Q \cdot L}{A \cdot \Delta T} $

Where:

$ Q $ is the rate of heat flow.

$ L $ is the length of the rod.

$ A $ is the cross-sectional area.

$ \Delta T $ is the temperature difference across the rod.

Since the rods $ A $ and $ B $ are made of the same material, their thermal conductivity, $ K $, is equal. This gives us:

$ K_A = \frac{Q_A \cdot L_A}{A_A \cdot \Delta T_A} \quad \text{and} \quad K_B = \frac{Q_B \cdot L_B}{A_B \cdot \Delta T_B} $

Given that the rods have the same mass and material, their volumes are equal:

$ V_A = V_B $

Thus, the equation for equal volumes becomes:

$ A_A \cdot L_A = A_B \cdot L_B $

From this, we derive:

$ \frac{L_A}{L_B} = \frac{A_B}{A_A} $

Since $ K_A = K_B $, we equate the expressions:

$ \frac{Q_A \cdot L_A}{A_A \cdot \Delta T_A} = \frac{Q_B \cdot L_B}{A_B \cdot \Delta T_B} $

Substituting the ratio of volumes, we get:

$ \frac{L_A}{L_B} = \frac{Q_B}{Q_A} \times \frac{A_A}{A_B} \times \frac{\Delta T_A}{\Delta T_B} $

Using the given values:

$ \frac{L_A}{L_B} = \frac{8}{3} \times \frac{L_B}{L_A} \times \frac{40}{60} $

Solving the equation:

$ \frac{L_A^2}{L_B^2} = \frac{8}{3} \times \frac{4}{6} = \frac{32}{18} = \frac{16}{9} $

Taking the square root gives:

$ \frac{L_A}{L_B} = \sqrt{\frac{16}{9}} = \frac{4}{3} $

Thus, the ratio of the lengths of rods $ A $ and $ B $ is $ \frac{4}{3} $.

Heat And Thermodynamics

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