Question

A spherical body of radius $4\text{m}$ has a concentric cavity of radius $2\text{m}$ as shown in the figure. Thermal conductivity of the material is $\frac{100}{\pi }{\text{W/m}}^{\circ }\text{C}$. If the temperature of inner surface is kept at ${120}^{\circ }\text{C}$ and of the outer surface at ${40}^{\circ }\text{C}$. Then, find the temperature $T$ at a distance $3\text{m}$ from centre. (in ${}^{\circ }\text{C}$).

• A. ${66.66}^{\circ }\text{C}$
• B. ${33.66}^{\circ }\text{C}$
• C. ${23}^{\circ }\text{C}$
• D. ${63.32}^{\circ }\text{C}$

Answer 1

The correct option is A ${66.66}^{\circ }\text{C}$

Radius of spherical body $\left(b\right)=4\text{m}$
Radius of spherical cavity $\left(a\right)=2\text{m}$
Thermal conductivity of material $\left(K\right)=\frac{100}{\pi }{\text{W/m}}^{\circ }\text{C}$
Inner surface temperature $\left(T_1\right)={120}^{\circ }\text{C}$
Outer surface temperature $\left(T_2\right)={40}^{\circ }\text{C}$

At $r=3\text{m}$
Heat current $H_1$ = Heat current $H_2$
$\frac{(\Delta T{)}_{PM}}{R_{PM}}=\frac{(\Delta T{)}_{MQ}}{R_{MQ}}...\left(1\right)$
We know that, $R_{PM}=\frac{1}{4\pi K}(\frac{1}{a}-\frac{1}{r});R_{MQ}=\frac{1}{4\pi K}(\frac{1}{r}-\frac{1}{b})$
From (1)
$\frac{T_1-T}{\frac{1}{4\pi K}(\frac{1}{a}-\frac{1}{r})}=\frac{T-T_2}{\frac{1}{4\pi K}(\frac{1}{r}-\frac{1}{b})}$
$\Rightarrow \frac{120-T}{(\frac{1}{2}-\frac{1}{3})}=\frac{T-40}{(\frac{1}{3}-\frac{1}{4})}$
$\Rightarrow 6(120-T)=12(T-40)$
$T={66.66}^{\circ }\text{C}$