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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
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The correct answer is:
$2 K_{1} K_{2} /\left(K_{1}+K_{2}\right)$
The thermal resistance of a rod is given by
$R=\frac{l}{K A}$
As, length and area of two rods are same, then in series combination,
$\begin{aligned} R &=R_{1}+R_{2} \\ \Rightarrow & \frac{2 l}{K_{s} A} &=\frac{l}{K_{1} A}+\frac{l}{K_{2} A} \\ \Rightarrow & \frac{2}{K_{s}} &=\frac{1}{K_{1}}+\frac{1}{K_{2}}=\frac{K_{1}+K_{2}}{K_{1} K_{2}} \\ \Rightarrow & K_{s} &=\frac{2 K_{1} K_{2}}{K_{1}+K_{2}} \end{aligned}$
$R=\frac{l}{K A}$
As, length and area of two rods are same, then in series combination,
$\begin{aligned} R &=R_{1}+R_{2} \\ \Rightarrow & \frac{2 l}{K_{s} A} &=\frac{l}{K_{1} A}+\frac{l}{K_{2} A} \\ \Rightarrow & \frac{2}{K_{s}} &=\frac{1}{K_{1}}+\frac{1}{K_{2}}=\frac{K_{1}+K_{2}}{K_{1} K_{2}} \\ \Rightarrow & K_{s} &=\frac{2 K_{1} K_{2}}{K_{1}+K_{2}} \end{aligned}$
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