JEE Main 20199 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 9 January
Shift: Morning Shift
Subject: Physics
Chapter: Thermal Properties of Matter

JEE Main 2019 · 9 January · Morning Shift · Physics · Thermal Properties of Matter

Temperature difference of 120 ^ o C is maintained between two ends...

Temperature difference of ${120}^{o}C$ is maintained between two ends of a uniform rod $AB$ of length $2L .$ Another bent rod $PQ,$ of same cross-section as $AB$ and length $\dfrac{3L}{2},$ is connected across $AB$ (See figure). In steady state, temperature difference between $P$ and $Q$ will be close to:
Temperature difference of 120 ^ o C is maintained between two ends of a uniform rod AB of length 2L . Another bent rod

Options

A ${45}^{o}C$
B ${35}^{o}C$
C ${60}^{o}C$
D ${75}^{o}C$

Correct answer

A. ${45}^{o}C$

Step-by-step solution

${T}_{1} - {T}_{2} = {120}^{o}C$

We define thermal resistance

$R = \dfrac{L}{KA}$

Given circuit can be reduced to

Temperature difference of 120 ^ o C is maintained between two ends of a uniform rod AB of length 2L . Another bent rod

Equivalent resistance ${R}_{eq} = \dfrac{8R}{5}$

Thermal current $i = \dfrac{{T}_{1} - {T}_{2}}{\dfrac{8R}{5}}$

${T}_{P} - {T}_{Q} = \dfrac{{T}_{1} - {T}_{2}}{\left(\dfrac{8R}{5}\right)}\left(\dfrac{3R}{5}\right)$

$= \dfrac{3}{8} \times 120 = 45$

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