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Five rods of same dimensions are arranged as shown in the 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
${K_1} = {K_4}\,{\rm{and}}\,\;{K_2} = {K_3}$
${K_1}{K_4} = {K_2}{K_3}$
${K_1}{K_2} = {K_3}{K_4}$
$\frac{{{K_1}}}{{{K_4}}} = \frac{{{K_2}}}{{{K_3}}}$
Solution
(b) For no current flow between C and D
${\left( {\frac{Q}{t}} \right)_{AC}} = {\left( {\frac{Q}{t}} \right)_{CB}}$
==> $\frac{{{K_1}A({\theta _A} – {\theta _C})}}{l} = \frac{{{K_2}A({\theta _C} – {\theta _B})}}{l}$
==> $\frac{{{\theta _A} – {\theta _C}}}{{{\theta _C} – {\theta _B}}} = \frac{{{K_2}}}{{{K_1}}}$…$(i)$
Also ${\left( {\frac{Q}{t}} \right)_{AD}} = {\left( {\frac{Q}{t}} \right)_{DB}}$
==>$\frac{{{K_3}A({\theta _A} – {\theta _D})}}{l} = \frac{{{K_4}A({\theta _D} – {\theta _B})}}{l}$
==>$\frac{{{\theta _A} – {\theta _D}}}{{{\theta _D} – {\theta _B}}} = \frac{{{K_4}}}{{{K_3}}}$ …$(ii)$
It is given that ${\theta _C} = {\theta _D},$ hence from equation $(i)$ and $(ii)$ we get $\frac{{{K_2}}}{{{K_1}}} = \frac{{{K_4}}}{{{K_3}}}$
==> ${K_1}{K_4} = {K_2}{K_3}$
