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Practice question

MCQ

Question

A brass ring has an inner diameter of 5cm at 30∘C. To what temperature must it be heated to increase the diameter to 5.009cm? (αl\=1.8×10−5K−1)

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Explanation

Given: L0 = 5cm, L = 5.009cm, αl = 1.8×10−5K−1, T1 = 30∘C. ΔL = 5.009−5 = 0.009cm. ΔL = L0αlΔT⇒0.009 = 5×1.8×10−5×ΔT. ΔT = 0.0095×1.8×10−5 = 100K. T2 = 30+100 = 130∘C.

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