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Conduction Law

A spherical s...

Question

A spherical shell of inner radius ‘ $R_{1}$ ’ and outer radius ‘ $R_{2}$ ’ is having variable thermal conductivity given by K = $(a_{0} T) r$ 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

\frac{4 π a_{0} (T_{1}^{2} - T_{2}^{2}) \times R_{1} R_{2}}{R_{2} - R_{1}}

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\frac{4 π a_{0} (T_{1} - T_{2}) R_{1} R_{2}}{R_{2} - R_{1}}

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\frac{4 π a_{0} R_{1}^{2} R_{2}^{2} (T_{1}^{2} - T_{2}^{2})}{R_{2}^{2} - R_{1}^{2}}

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\frac{4 π a_{0} (T_{1}^{2} - T_{2}^{2}) (R_{1} + R_{2})^{2}}{R_{2} - R_{1}}

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Solution

The correct option is C $\frac{4 π a_{0} R_{1}^{2} R_{2}^{2} (T_{1}^{2} - T_{2}^{2})}{R_{2}^{2} - R_{1}^{2}}$

Temperature is decreasing we are going out, let at a distance ‘x’ from the centre, the temperature gradient is $\frac{- d T}{d x}$

At this location, $K = a_{0} T x$
From $H = - K A (\frac{- d T}{d x})$
$\Rightarrow H = (- a_{0} T x) (4 π x^{2}) (\frac{d T}{d x})$
$\Rightarrow \int_{R_{1}}^{R_{2}} \frac{H d x}{x^{3}} = - \int_{T_{1}}^{T_{2}} 4 π a_{0} T d T$
$\Rightarrow H = \frac{4 π a_{0} R_{1}^{2} R_{2}^{2} (T_{1}^{2} - T_{2}^{2})}{R_{2}^{2} - R_{1}^{2}}$

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