At 20∘C20 ∘ C, the hole in an aluminum ring is 2.500 cm in diameter. You need to slip this ring over a steel shaft that has a room-temperature diameter of 2.506 cm. To what common temperature should the ring and the shaft be heated so that the ring will just fit onto the shaft?

238.75⁰C .

Explanation:

coefficient of linear thermal expansion of aluminum and steel is  23 x 10⁻⁶ K⁻¹ and 12 x 10⁻⁶ K⁻¹ respectively .

Rise in temperature be Δ t .

Formula for linear expansion due to heat is as follows

l = l₀ ( 1 + α x Δt )

l is expanded length , l₀ is initial length , α is coefficient of linear expansion and  Δt is increase in temperature .

For aluminum

l = 2.5 ( 1 +  23 x 10⁻⁶ Δt )

For steel

l = 2.506 ( 1 + 12 x 10⁻⁶ Δt )

Given ,

2.5 ( 1 +  23 x 10⁻⁶ Δt ) = 2.506 ( 1 + 12 x 10⁻⁶ Δt )

1 +  23 x 10⁻⁶ Δt = 1.0024 ( 1 + 12 x 10⁻⁶ Δt )

1 +  23 x 10⁻⁶ Δt = 1.0024   + 12.0288 x 10⁻⁶ Δt

10.9712 x 10⁻⁶ Δt = .0024

Δt = 218.75

Initial temperature = 20⁰C

final temperature = 218.75 + 20 = 238.75⁰C .

which piece of furniture will have the most inertia and give the furniture movers the most difficulty in moving?  

answer

a 75kg dining table