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(C) The rate of heat flow through a rod is given by $H = \frac{KA\Delta 	heta}{l}$.For the rod $PQ$ with thermal conductivity $K_3$,length $l$,and cr

Vedclass · Accepted Answer

${K_3} = \frac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}$

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(C) The rate of heat flow through a rod is given by $H = \frac{KA\Delta 	heta}{l}$.For the rod $PQ$ with thermal conductivity $K_3$,length $l$,and cross-sectional area $A$,the heat flow rate is:$H_{PQ} = \frac{K_3 A \Delta 	heta}{l} \quad ...(i)$For the path $PRQ$,the rods $PR$ and $RQ$ are in series. Their equivalent thermal conductivity $K_s$ for a total length $2l$ is given by $\frac{2l}{K_s A} = \frac{l}{K_1 A} + \frac{l}{K_2 A}$,which simplifies to $K_s = \frac{2K_1 K_2}{K_1 + K_2}$.The heat flow rate through $PRQ$ is:$H_{PRQ} = \frac{K_s A \Delta 	heta}{2l} = \left( \frac{2K_1 K_2}{K_1 + K_2} ight) \frac{A \Delta 	heta}{2l} = \frac{K_1 K_2}{K_1 + K_2} \frac{A \Delta 	heta}{l} \quad ...(ii)$Given that $H_{PQ} = H_{PRQ}$,we equate $(i)$ and $(ii)$:$\frac{K_3 A \Delta 	heta}{l} = \frac{K_1 K_2}{K_1 + K_2} \frac{A \Delta 	heta}{l}$Therefore,${K_3} = \frac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}$.

Three rods of identical dimensions are connected as shown in the figure. Points $P$ and $Q$ are maintained at different temperatures. If the rate of heat flow through $PRQ$ and $PQ$ is the same,then...

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