Heat flows radially outward through a spherical shell of outside radius `R_(2)` and inner radius `R_(1)`. The temperature of inner surface of shell is `theta_(1)` and that of outer is `theta_(2)`.The radial distance from centre of shell where the temperature is just half way between `theta_(1)` and `theta_(2)` is "

Options

A. `(R_(1)+R_(2))/(2)`
B. `(R_(1)R_(2))/(R_(1)+R_(2))`
C. `(2R_(1)R_(2))/(R_(1)+R_(2))`
D. `R_(1)+(R_(2))/(2)`

Answer

C

Explanation

`theta_(1) - theta_(2) = Delta theta (theta_(1)-theta)/(overset(R)underset(R_(1))int(dr)/(K4pir^(2))) =(theta_(1)-theta_(2))/(overset(R_(2))underset(R_(1))int(dr)/(K4pir^(2)))`
`(Delta theta//2)/((1)/(4piK_(1))[(1)/(R_(1))-(1)/(R)]) =(Delta theta)/((1)/(4piK_(1))((1)/(R_(1))-(1)/(R_(2))))`
`rArr R = (2R_(1)R_(2))/(R_(1)+R_(2))`