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Practice question

MCQ

Question

A brass ring has an inner diameter of 10cm at 40∘C. What temperature must it be cooled to for the diameter to decrease by 0.009cm? (αl\=1.8×10−5K−1)

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Explanation

Given: L0 = 10cm, ΔL = −0.009cm, αl = 1.8×10−5K−1, T1 = 40∘C. ΔL = L0αlΔT⇒−0.009 = 10×1.8×10−5×ΔT. ΔT = −0.00910×1.8×10−5 = −0.0091.8×10−4 = −50K. T2 = 40−50 = −10∘C.

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