A steel sphere sits on top of an aluminum ring. The steel sphere (a = 1.1 x 10^-5/degrees celsius) has a diameter of 4.000 cm at 0 degrees celsius. The aluminum ring (a = 2.4 x 10^-5/degrees celsius) has an inside diamter of 3.994 cm at 0 degrees celsius. At what temperature will teh sphere fall through teh ring? Mulitple Choice Questiona, 461.5 degrees celsiusb. 208.0 degrees celsiusc. 115.7 degrees celsiusd. 57.7 degrees celsius

C

Explanation:

To solve this question, we will need to develop an expression that relates the diameter 'd', at temperature T equals the original diameter d₀ (at 0 degrees) plus the change in diameter from the temperature increase (ΔT = T):

d = d₀ + d₀αT

for the sphere, we were given

D₀ = 4.000 cm

α = 1.1 x 10⁻⁵/degrees celsius

we have D = 4 + (4x (1.1 x 10⁻⁵) T = 4 + (4.4x10⁻⁵) T EQN 1

Similarly for the Aluminium ring we have

we were given

d₀ = 3.994 cm

α = 2.4 x 10⁻⁵/degrees celsius

we have d = 3.994 + (3.994x (2.4 x 10⁻⁵) T = 3.994 + (9.58x10⁻⁵) T EQN 2

Since @ the temperature T at which the sphere fall through the ring, d=D

Eqn 1 = Eqn 2

4 + (4.4x10⁻⁵) T = 3.994 + (9.58x10⁻⁵) T, collect like terms

0.006=5.18x10⁻⁵T

T=115.7K