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A brass ring of diameter 10.00 cm at 22.0°C is heated and slipped over an aluminum rod of diameter 10.01 cm at 22.0°C. Assume

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b) To separate the two metals difference between their non expanded diameter must be equal to the change in expanded diameters for a given temperature difference i.e.,

LAL LBr = (Br& Br-LaiaAi)AT

LAl-Br pr, ΔΤ- - Liaai BraBr-aiaA

l denotes the length and \alpha denotes the coefficient of linear expansion

Putting the values,

10.02 10 AT = C 396.2 10 x 19 x 10-6 - 10.02 x 24 x 10-6

So, the temperature they must be cooled to is

T = (22 396.2)° C = _374.2°C

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