JEE MAIN - Physics (2021 - 31st August Evening Shift - No. 6)
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 $$\theta$$1 and the outer shell at temperature $$\theta$$2($$\theta$$1 < $$\theta$$2). The rate at which heat flows radially through the material is :-
$${{4\pi K{r_1}{r_2}({\theta _2} - {\theta _1})} \over {{r_2} - {r_1}}}$$
$${{\pi {r_1}{r_2}({\theta _2} - {\theta _1})} \over {{r_2} - {r_1}}}$$
$${{K({\theta _2} - {\theta _1})} \over {{r_2} - {r_1}}}$$
$${{K({\theta _2} - {\theta _1})({r_2} - {r_1})} \over {4\pi {r_1}{r_2}}}$$
Explanation
_31st_August_Evening_Shift_en_6_2.png)
Thermal resistance of spherical sheet of thickness dr and radius r is
$$dR = {{dr} \over {K(4\pi {r^2})}}$$
$$R = \int\limits_{{r_1}}^{{r_2}} {{{dr} \over {K(4\pi {r^2})}}} $$
$$R = {1 \over {4\pi K}}\left( {{1 \over {{r_1}}} - {1 \over {{r_2}}}} \right) = {1 \over {4\pi K}}\left( {{{{r_2} - {r_1}} \over {{r_1}{r_2}}}} \right)$$
Thermal current (i) $$ = {{{\theta _2} - {\theta _1}} \over R}$$
$$i = {{4\pi K{r_1}{r_2}} \over {{r_2} - {r_1}}}({\theta _2} - {\theta _1})$$
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