Question:
Three identical thermal conductors are connected as shown in figure. Considering no heat loss due to radiation, temperature at the junction will be
Three identical thermal conductors are connected as shown in figure. Considering no heat loss due to radiation, temperature at the junction will be
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Three identical thermal conductors are connected as shown in figure. Considering no heat loss due to radiation, temperature at the junction will be \includegraphics[width=0.5\linewidth]3phy.png \labelfig:placeholder
Updated On: Apr 15, 2026
- $40^{\circ}C$
- $60^{\circ}C$
- $50^{\circ}C$
- $35^{\circ}C$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
In equilibrium, the rate of flow of heat entering the junction equals the rate of flow of heat leaving it.
Step 2: Equation
$$(\frac{dQ}{dt})_{1} = (\frac{dQ}{dt})_{2} + (\frac{dQ}{dt})_{3}$$ $$\frac{KA(\theta - 20)}{l} = \frac{KA(60 - \theta)}{l} + \frac{KA(70 - \theta)}{l}$$
Step 3: Solving
$$\theta - 20 = 130 - 2\theta$$ $$3\theta = 150 \Rightarrow \theta = 50^{\circ}C$$
Final Answer: (C)
In equilibrium, the rate of flow of heat entering the junction equals the rate of flow of heat leaving it.
Step 2: Equation
$$(\frac{dQ}{dt})_{1} = (\frac{dQ}{dt})_{2} + (\frac{dQ}{dt})_{3}$$ $$\frac{KA(\theta - 20)}{l} = \frac{KA(60 - \theta)}{l} + \frac{KA(70 - \theta)}{l}$$
Step 3: Solving
$$\theta - 20 = 130 - 2\theta$$ $$3\theta = 150 \Rightarrow \theta = 50^{\circ}C$$
Final Answer: (C)
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