Question

Heat flows radially outward through a spherical shell of outside radius $R_2$ and inner radius $R_1$. The temperature of inner surface is ${\Theta }_1$ and that outer is ${\Theta }_2$. At what radial distance fro m center of shell the temperature is just half way between ${\Theta }_1$ and ${\Theta }_2$

• A. $\frac{R_1+R_2}{2}$
• B. $\frac{4R_2+3R_1}{2}$
• C. $\frac{8R_2{R}_1}{4R_1+3R_2}$
• D. $\frac{2R_2{R}_1}{R_1+R_2}$

Answer 1

The correct option is A $\frac{R_1+R_2}{2}$

Change in temperature $=$ potential difference

Conductivity $=$ Thermal conductivity

Resistance $=$ Thermal resistance

$\frac{{\theta }_1-{\theta }_2}{Thermal{R}_{shell}}=H$

$\Rightarrow$ $\frac{{\theta }_1-{\theta }_2}{l/kA}=H$

$\Rightarrow$ $\frac{kA({\theta }_1-{\theta }_2)}{R_2-R_1}=H$

$\frac{V_2-V_1}{R}=I$

$\frac{V_2-V_1}{1/\rho A}=I$

Let $x$ is the distance from centre where we are required to find out term

$\frac{\left(\frac{{\theta }_2+{\theta }_1}{2}\right){-\theta }_2}{\frac{R_2-x}{kA}}=H\Rightarrow \frac{\left(\frac{{\theta }_2+{\theta }_1}{2}\right)-\theta }{R_2-\frac{x}{kA}}=\frac{kA({\theta }_1-{\theta }_2)}{R_2-R_1}$

$\frac{kA\left(\frac{{\theta }_2-{\theta }_1}{2}\right)}{R_2-x}=\frac{kA({\theta }_1-{\theta }_2)}{R_2-R_1}$

$R_2-R_1=2(R_2-x)\Rightarrow R_2-R_1=2R_2-2x$

$-R_2-R_1=-2x$

$\therefore x=\frac{R_1+R_2}{2}$