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Question #42526

Difficulty: 8/10 | 🟢 Free

The Question

Three identical cylindrical rods, each of length $L$ and cross-sectional area $A$ , are joined at a common junction $J$ to form a star network. The free ends of the rods are maintained at constant temperatures $T_1 = 500 \text{ K}$ , $T_2 = 100 \text{ K}$ , and $T_3 = 100 \text{ K}$ . The thermal conductivity of the material of the rods varies with absolute temperature $T$ as $k = \alpha T$ , where $\alpha$ is a positive constant. Assuming steady-state heat flow and no heat loss from the lateral surfaces of the rods, what is the temperature of the junction $J$ ?

Options

A.
300 \text{ K}
B.
$\frac{700}{3} \text{ K}$
C.
250 \text{ K}
D.
$100\sqrt{23} \text{ K}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Multi-Concept_Bridge

Master problems that require linking physics and math concepts to solve complex competitive questions.

Concepts Behind This Question

Steady State Temperature Gradient|Kirchhoff’s Loop Rule|Integration

Why This Question Is Tricky

Common Mistakes

Students typically treat thermal resistance as a constant $R = L/(kA)$ and apply linear nodal analysis $\sum (T_i - T_J)/R = 0$ .

This question tests the fundamental application of Fourier's Law when thermal conductivity is temperature-dependent ( $k = \alpha T$ ).

By mapping the thermal circuit to a non-linear equivalent, students must integrate $k(T)dT$ to find that heat current is proportional to $\Delta(T^2)$ , not $\Delta T$ .

Failing to recognize this leads to the classic linear average trap of $700/3$ K.

Quick Facts About This Question

Question Information

Subject	Physics
Chapter	Thermal Properties of Matter - Class 11 Part 2
Concept	Steady State Temperature Gradient
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free