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JEE Main 202126 FebruaryMorning Shift
Exam
JEE Main
Year
2021
Date
26 February
Shift
Morning Shift
Subject
Physics
Chapter
Thermal Properties of Matter

JEE Main 2021 · 26 February · Morning Shift · Physics · Thermal Properties of Matter

The temperature θ at the junction of two insulating sheets, having...

The temperature θ\theta at the junction of two insulating sheets, having thermal resistances R1{R}_{1} and R2{R}_{2} as well as top and bottom temperatures θ1{\theta}_{1} and θ2{\theta}_{2} (as shown in figure) is given by :

The temperature θ at the junction of two insulating sheets, having thermal resistances R _ 1 and R _ 2 as well as top

Options

  1. Aθ1R2−θ2R1R2−R1\dfrac{{\theta}_{1}{R}_{2} - {\theta}_{2}{R}_{1}}{{R}_{2} - {R}_{1}}
  2. Bθ2R2−θ1R1R2−R1\dfrac{{\theta}_{2}{R}_{2} - {\theta}_{1}{R}_{1}}{{R}_{2} - {R}_{1}}
  3. Cθ1R2+θ2R1R1+R2\dfrac{{\theta}_{1}{R}_{2} + {\theta}_{2}{R}_{1}}{{R}_{1} + {R}_{2}}
  4. Dθ1R1+θ2R2R1+R2\dfrac{{\theta}_{1}{R}_{1} + {\theta}_{2}{R}_{2}}{{R}_{1} + {R}_{2}}

Correct answer

C. θ1R2+θ2R1R1+R2\dfrac{{\theta}_{1}{R}_{2} + {\theta}_{2}{R}_{1}}{{R}_{1} + {R}_{2}}

Step-by-step solution

The temperature θ at the junction of two insulating sheets, having thermal resistances R _ 1 and R _ 2 as well as top

Heat flow rate will be same through both

∴θ1−θR1=θ−θ2R2∴ \dfrac{{\theta}_{1} - \theta}{{R}_{1}} = \dfrac{\theta - {\theta}_{2}}{{R}_{2}}

R2θ1−R2θ=R1θ−R1θ2{R}_{2}{\theta}_{1} - {R}_{2}\theta = {R}_{1}\theta - {R}_{1}{\theta}_{2}

θ=R2θ1+R1θ2R1+R2\theta = \dfrac{{R}_{2}{\theta}_{1} + {R}_{1}{\theta}_{2}}{{R}_{1} + {R}_{2}}

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