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1 April, 13:14

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?

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  1. 1 April, 17:02
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    -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
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