AIPMT PRELIMS 2001 Physics Heat Transfer MCQ Question

The rate of heat flow through a cylindrical rod can be described by Fourier's law of heat conduction, which states that the rate of heat transfer (Q) is proportional to the temperature difference (ΔT) and the cross-sectional area (A) of the rod, and inversely proportional to its length (L): $Q = k \frac{A \Delta T}{L}$ where k is the thermal conductivity. When we double the linear dimensions of the rod, the new length becomes 2L and the new cross-sectional area becomes $A' = 4A$ (since area is proportional to the square of the linear dimensions). Thus, the new rate of heat flow (Q₂) can be expressed as: $Q_2 = k \frac{A' \Delta T}{L'} = k \frac{4A \Delta T}{2L} = 2 \left( k \frac{A \Delta T}{L} \right) = 2Q_1$ Therefore, the correct answer is B) 2Q₁. The other options are incorrect because they do not account for the changes in both area and length correctly.