Five rods of same dimensions c are arranged as shown in figure. They have thermal conductivities K₁, K₂, K₃, K₄ and K₅. When points A and B are maintained at different temperatures, no heat flows through the central rod, if

For no current flow between C and D,

${\left(\frac{Q}{t}\right)}_{AC}={\left(\frac{Q}{t}\right)}_{CB}$

$\Rightarrow \frac{K_{1}A\left({\theta }_A-{\theta }_C\right)}{l}=\frac{K_{2}A\left({\theta }_C-{\theta }_B\right)}{l}\Rightarrow \frac{{\theta }_A-{\theta }_C}{{\theta }_C-{\theta }_B}=\frac{K_2}{K_1}\dots \left(\mathrm{i}\right)$

Also,   ${\left(\frac{Q}{t}\right)}_{AD}={\left(\frac{Q}{t}\right)}_{DB}$

$\Rightarrow \frac{K_{3}A\left({\theta }_A-{\theta }_D\right)}{l}=\frac{K_{4}A\left({\theta }_D-{\theta }_B\right)}{l}$

$\Rightarrow \frac{{\theta }_A-{\theta }_D}{{\theta }_D-{\theta }_B}=\frac{K_4}{K_3}$                      …(ii)

It is given that    ${\theta }_C={\theta }_D$  hence from Eqs. (i) and (ii), we get

$\frac{K_2}{K_1}=\frac{K_4}{K_3}\Rightarrow K_1{K}_4=K_2{K}_3$