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 leng...

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

Answer Details

The increase in length of a material due to heating is proportional to its linear expansivity and the change in temperature. The linear expansivity of a material is defined as the change in length per unit length per degree Celsius change in temperature. Let the increase in length of metal P be x, and the increase in length of metal Q be y. Let the temperature difference be ΔT. Then, we can write: x = αP ΔT Lp ...(1) y = αQ ΔT Lq ...(2) where αP and αQ are the linear expansivities of metals P and Q respectively, and Lp and Lq are their original lengths. We are given that the ratio of the linear expansivities of P to Q is 2:3. So, we can write: αP/αQ = 2/3 Cross-multiplying, we get: 3αP = 2αQ αP/αQ = 2/3 = 2k/3k ...(3) where k is a constant. We are also given that the ratio of the lengths of P to Q is 3:4. So, we can write: Lp/Lq = 3/4 Cross-multiplying, we get: 4Lp = 3Lq Lp/Lq = 3/4 = 3m/4m ...(4) where m is a constant. Substituting equations (3) and (4) into equations (1) and (2), we get: x = (2k/3k)αQ ΔT (3m) y = αQ ΔT (4m) Simplifying, we get: x/y = (2k/3k) (3/4) x/y = 1/2 Therefore, the ratio of the increase in lengths of P to Q is 1:2. Hence, the correct answer is 1:2.