Verified Solution • 2025–26 Session moderate

QJEEM21Ph563d: JEE Main 2021 Physics - Class 11 JEEM21Ph563d Solution: Two thin metallic spherical shells of radii r 1 and r 2 (r 1 <…

Two thin metallic spherical shells of radii r1 and r2 (r1 < r2) are placed with their centres coinciding. A material of thermal conductivity K is filled in the space between the shells. The inner shell is maintained at temperature θ\theta 1 and the outer shell at temperature θ\theta 2( θ\theta 1 < θ\theta 2). The rate at which heat flows radially through the material is :-

Multiple Choice Options

  • A 4πKr1r2(θ2−θ1)r2−r1{{4\pi K{r_1}{r_2}({\theta _2} - {\theta _1})} \over {{r_2} - {r_1}}}
  • B πr1r2(θ2−θ1)r2−r1{{\pi {r_1}{r_2}({\theta _2} - {\theta _1})} \over {{r_2} - {r_1}}}
  • C K(θ2−θ1)r2−r1{{K({\theta _2} - {\theta _1})} \over {{r_2} - {r_1}}}
  • D K(θ2−θ1)(r2−r1)4πr1r2{{K({\theta _2} - {\theta _1})({r_2} - {r_1})} \over {4\pi {r_1}{r_2}}}

Step-by-Step Verified Solution

JEE Main 2021 (Online) 31st August Evening Shift Physics - Heat and Thermodynamics Question 185 English Explanation

Thermal resistance of spherical sheet of thickness dr and radius r is

dR=drK(4πr2)dR = {{dr} \over {K(4\pi {r^2})}}

R=∫r1r2drK(4πr2)R = \int\limits_{{r_1}}^{{r_2}} {{{dr} \over {K(4\pi {r^2})}}}

R=14πK(1r1−1r2)=14πK(r2−r1r1r2)R = {1 \over {4\pi K}}\left( {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right) = {1 \over {4\pi K}}\left( {{{{r_2} - {r_1}} \over {{r_1}{r_2}}}} \right)

Thermal current (i) =θ2−θ1R= {{{\theta _2} - {\theta _1}} \over R}

i=4πKr1r2r2−r1(θ2−θ1)i = {{4\pi K{r_1}{r_2}} \over {{r_2} - {r_1}}}({\theta _2} - {\theta _1})

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