A brass ring of diameter 10.00 cm at 20.0 ∘C is heated and slipped over an aluminum rod of diameter 10.01 cm at 20.0 ∘C. Assuming the average coefficients of linear expansion are constant, What if the aluminum rod were 10.02 cm in diameter?

-376°

Explanation:

ΔL = L (f) - L (i)

ΔL = α. L (i).[Δθ]

To remove the ring from the rod, then the final diameter of the ring must be as large as the final diameter of the rod.

Thus, we could write that

L (f-br) = L (f-al) = L (br) + α (br). L (br) Δθ = L (al) + α (al). L (al) Δθ

If we collect like terms and make Δθ subject of the formula, then we have

Δθ = [L (al) - L (br) ] / [α (br). L (br) - α (al). L (al) ]

Now, at 10.02 cm, we then have

Δθ = [10.02 - 10] / [ (1.9*10^-6 * 10.01) - (24*10^-6 * 10.02) ]

Δθ = - 396°

T (f) = T (i) + ΔT

T (f) = 20 - 396

T (f) = - 376° which can't be attained