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

To solve this problem, we will use the formula for the rate of heat flow through a cylindrical rod in steady state:

Q/t = (kAΔT)/L

Where: 

• Q/t is the rate of heat flow.
• k is the thermal conductivity of the material.
• A is the cross-sectional area of the rod.
• ΔT is the temperature difference across the rod.
• L is the length of the rod.

Since the rods A and B have the same material and mass, their densities and volumes are the same, implying equal cross-sectional areas for equal length rods since they are cylindrical. Thus, the cross-sectional area A can be considered equal for both rods.

For rod A:

Q_A/t = (kAΔT_A)/L_A

For rod B:

Q_B/t = (kAΔT_B)/L_B

Given:

• ΔT_A = 40°C
• ΔT_B = 60°C
• Q_A/t : Q_B/t = 3 : 8

To find L_A/L_B}, we set up the ratio using the heat flow formula:

(Q_A/t) / (Q_B/t) = (ΔT_A * L_B) / (ΔT_B * L_A)

Substituting the given values:

3/8 = (40 * L_B) / (60 * L_A)

Simplifying the fractions, we get:

3/8 = (2/3) * (L_B/L_A)

Cross-multiplying to solve for the length ratio:

3 * 3 = 8 * 2 * (L_B/L_A)

9 = 16 * (L_B/L_A)

L_A/L_B = 16/9

Reversing the order for ratios:

L_B/L_A = 9/16

The ratio of the lengths of rods A to B is therefore:

4 : 3

Answer 2

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).