the hole in each arrangement?

Answer 1

Answer:

The answer is 41, 7° C

Explanation:

Since the question didn't give me the coefficient of expansion of gold, I looked up in the data table.

Gold= 1.42x10^-5/°C

Supposed the coefficient of expansion of the material the plate is made of is 2.00x10^-5/°C

Diameter of hole at 26°C= 0,11m

Diameter of gold ball at 26°C =0.11+1.0x10^-5= 0.11001m

When the temperature has increased by ΔT:

hole's diameter: 0.11[1+(2.00x10^-5) ΔT]

ball's diameter: 0.11001[1+(1.42x10^-5) ΔT]

The ball passes through the hole when these 2 diameters are equal:

0.11[1+(2.00x10^-5) ΔT]= 0.11001[1+(1.42x10^-5) ΔT]

0.11+0.22x10^-5 ΔT= 0.11001+0.1562x10^-5 ΔT

0.0638x10^-5 ΔT= 0.00001

ΔT=15.7°C

T= 26.0+15.7= 41.7°C