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(B) The sensors $S_1$ and $S_2$ are $0$ when the water level is low and $1$ when the water level is sufficient.We want the pump $P$ to be $1$ if at le

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(B) The sensors $S_1$ and $S_2$ are $0$ when the water level is low and $1$ when the water level is sufficient.We want the pump $P$ to be $1$ if at least one tank is low. 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 corresponds to a $NAND$ gate, where the inputs are $S_1$ and $S_2$.Looking at the options, the circuit in image $814-$b389 consists of two $NAND$ gates whose outputs are fed into an $OR$ gate. However, let's re-evaluate the requirement: $P = 1$ if $(S_1 = 0)$ or $(S_2 = 0)$.This is equivalent to $P = 	ext{NOT}(S_1 	ext{ AND } S_2)$.Option $(B)$ shows two $NAND$ gates with inputs $(S_1, S_2)$ and $(S_1, S_2)$ respectively, feeding into an $OR$ gate. This is not correct.Let's check the logic for $P = \overline{S_1} + \overline{S_2}$. This is exactly what a $NAND$ gate does. If we look at the provided options, none of the circuits directly represent a single $NAND$ gate. Let's re-examine the logic: we need $P = 1$ when $S_1=0$ or $S_2=0$. This is the definition of a $NAND$ gate output. If we look at option $(B)$, it is a $NAND$ gate followed by an $OR$ gate, which is not standard. Wait, let's re-read the circuit in $(B)$. It is two $NAND$ gates with inputs $(S_1, S_2)$ and $(S_1, S_2)$. This is just a $NAND$ gate. Actually, the circuit in $(B)$ is a $NAND$ gate. The output of a $NAND$ gate is $P = \overline{S_1 \cdot S_2} = \overline{S_1} + \overline{S_2}$. This matches our requirement.

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