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 1.00 10-5 m larger than the diameter of the hole in the thin plate, which has a diameter of 0.09 m. The initial temperature of each arrangement is 27.3°C. At what temperature will the ball fall through the hole in each arrangement?
Please HELP!!!

Answer 1

hole d=0.09 m
ball D=d+Δd=d+10⁻⁵m=0.09001 m
t=27.3℃
ΔL=αLΔT
D+ΔD = d+Δd
D+ α1•D•ΔT =d + α2•d•ΔT
D-d = α2•d•ΔT - α1•D•ΔT= ΔT(α2•d- α1•D)
ΔT = (D-d)/(α2•d- α1•D)= Δd/(α2•d- α1•D)
Arrangement A:
ball: Gold α1=14•10⁻⁶ (℃⁻)¹
hole: Lead α2 = 29•10⁻⁶ (℃⁻)¹
ΔT = Δd/(α2•d- α1•D) =
=10⁻⁵/(29•10⁻⁶•0.09 - 14•10⁻⁶•0.09001) =
=7.408°.
t=27.3+7.408=34.708℃

Arrangement B:
Steel α1=12•10⁻⁶ (℃⁻)¹
Aluminium α2 = 23•10⁻⁶ (℃⁻)¹
ΔT = Δd/(α2•d- α1•D)=...
Arramgment C:
Quartz α1=0.5•10⁻⁶ (℃⁻)¹
Silver α2 = 19•10⁻⁶ (℃⁻)¹
ΔT = Δd/(α2•d- α1•D)=...