Question

The temperature difference across two cylindrical rods A and B of same material and same mass are 40°C and 60°C respectively. In steady state, if the rates of flow of heat through the rods A and B are in the ratio 3 : 8, the ratio of the lengths of the rods A and B is:

• A. 1 : 3
• B. 5 : 3
• C. 4 : 3
• D. 2 : 3

Hint:

N/A

Answer 1

The Correct Option is C

We are given the following data: - Temperature difference across rod A, \( \Delta T_A = 40^\circ C \), - Temperature difference across rod B, \( \Delta T_B = 60^\circ C \), - The rates of heat flow through the rods A and B are in the ratio 3 : 8. In steady state, the rate of heat flow \( Q \) through a rod is given by the formula: \[ Q = \frac{kA\Delta T}{L}, \] where: - \( k \) is the thermal conductivity, - \( A \) is the cross-sectional area, - \( \Delta T \) is the temperature difference across the rod, - \( L \) is the length of the rod. Since both rods are made of the same material and have the same cross-sectional area, the formula for the rate of heat flow simplifies to: \[ Q \propto \frac{\Delta T}{L}. \] Let the lengths of rods A and B be \( L_A \) and \( L_B \), respectively. The ratio of the rates of heat flow through rods A and B is: \[ \frac{Q_A}{Q_B} = \frac{\frac{\Delta T_A}{L_A}}{\frac{\Delta T_B}{L_B}} = \frac{\Delta T_A}{\Delta T_B} \times \frac{L_B}{L_A}. \] Substitute the given values: \[ \frac{Q_A}{Q_B} = \frac{40}{60} \times \frac{L_B}{L_A} = \frac{2}{3} \times \frac{L_B}{L_A}. \] We are told that the ratio of the heat flow is 3 : 8, so: \[ \frac{2}{3} \times \frac{L_B}{L_A} = \frac{3}{8}. \] Solving for \( \frac{L_B}{L_A} \): \[ \frac{L_B}{L_A} = \frac{3}{8} \times \frac{3}{2} = \frac{9}{16}. \] Thus, the ratio of the lengths of rods A and B is: \[ \frac{L_A}{L_B} = \frac{4}{3}. \] Therefore, the correct answer is option (3).