Two rods of same area of cross section and lengths, and conductivities $K_{1}$ and $K_{2}$ are connected in series. Then in steady state conductivity of the combination is

(K_{1} + K_{2}) / (K_{1} K_{2})

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2 K_{1} K_{2} / (K_{1} + K_{2})

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(K_{1} + K_{2}) / 2

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K_{1} K_{2} / (K_{1} + K_{2})

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Solution

The correct option is D $2 K_{1} K_{2} / (K_{1} + K_{2})$
Rods are in series $R = R_{1} + R_{2}$
$\frac{2 l}{K A} = \frac{l}{K_{1} A} + \frac{l}{K_{2} A}$
$K = \frac{2 K_{1} K_{2}}{K_{1} + K_{2}}$

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