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(C) Let the thermal resistance of each rod be $R$. The square consists of four rods $AC, CB, BD, DA$. The points $A$ and $B$ are maintained at tempera

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(C) Let the thermal resistance of each rod be $R$. The square consists of four rods $AC, CB, BD, DA$. The points $A$ and $B$ are maintained at temperatures $T_A = \sqrt{2}T$ and $T_B = T$. Due to the symmetry of the circuit,the heat current $H$ flowing from $A$ splits equally into two paths: $AC-CB$ and $AD-DB$. Let $T_C$ and $T_D$ be the temperatures at points $C$ and $D$ respectively. For path $AC-CB$: The heat current is $H_1 = \frac{T_A - T_C}{R} = \frac{T_C - T_B}{R}$. This implies $T_A - T_C = T_C - T_B$,so $T_C = \frac{T_A + T_B}{2}$. For path $AD-DB$: The heat current is $H_2 = \frac{T_A - T_D}{R} = \frac{T_D - T_B}{R}$. This implies $T_A - T_D = T_D - T_B$,so $T_D = \frac{T_A + T_B}{2}$. Since $T_C = T_D$,the temperature difference between $C$ and $D$ is $T_C - T_D = 0$.

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