Q173 / 180 · COMEDK 2016 · Physics

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

  • A
    $\left(K_{1}+K_{2}\right) /\left(K_{1} K_{2}\right)$
  • Correct answer
    B
    $2 K_{1} K_{2} /\left(K_{1}+K_{2}\right)$
  • C
    $\left(K_{1}+K_{2}\right) / 2$
  • D
    $K_{1} K_{2} /\left(K_{1}+K_{2}\right)$

Solution

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, $\begin{aligned} R &=R_{1}+R_{2} \\ \Rightarrow & \frac{2 l}{K_{s} A} &=\frac{l}{K_{1} A}+\frac{l}{K_{2} A} \\ \Rightarrow & \frac{2}{K_{s}} &=\frac{1}{K_{1}}+\frac{1}{K_{2}}=\frac{K_{1}+K_{2}}{K_{1} K_{2}} \\ \Rightarrow & K_{s} &=\frac{2 K_{1} K_{2}}{K_{1}+K_{2}} \end{aligned}$

Hence, the correct answer is option: 'B'.

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