Question

A spherical body of radius $b=\frac{4}{\pi }\text{m}$ has a concentric cavity of radius $a=\frac{2}{\pi }\text{m}$ as shown. Thermal conductivity of the material is $100{\text{W/m}}^{\circ }\text{C}$. Find the thermal resistance between inner and outer surface.

• A. ${6.25\times {10}^{-4}}^{\circ }\text{C/W}$
• B. ${1.25\times {10}^{-4}}^{\circ }\text{C/W}$
• C. ${6.5\times {10}^{-4}}^{\circ }\text{C/W}$
• D. ${3.75\times {10}^{-4}}^{\circ }\text{C/W}$

Answer 1

The correct option is A ${6.25\times {10}^{-4}}^{\circ }\text{C/W}$

Given,
Outer radius of spherical body $\left(b\right)=\frac{4}{\pi }\text{m}$
Inner radius of spherical body $\left(a\right)=\frac{2}{\pi }\text{m}$
Thermal conductivity of material $\left(k\right)=100{\text{W/m}}^{\circ }\text{C}$
Then, thermal resistance $\left(R\right)=\frac{1}{4\pi k}(\frac{1}{a}-\frac{1}{b})$
$=\frac{1}{4\pi \times 100}(\frac{\pi }{2}-\frac{\pi }{4})$
$=\frac{1}{1600}=$ ${6.25\times {10}^{-4}}^{\circ }\text{C/W}$