Verified Solution • 2025–26 Session moderate

QJEEM21Ph7a1e: JEE Main 2021 Physics - Class 11 JEEM21Ph7a1e Solution: The temperature at the junction of two insulating sheets,…

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

JEE Main 2021 (Online) 26th February Morning Shift Physics - Heat and Thermodynamics Question 240 English

Multiple Choice Options

  • A θ1R2+θ2R1R1+R2{{{\theta _1}{R_2} + {\theta _2}{R_1}} \over {{R_1} + {R_2}}}
  • B θ1R1+θ2R2R1+R2{{{\theta _1}{R_1} + {\theta _2}{R_2}} \over {{R_1} + {R_2}}}
  • C θ1R2−θ2R1R2−R1{{{\theta _1}{R_2} - {\theta _2}{R_1}} \over {{R_2} - {R_1}}}
  • D θ2R2−θ1R1R2−R1{{{\theta _2}{R_2} - {\theta _1}{R_1}} \over {{R_2} - {R_1}}}

Step-by-Step Verified Solution

Temperature at the junction is θ\theta .

so using the formula

T2−TR1=T−T1R2{{{T_2} - T} \over {{R_1}}} = {{T - {T_1}} \over {{R_2}}}

θ2−θR2=θ−θ1R1{{{\theta _2} - \theta } \over {{R_2}}} = {{\theta - {\theta _1}} \over {{R_1}}} R1(θ2−θ)=R2(θ−θ1){R_1}({\theta _2} - \theta ) = {R_2}(\theta - {\theta _1})

R1θ2−R1θ=R2θ−R2θ1{R_1}{\theta _2} - {R_1}\theta = {R_2}\theta - {R_2}{\theta _1}

R1θ+R2θ=R1θ2+R2θ1{R_1}\theta + {R_2}\theta = {R_1}{\theta _2} + {R_2}{\theta _1}

θ=R1θ2+R2θ1R1+R2\theta = {{{R_1}{\theta _2} + {R_2}{\theta _1}} \over {{R_1} + {R_2}}}

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