Question:
Two semicircular rods AB and CD each of radius of curvature 14 cm and a straight rod BC of length 22 cm are connected in series. The three rods have equal area of cross-section and the thermal conductivities of the materials of the rods AB, BC and CD are in the ratio 1:2:3. In steady state, if the temperature difference between the ends of the middle rod BC is 30°C, then the temperature difference between the ends of the rods AB and CD are respectively:
Two semicircular rods AB and CD each of radius of curvature 14 cm and a straight rod BC of length 22 cm are connected in series. The three rods have equal area of cross-section and the thermal conductivities of the materials of the rods AB, BC and CD are in the ratio 1:2:3. In steady state, if the temperature difference between the ends of the middle rod BC is 30°C, then the temperature difference between the ends of the rods AB and CD are respectively:
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For series heat flow, the temperature drop across each segment is proportional to its thermal resistance, \( R_{th} = \frac{L}{KA} \).
Updated On: Jun 9, 2026
- \( 120^\circ\text{C}, 40^\circ\text{C} \)
- \( 60^\circ\text{C}, 20^\circ\text{C} \)
- \( 120^\circ\text{C}, 60^\circ\text{C} \)
- \( 60^\circ\text{C}, 40^\circ\text{C} \)
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The Correct Option is A
Solution and Explanation
Concept:
In steady-state heat conduction through rods connected in series, the rate of heat flow (\( \dot{Q} \)) is the same for all rods. The formula for heat flow is \( \dot{Q} = \frac{KA \Delta T}{L} \). Since the rods are in series, \( \dot{Q} = \text{constant} \).
Step 1: Calculate lengths and determine heat flow relationships.
Length of semicircular rod (\( L_{AB} = L_{CD} \)) = \( \pi R = \frac{22}{7} \times 14 = 44 \text{ cm} \). Length of straight rod (\( L_{BC} \)) = \( 22 \text{ cm} \). Conductivities (\( K_{AB}:K_{BC}:K_{CD} \)) = \( 1:2:3 \). Let \( K_{AB} = k, K_{BC} = 2k, K_{CD} = 3k \).
Step 2: Apply the steady-state heat flow condition.
\( \dot{Q}_{AB} = \dot{Q}_{BC} = \dot{Q}_{CD} \). Since \(\dot{Q} = \frac{KA \Delta T}{L}\), we have \(\Delta T = \frac{\dot{Q} L}{KA}\). Given \(\Delta T_{BC} = 30^\circ\text{C}\): $$ 30 = \frac{\dot{Q} \times 22}{2k \times A} \implies \frac{\dot{Q}}{kA} = \frac{30 \times 2}{22} = \frac{30}{11} $$
Step 3: Find temperature differences for AB and CD.
For AB: $$ \Delta T_{AB} = \frac{\dot{Q} \times 44}{k \times A} = \left( \frac{\dot{Q}}{kA} \right) \times 44 = \frac{30}{11} \times 44 = 120^\circ\text{C} $$ For CD: $$ \Delta T_{CD} = \frac{\dot{Q} \times 44}{3k \times A} = \left( \frac{\dot{Q}}{kA} \right) \times \frac{44}{3} = \frac{30}{11} \times \frac{44}{3} = 40^\circ\text{C} $$ $$\boxed{120^\circ\text{C}, 40^\circ\text{C}}$$
Step 1: Calculate lengths and determine heat flow relationships.
Length of semicircular rod (\( L_{AB} = L_{CD} \)) = \( \pi R = \frac{22}{7} \times 14 = 44 \text{ cm} \). Length of straight rod (\( L_{BC} \)) = \( 22 \text{ cm} \). Conductivities (\( K_{AB}:K_{BC}:K_{CD} \)) = \( 1:2:3 \). Let \( K_{AB} = k, K_{BC} = 2k, K_{CD} = 3k \).
Step 2: Apply the steady-state heat flow condition.
\( \dot{Q}_{AB} = \dot{Q}_{BC} = \dot{Q}_{CD} \). Since \(\dot{Q} = \frac{KA \Delta T}{L}\), we have \(\Delta T = \frac{\dot{Q} L}{KA}\). Given \(\Delta T_{BC} = 30^\circ\text{C}\): $$ 30 = \frac{\dot{Q} \times 22}{2k \times A} \implies \frac{\dot{Q}}{kA} = \frac{30 \times 2}{22} = \frac{30}{11} $$
Step 3: Find temperature differences for AB and CD.
For AB: $$ \Delta T_{AB} = \frac{\dot{Q} \times 44}{k \times A} = \left( \frac{\dot{Q}}{kA} \right) \times 44 = \frac{30}{11} \times 44 = 120^\circ\text{C} $$ For CD: $$ \Delta T_{CD} = \frac{\dot{Q} \times 44}{3k \times A} = \left( \frac{\dot{Q}}{kA} \right) \times \frac{44}{3} = \frac{30}{11} \times \frac{44}{3} = 40^\circ\text{C} $$ $$\boxed{120^\circ\text{C}, 40^\circ\text{C}}$$
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