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(B) The problem states that the sensor output is $0$ when the water level is below the set value, and $1$ otherwise.We want the pump to turn on $(P =

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(B) The problem states that the sensor output is $0$ when the water level is below the set value, and $1$ otherwise.We want the pump to turn on $(P = 1)$ if at least one tank has a water level below the set value.This means $P = 1$ if $S_1 = 0$ OR $S_2 = 0$.Using Boolean algebra, this is equivalent to $P = \overline{S_1} + \overline{S_2}$.By De Morgan's Law, $\overline{S_1} + \overline{S_2} = \overline{S_1 \cdot S_2}$.This is the Boolean expression for a $NAND$ gate.Looking at the options:Option $A$ is an $OR$ gate fed by two $AND$ gates: $P = (S_1 \cdot S_2) + (S_1 \cdot S_2) = S_1 \cdot S_2$.Option $B$ is an $OR$ gate fed by two $NAND$ gates: $P = \overline{S_1 \cdot S_2} + \overline{S_1 \cdot S_2} = \overline{S_1 \cdot S_2}$.This matches our requirement $P = \overline{S_1 \cdot S_2}$.Therefore, the correct circuit is $B$.

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