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(C) Let the temperature of junction $D$ be $T_D$. In steady state, the net heat current at junction $D$ is zero: $i_{AD} + i_{CD} = 0$.$\frac{KA}{L}(1

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Heat current in $AB$ is equal to the heat current in $BC$.

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(C) Let the temperature of junction $D$ be $T_D$. In steady state, the net heat current at junction $D$ is zero: $i_{AD} + i_{CD} = 0$.$\frac{KA}{L}(100 - T_D) + \frac{KA}{L}(0 - T_D) = 0 \implies 100 - T_D - T_D = 0 \implies 2T_D = 100 \implies T_D = 50^{\circ}C$. (Statement $B$ is correct).Now, calculate heat currents:$i_{AB} = \frac{KA}{L}(100 - 40) = 1 	imes 60 = 60 	ext{ J/s}$.$i_{BC} = \frac{KA}{L}(40 - 0) = 1 	imes 40 = 40 	ext{ J/s}$.$i_{AD} = \frac{KA}{L}(100 - 50) = 50 	ext{ J/s}$.$i_{DC} = \frac{KA}{L}(50 - 0) = 50 	ext{ J/s}$.Comparing $i_{AB}$ and $i_{BC}$: $i_{AB} = 60 	ext{ J/s}$ and $i_{BC} = 40 	ext{ J/s}$. Thus, $i_{AB} = 1.5 	imes i_{BC}$. (Statement $A$ is correct, Statement $C$ is incorrect).Heat current withdrawn at $B$: The junction $B$ receives heat from $A$ $(i_{AB} = 60 	ext{ J/s})$ and loses heat to $C$ $(i_{BC} = 40 	ext{ J/s})$. The remaining heat must be withdrawn from $B$: $i_{withdrawn} = i_{AB} - i_{BC} = 60 - 40 = 20 	ext{ J/s}$. (Statement $D$ is correct).

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