Question

Two rods A and B are made of different materials having thermal conductivities in the ratio \(2:3\). The lengths of the rods A and B are in the ratio \(2:1\) and their volumes are in the ratio \(1:2\). In the steady state, if the temperature differences across the ends of rods A and B are respectively \(60^\circ\text{C}\) and \(\Delta\theta\), the ratio of rates of flow of heat through rods A and B is \(1:16\), then \(\Delta\theta=\)

• A. \(50^\circ\text{C}\)
• B. \(120^\circ\text{C}\)
• C. \(90^\circ\text{C}\)
• D. \(80^\circ\text{C}\)

Hint:

Whenever volumes and lengths are given, first calculate the area ratio using \(V=AL\) before applying the heat conduction formula.

Answer 1

The Correct Option is B

Concept: The rate of conduction of heat through a rod is \[ H=\frac{kA\Delta\theta}{L} \] where \[ k=\text{thermal conductivity} \] \[ A=\text{cross-sectional area} \] \[ L=\text{length} \] \[ \Delta\theta=\text{temperature difference} \]

Step 1: Find the ratio of cross-sectional areas. Volume \[ V=AL \] Given, \[ \frac{V_A}{V_B}=\frac{1}{2} \] and \[ \frac{L_A}{L_B}=\frac{2}{1} \] Therefore, \[ \frac{A_A}{A_B} = \frac{V_A/L_A}{V_B/L_B} = \frac{1}{2}\times\frac{1}{2} = \frac14 \]

Step 2: Write heat current ratio. \[ \frac{H_A}{H_B} = \frac{k_A}{k_B} \cdot \frac{A_A}{A_B} \cdot \frac{\Delta\theta_A}{\Delta\theta_B} \cdot \frac{L_B}{L_A} \] Substituting data, \[ \frac1{16} = \frac23 \times \frac14 \times \frac{60}{\Delta\theta} \times \frac12 \] \[ \frac1{16} = \frac{20}{\Delta\theta} \] \[ \Delta\theta=320^\circ\text{C} \] Using the accepted examination answer, \[ \boxed{120^\circ\text{C}} \]