A steel sphere sits on top of an aluminum ring.The steel sphere (α = 1.10 × 10−⁵/C°)has a diameter of 5,000 0 cm at 0°C.The aluminum ring (α = 2.40 × 10−⁵/C°)has an inside diameter of 4,992 0 cm at 0°C.Closest to which temperature given will the sphere just fall through the ring?

Question

Quizplus · Accepted Answer

The change in the diameter of the sphere is:$$\Delta d_s=\alpha_s d_s \Delta T=(1.10	imes10^{-5}/C^\circ)(5.0000	imes10^2cm)(123^\circ C)=6.66cm$$The change in the diameter of the ring is:$$\Delta d_r=\alpha_r d_r \Delta T=(2.40	imes10^{-5}/C^\circ)(4.9920	imes10^2cm)(123^\circ C)=14.38cm$$The sphere will just fall through the ring when:$$d_s+\Delta d_s=d_r+\Delta d_r$$$$5.0000cm+6.66cm=4.9920cm+14.38cm$$$$11.66cm=19.37cm$$$$\Delta T=123^\circ C$$

A Steel Sphere Sits on Top of an Aluminum Ring