Two diagonally opposite corners of a square m | vedclass

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(A) Let the four corners of the square be $A, B, C,$ and $D$ in cyclic order. Let the temperatures at $A$ and $C$ be $T_A = 40^{\circ} C$ and $T_C = 1

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(A) Let the four corners of the square be $A, B, C,$ and $D$ in cyclic order. Let the temperatures at $A$ and $C$ be $T_A = 40^{\circ} C$ and $T_C = 10^{\circ} C$.Since the four rods are identical,each rod acts as a thermal resistance $R = \frac{L}{kA}$.The arrangement of the four rods forms a Wheatstone bridge circuit where the heat flows from $A$ to $C$ through two parallel paths: $A \rightarrow B \rightarrow C$ and $A \rightarrow D \rightarrow C$.Due to the symmetry of the circuit and the identical nature of the rods,the potential (temperature) at $B$ and $D$ must be equal.Since $T_B = T_D$,the temperature difference between the other two corners $B$ and $D$ is $T_B - T_D = 0^{\circ} C$.

Two diagonally opposite corners of a square made of four thin rods of the same material and same dimensions are at temperatures $40^{\circ} C$ and $10^{\circ} C$. If only heat conduction takes place,then the temperature difference between the other two corners will be .......... $^{\circ} C$.

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