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Two thin metallic spherical shells of radii $20 \mathrm{~cm}$ and $30 \mathrm{~cm}$, respectively are placed with their centres coinciding. A material of thermal conductivity $\alpha$ is filled in the space between the shells. The inner shell is maintained at $300 \mathrm{~K}$ and the outer shell at $310 \mathrm{~K}$. If the rate at which heat flows radially through the material is $40 \mathrm{~W}$, find the value of $\alpha$ (in units of $\mathrm{J} \mathrm{s}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$ ).
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The correct answer is:
$\frac{5}{3 \pi}$
Given, radius of two thin metallic spherical
shells,
$\begin{aligned} & r_1=20 \mathrm{~cm}=0.2 \mathrm{~m} \\ & r_2=30 \mathrm{~cm}=0.3 \mathrm{~m}\end{aligned}$
temperature of inner shell, $T_1=300 \mathrm{~K}$, temperature of outer shell, $T_2=310 \mathrm{~K}$, and rate of heat flow, $H=40 \mathrm{~W}$.
Radial rate of flow of heat through the shell in the steady state,
$\begin{aligned} H & =\frac{d Q}{d t}=\alpha A \frac{d T}{d r}=\alpha\left(4 \pi r^2\right) \frac{d T}{d r} \\ \frac{d r}{r^2} & =\frac{4 \pi \alpha}{H} \cdot d T\end{aligned}$
Integrate on both sides, we get
$\begin{array}{rlrl}\Rightarrow & \quad \int_{r_1}^{r_2} \frac{d r}{r^2} & =\frac{4 \pi \alpha}{H} \int_{T_1}^{T_2} d T \\ \frac{r_2-r_1}{r_1 r_2} & =\frac{4 \pi \alpha}{H}\left(T_2-T_1\right) \\ \therefore \quad & \alpha & =\frac{H\left(r_2-r_1\right)}{4 \pi r_1 r_2\left(T_2-T_1\right)}\end{array}$
Putting the given values, we get
$\alpha=\frac{40(0.3-0.2)}{4 \pi(0.3)(0.2)(310-300)}=\frac{4}{2.4 \pi}=\frac{5}{3 \pi}$
Hence, the value of $\alpha=\frac{5}{3 \pi} \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$.
shells,
$\begin{aligned} & r_1=20 \mathrm{~cm}=0.2 \mathrm{~m} \\ & r_2=30 \mathrm{~cm}=0.3 \mathrm{~m}\end{aligned}$
temperature of inner shell, $T_1=300 \mathrm{~K}$, temperature of outer shell, $T_2=310 \mathrm{~K}$, and rate of heat flow, $H=40 \mathrm{~W}$.
Radial rate of flow of heat through the shell in the steady state,
$\begin{aligned} H & =\frac{d Q}{d t}=\alpha A \frac{d T}{d r}=\alpha\left(4 \pi r^2\right) \frac{d T}{d r} \\ \frac{d r}{r^2} & =\frac{4 \pi \alpha}{H} \cdot d T\end{aligned}$
Integrate on both sides, we get
$\begin{array}{rlrl}\Rightarrow & \quad \int_{r_1}^{r_2} \frac{d r}{r^2} & =\frac{4 \pi \alpha}{H} \int_{T_1}^{T_2} d T \\ \frac{r_2-r_1}{r_1 r_2} & =\frac{4 \pi \alpha}{H}\left(T_2-T_1\right) \\ \therefore \quad & \alpha & =\frac{H\left(r_2-r_1\right)}{4 \pi r_1 r_2\left(T_2-T_1\right)}\end{array}$
Putting the given values, we get
$\alpha=\frac{40(0.3-0.2)}{4 \pi(0.3)(0.2)(310-300)}=\frac{4}{2.4 \pi}=\frac{5}{3 \pi}$
Hence, the value of $\alpha=\frac{5}{3 \pi} \mathrm{Js}^{-1} \mathrm{~m}^{-1} \mathrm{~K}^{-1}$.
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