JAMB 2012 Physics · Question 39

Two metals P and Q are heated through the same temperature difference. If the ratio of the linear…

Question

Two metals P and Q are heated through the same temperature difference. If the ratio of the linear expansivities of P to Q is 2:3 and the ratio of their length is 3: 4 respectively, the ratio of the increase in lengths of P to Q is

A.1:2
B.2:1
C.8:9
D.9:8

Answer: A. 1:2

Explanation written with AI and checked against the answer key

Analyze the question Two metals, P and Q, are heated through the same temperature change. We are given the ratio of their linear expansivities (α) and the ratio of their original lengths, and we need the ratio of their increases in length. Concept Linear expansion is given by: ΔL = α · L · Δθ where α is the linear expansivity, L is the original length, and Δθ is the temperature change. Solution For P: ΔL_P = α_P · L_P · Δθ For Q: ΔL_Q = α_Q · L_Q · Δθ Dividing, Δθ cancels because it is the same for both: ΔL_P / ΔL_Q = (α_P / α_Q) × (L_P / L_Q) Substituting the values: α_P : α_Q = 2 : 3, so α_P/α_Q = 2/3 L_P : L_Q = 3 : 4, so L_P/L_Q = 3/4 ΔL_P / ΔL_Q = (2/3) × (3/4) = 6/12 = 1/2 So the ratio is 1 : 2. Tip: In ratio problems like this, write the formula, divide one side by the other, and cancel anything that is the same for both (here Δθ). Then simply multiply the given ratios. Always simplify at the end, since 6:12 becomes 1:2.

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