Question

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. $\frac{\left(K_{1} + K_{2}\right)}{\left(K_{1} K_{2} \right)}$
• B. $\frac{2 K_{1} + K_{2}}{\left(K_{1} + K_{2} \right)}$
• C. $\frac{\left(K_{1} + K_{2}\right)}{2}$
• D. $\frac{K_{1} K_{2} }{\left(K_{1} + K_{2}\right)}$

Answer 1

Answer: (B)

$\frac{2 K_{1} + K_{2}}{\left(K_{1} + K_{2} \right)}$

Answer 2

Let $T_0$ be the temperature of the junction of the two rods and $A$ be the area of cross-section of each rod. Under steady condition, the heat conducted per second through rod 1 is equal to the heat conducted per second through rod 2,
$H = \frac{K_{1} A \left(T_{1 } -T_{0}\right)}{L} = \frac{K_{2}A\left(T_{0} -T_{2}\right)}{L} \, \, \, \, \,$ ...(i)
where L is the length of each rod.
or $K_{1}\left(T_{1} -T_{0}\right) = K_{2}\left(T_{0} -T_{2}\right)$
or $T_{0 } = \frac{K_{1}T_{1} + K_{2} T_{2}}{K_{1} +K_{2}} \, \, \, \, \, \,$ ....(ii)
Put equation (ii) in (i), we get
$H = \frac{K_{1}A}{L} \left(T_{1} - \frac{K_{1}T_{1} +K_{2}T_{2}}{K_{1} + K_{2}}\right)$
$= \frac{K_{1}K_{2}}{\left(K_{1} + K_{2}\right)} \frac{A}{L} \left(T_{1} -T_{2}\right) \, \, \, \,$ ...(iii)
If $K$ is the equivalent thermal conductivity of the two rods, then
$H= \frac{KA\left(T_{1} -T_{2}\right) }{2L} \, \, \, \, \,$ ....(iv)
From (iii) and (iv)
$\frac{K}{2} = \frac{K_{1}K_{2}}{K_{1 } +K_{2}}$ or $K= \frac{2K_{1 }K_{2}}{K_{1} +K_{2}}$