A point source of heat of power P is placed at the centre of a spherical shell of mean radius fi. The material of the shell has thermal conductivity K. If the temperature difference between the outer and inner surface of the shell is not to exceed L the thickness of the shell should not be less than

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$4 {πKTR}^{2} / P$

$4 {πK}^{2} TR / P$

$4 {πKT}^{2} R / P$

$4 πKTR / P^{2}$

answer is A.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

$P = - KA \frac{dθ}{dr} = - K 4 {πr}^{2} (\frac{dθ}{dr})$
or $dθ = - \frac{Pdr}{4 {πKr}^{2}}$
Integrating, we have
$\int_{θ_{1}}^{θ_{2}} dθ = - \frac{P}{4 πK} \int_{r_{1}}^{T_{2}} (\frac{dr}{r^{2}})$
$∴ (θ_{1} - θ_{2}) = \frac{P}{4 πK} (\frac{1}{I_{1}} - \frac{1}{r_{2}}) = \frac{P}{4 πK} \frac{(I_{2} - r_{1})}{r_{1} I_{2}}$
So thickness
$I_{2} - I_{1} = \frac{4 πK (θ_{1} - θ_{2}) r_{1} r_{2}}{P} = \frac{4 {πKTR}^{2}}{P}$