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$ABCDE$ is a regular pentagon of uniform wire. The rate of heat entering at $A$ and leaving at $C$ is equal. $T_B$ and $T_D$ are temperature of $B$ and $D$ . Find the temperature $T_C$

$\frac{{3{T_B} + 2{T_D}}}{5}$
$3T_D -2T_B$
$3T_D + 2T_B$
Can have any value
Solution
$\mathrm{R}_{1}=\frac{2 \ell}{\mathrm{k} \mathrm{A}}$
$R_{2}=\frac{3 \ell}{k A}$
heat goes in inverse ratio of resistance
$\mathrm{i}_{1}=\frac{\mathrm{T}_{\mathrm{A}}-\mathrm{T}_{\mathrm{C}}}{\mathrm{R}_{1}}=\frac{\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{C}}}{\mathrm{R}}$
$\mathrm{i}_{2}=\frac{\mathrm{T}_{\mathrm{A}}-\mathrm{T}_{\mathrm{C}}}{\mathrm{R}_{2}}=\frac{\mathrm{T}_{\mathrm{D}}-\mathrm{T}_{\mathrm{C}}}{\mathrm{R}}$
dividing, $\frac{\mathrm{R}_{2}}{\mathrm{R}_{1}}=\frac{\mathrm{T}_{\mathrm{B}}-\mathrm{T}_{\mathrm{C}}}{\mathrm{T}_{\mathrm{D}}-\mathrm{T}_{\mathrm{C}}}$
$3 \mathrm{T}_{\mathrm{D}}-3 \mathrm{T}_{\mathrm{C}}=2 \mathrm{T}_{\mathrm{B}}-2 \mathrm{T}_{\mathrm{C}}$
$\mathrm{T}_{\mathrm{C}}=3 \mathrm{T}_{\mathrm{D}}-2 \mathrm{T}_{\mathrm{B}}$