A slender rod of length L, diameter d (L>>d)and thermal conductivity $k_{1}$ is joined with another rod of identical dimensions, but of thermal conductivity $k_{2}$ , to form a composite cylindrical rod of length 2L. The heat transfer in radial direction and contact resistance are negligible. The effective thermal conductivity of the composite rod is

\frac{k_{1} k_{2}}{k_{1} + k_{2}}

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\frac{2 k_{1} k_{2}}{k_{1} + k_{2}}

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k_{1} + k_{2}

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\sqrt{k_{1} k_{2}}

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Solution

The correct option is B $\frac{2 k_{1} k_{2}}{k_{1} + k_{2}}$
$(R_{t h})_{I} = \frac{L}{k_{1} A} + \frac{L}{k_{2} A}$

$(R_{t h})_{I I} = \frac{2 L}{k_{e q} A}$

Equivalent resistance will be the same in both cases.

$(R_{t h})_{I} = (R_{t h})_{I I}$

$\Rightarrow \frac{2 L}{k_{e q} A} = \frac{L}{k_{1} A} + \frac{L}{k_{2} A}$

$\Rightarrow k_{e q} = \frac{2 k_{1} k_{2}}{k_{1} + k_{2}}$

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The correct option is B 2k1k2k1+k2(Rth)I=Lk1A+Lk2A (Rth)II=2LkeqA Equivalent resistance will be the same in both cases. (Rth)I=(Rth)II ⇒2LkeqA=Lk1A+Lk2A ⇒keq=2k1k2k1+k2