Question

A spherical shell of inner radius ‘$R_1$’ and outer radius ‘$R_2$’ is having variable thermal conductivity given by K = $\left(a_{0}T\right)r$ where $a_0$ is constant, ‘T’ is temperature in Kelvin and ‘r’ is the distance from the centre. Two surfaces of shell are maintained at temperature $T_1$ (inner surface) and $T_2$ (outer surface), respectively $(T_1>T_2)$. The heat current flowing through the shell would be

• A. $\frac{4\pi a_0(T_1^2-T_2^2)\times R_1{R}_2}{R_2-R_1}$
  
• B. $\frac{4\pi a_0(T_1-T_2)R_1{R}_2}{R_2-R_1}$
  
• C. $\frac{4\pi a_0{R}_1^2{R}_2^2(T_1^2-T_2^2)}{R_2^2-R_1^2}$
  
• D. $\frac{4\pi a_0(T_1^2-T_2^2)(R_1+R_2{)}^2}{R_2-R_1}$
  

Answer 1

The correct option is C $\frac{4\pi a_0{R}_1^2{R}_2^2(T_1^2-T_2^2)}{R_2^2-R_1^2}$


Temperature is decreasing we are going out, let at a distance ‘x’ from the centre, the temperature gradient is $\frac{-dT}{dx}$

At this location, $K=a_{0}Tx$
From $H=–KA\left(\frac{-dT}{dx}\right)$
$\Rightarrow H=(-a_{0}Tx)\left(4\pi x^2\right)\left(\frac{dT}{dx}\right)$
$\Rightarrow {\int }_{R_1}^{R_2}\frac{Hdx}{x^3}=-{\int }_{T_1}^{T_2}4\pi a_{0}TdT$
$\Rightarrow H=\frac{4\pi a_0{R}_1^2{R}_2^2(T_1^2-T_2^2)}{R_2^2-R_1^2}$