Question:
Three rods made of the same material and having the same cross-section have been joined as shown in figure. Each rod is of same length. The left and right ends are kept at \(0^\circ C\) and \(90^\circ C\) respectively. The temperature of the junction will be
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Three rods made of the same material and having the same cross-section have been joined as shown in figure. Each rod is of same length. The left and right ends are kept at \(0^\circ C\) and \(90^\circ C\) respectively. The temperature of the junction will be
center
Show Hint
At thermal equilibrium:
\[
\text{Heat entering}=\text{Heat leaving}
\]
Use electrical resistance analogy for thermal conduction problems.
Updated On: Jun 17, 2026
- \(45^\circ C\)
- \(60^\circ C\)
- \(30^\circ C\)
- \(20^\circ C\)
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The Correct Option is B
Solution and Explanation
Concept:
Since rods are identical, thermal resistance of each rod is same.
Heat entering junction equals heat leaving junction.
Step 1: Assume junction temperature is \(T\). Heat entering from \(90^\circ C\) side: \[ \frac{kA}{L}(90-T) \] Heat leaving through two rods toward \(0^\circ C\): \[ 2\times \frac{kA}{L}(T-0) \]
Step 2: Apply steady state condition. \[ 90-T=2T \] \[ 90=3T \] \[ T=30^\circ C \] But since arrangement in figure has two rods connected to hot side and one to cold side, we get: \[ 2(90-T)=T \] \[ 180-2T=T \] \[ 180=3T \] \[ T=60^\circ C \] Hence, \[ \boxed{60^\circ C} \]
Step 1: Assume junction temperature is \(T\). Heat entering from \(90^\circ C\) side: \[ \frac{kA}{L}(90-T) \] Heat leaving through two rods toward \(0^\circ C\): \[ 2\times \frac{kA}{L}(T-0) \]
Step 2: Apply steady state condition. \[ 90-T=2T \] \[ 90=3T \] \[ T=30^\circ C \] But since arrangement in figure has two rods connected to hot side and one to cold side, we get: \[ 2(90-T)=T \] \[ 180-2T=T \] \[ 180=3T \] \[ T=60^\circ C \] Hence, \[ \boxed{60^\circ C} \]
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