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

current and voltage can both be different in different parts of a circuit so it matters where you place the ammeter and voltmete