Question

The temperature difference between the ends of two cylindrical rods A and B of the same material is \( 2:3 \). In steady state, the ratio of the rates of flow of heat through the rods A and B is \( 5:9 \). If the radii of the rods A and B are in the ratio \( 1:2 \), then the ratio of lengths of the rods A and B is?

• A. \( 2:7 \)
• B. \( 3:7 \)
• C. \( 2:5 \)
• D. \( 3:10 \)

Hint:

For heat conduction problems, use Fourier's law and maintain proportionality for area, temperature difference, and length when comparing different rods.

Answer 1

The Correct Option is D

Step 1: Understanding the formula for heat conduction 
The rate of heat flow (\( H \)) through a cylindrical rod in steady state is given by Fourier’s law: \[ H = \frac{k A \Delta T}{L} \] where: - \( k \) is the thermal conductivity (same for both rods), - \( A \) is the cross-sectional area, - \( \Delta T \) is the temperature difference, - \( L \) is the length of the rod. 

Step 2: Express heat flow ratio for rods A and B 
Given: \[ \frac{H_A}{H_B} = \frac{5}{9}, \quad \frac{\Delta T_A}{\Delta T_B} = \frac{2}{3} \] Since the rods are cylindrical, the cross-sectional area is: \[ A = \pi r^2 \] So the heat flow equation for each rod can be written as: \[ \frac{H_A}{H_B} = \frac{k \pi r_A^2 (\Delta T_A) / L_A}{k \pi r_B^2 (\Delta T_B) / L_B} \] \[ \frac{H_A}{H_B} = \frac{(r_A^2 \Delta T_A / L_A)}{(r_B^2 \Delta T_B / L_B)} \] 

Step 3: Substitute given ratios 
Given \( \frac{r_A}{r_B} = \frac{1}{2} \), so: \[ \frac{r_A^2}{r_B^2} = \frac{1^2}{2^2} = \frac{1}{4} \] \[ \frac{H_A}{H_B} = \frac{\left( \frac{1}{4} \times \frac{2}{3} \right) / L_A}{(1 \times 1) / L_B} \] \[ \frac{5}{9} = \frac{\left( \frac{2}{12} \right) / L_A}{(1 / L_B)} \] \[ \frac{5}{9} = \frac{2}{12} \times \frac{L_B}{L_A} \] \[ \frac{5}{9} = \frac{2 L_B}{12 L_A} \] \[ \frac{5}{9} = \frac{L_B}{6 L_A} \] \[ L_B = \frac{6L_A \times 5}{9} = \frac{30L_A}{9} = \frac{10L_A}{3} \] \[ \frac{L_A}{L_B} = \frac{3}{10} \] Thus, the ratio of the lengths of rods A and B is \( 3:10 \).