JAMB Physics 2012 Paper

Concept:The increase in length of a heated metal depends on its linear expansivity, original length, and temperature change.Explanation:The formula for linear expansion is:$$\Delta L = \alpha L \Delta \theta$$Since both metals are heated through the same temperature difference, $\Delta \theta$ cancels out.For P and Q, the ratio of increase in length is:$$\frac{\Delta L_P}{\Delta L_Q} = \frac{\alpha_P}{\alpha_Q} \times \frac{L_P}{L_Q}$$Substitute the given ratios:$$\frac{\Delta L_P}{\Delta L_Q} = \frac{2}{3} \times \frac{3}{4}$$$$\frac{\Delta L_P}{\Delta L_Q} = \frac{6}{12} = \frac{1}{2}$$Therefore, the ratio of the increase in lengths of P to Q is $1:2$.Answer:A. $1:2$