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(C) The rate of heat flow $H$ is given by $H = \frac{\Delta 	heta}{R_{th}}$,where $R_{th} = \frac{l}{KA}$ is the thermal resistance.For the path $PRQ

Vedclass · Accepted Answer

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

Vedclass · Answer

(C) The rate of heat flow $H$ is given by $H = \frac{\Delta 	heta}{R_{th}}$,where $R_{th} = \frac{l}{KA}$ is the thermal resistance.For the path $PRQ$,the rods with conductivities ${K_1}$ and ${K_2}$ are in series. Their equivalent thermal resistance is $R_{PRQ} = \frac{l}{{K_1}A} + \frac{l}{{K_2}A} = \frac{l}{A} \left( \frac{1}{{K_1}} + \frac{1}{{K_2}} ight) = \frac{l}{A} \left( \frac{{K_1} + {K_2}}{{K_1}{K_2}} ight)$.For the path $PQ$,the rod with conductivity ${K_3}$ has thermal resistance $R_{PQ} = \frac{l}{{K_3}A}$.Given that the heat flows at the same rate along both paths,the thermal resistances must be equal: $R_{PRQ} = R_{PQ}$.Therefore,$\frac{l}{A} \left( \frac{{K_1} + {K_2}}{{K_1}{K_2}} ight) = \frac{l}{{K_3}A}$.Simplifying this,we get $\frac{1}{{K_3}} = \frac{{K_1} + {K_2}}{{K_1}{K_2}}$,which implies ${K_3} = \frac{{{K_1}{K_2}}}{{{K_1} + {K_2}}}$.

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