Thermal conductivity of inner core of radius $2 cm$ is $k$ and of the outer one of radius $4 cm$ is $2 k$ . Find the equivalent thermal conductivity between its two ends $a$ and $b (k_{eq})$ .

\frac{7 k}{2}

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\frac{7 k}{8}

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\frac{7 k}{4}

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\frac{5 k}{4}

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Solution

The correct option is C $\frac{7 k}{4}$
Given,
Radius of inner core $(r_{1}) = 2 cm$
Radius of outer core $(r_{2}) = 4 cm$
Thermal conductivity of inner core $(k_{1}) = k$
Thermal conductivity of outer core $(k_{2}) = 2 k$
Between two ends $a$ and $b$ , two cylinders are parallel to each other
$∴ \frac{1}{R_{eq}} = \frac{1}{R_{1}} + \frac{1}{R_{2}}$ (where $R_{1}, R_{2}$ are thermal resistance of inner and outer core)
Since, $R = \frac{l}{k A}$ we can deduce that
$\frac{K k_{eq} A}{l} = \frac{k_{1} A_{1}}{l_{1}} + \frac{k_{2} A_{2}}{l_{2}}$
$\frac{k_{eq} (π (r_{2})^{2})}{l} = \frac{k_{1} (π r_{1}^{2})}{l} + \frac{k_{2} (π r_{2}^{2} - π r_{1}^{2})}{l}$
$16 k_{eq} = 4 k + 2 k (12)$
$k_{eq} = \frac{28}{16} k \Rightarrow k_{eq} = \frac{7}{4} k$

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