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 1.00 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 22.7°C. At what temperature will the ball fall through the hole in each arrangement?

(a) TA =  °C

(b) TB =  °C

(c) TC =  °C

Coefficient of Linear Expansion Arrangement A
Gold
Lead 1.40 10-5(C°)-1
2.90 10-5(C°)-1 Arrangement B
Steel
Aluminum 1.20 10-5(C°)-1
2.30 10-5(C°)-1 Arrangement C
Quartz
Silver

5.00 10-7(C°)-1
1.90 10-5(C°)-1

Explanation / Answer

for a sphere of radius R, hole of radius r
intiial temperature Ti
coeffieicnt of linear expansion of hole material = k
coefficient of linear expansion of the ball = K
now, coefficient of volumetric expansion = 3k, 3K
hence

at temperature T
R(1 - K(T - Ti)) = r(1 - k(T - Ti))

R - r = 0.5*10^-5 m
2r = 0.09 m
r = 0.045 m
hence
R = 0.045005 m

(R - r) - RK(T - Ti) + rk(T - Ti) = 0
0.5*10^-5/(0.045*k - 0.045005*K) + Ti= T
Ti = 22.7 C
hence
case 1
K = 1.4*10^-5
k = 2.9*10^-5
hence
T = 30.108175662 C

case 2
K = 1.2*10^-5
k = 2.3*10^-5
T = 32.8022346142 C

case 3
K = 5*10^-7
k = 1.9*10^-5
hence
T = 28.7060240421 C

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