Five rods of same dimensions are arranged as | vedclass

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Question

(b) For no current flow between C and D
${\left( {\frac{Q}{t}} ight)_{AC}} = {\left( {\frac{Q}{t}} ight)_{CB}}$

==> $\frac{{{K_1}A({	heta

Vedclass · Accepted Answer

${K_1}{K_4} = {K_2}{K_3}$

Vedclass · Answer

(b) For no current flow between C and D
${\left( {\frac{Q}{t}} ight)_{AC}} = {\left( {\frac{Q}{t}} ight)_{CB}}$

==> $\frac{{{K_1}A({	heta _A} - {	heta _C})}}{l} = \frac{{{K_2}A({	heta _C} - {	heta _B})}}{l}$

==> $\frac{{{	heta _A} - {	heta _C}}}{{{	heta _C} - {	heta _B}}} = \frac{{{K_2}}}{{{K_1}}}$...$(i)$

Also ${\left( {\frac{Q}{t}} ight)_{AD}} = {\left( {\frac{Q}{t}} ight)_{DB}}$

==>$\frac{{{K_3}A({	heta _A} - {	heta _D})}}{l} = \frac{{{K_4}A({	heta _D} - {	heta _B})}}{l}$

==>$\frac{{{	heta _A} - {	heta _D}}}{{{	heta _D} - {	heta _B}}} = \frac{{{K_4}}}{{{K_3}}}$ ...$(ii)$

It is given that ${	heta _C} = {	heta _D},$ hence from equation $(i)$ and $(ii)$ we get $\frac{{{K_2}}}{{{K_1}}} = \frac{{{K_4}}}{{{K_3}}}$ 

==> ${K_1}{K_4} = {K_2}{K_3}$

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

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