Boolean Algebra and Logic Gates Questions in English

(D) $NAND$ gate is a combination of an $AND$ gate followed by a $NOT$ gate.
First,the $AND$ operation is performed on inputs $A$ and $B$,which results in $A \cdot B$.
Then,the $NOT$ operation is applied to this result,which inverts the output.
Therefore,the output $Y$ of a $NAND$ gate is given by the Boolean expression $Y = \overline{A \cdot B}$.

(C) The given circuit consists of a $NAND$ gate and a $NOT$ gate,whose outputs are fed into an $OR$ gate.
$1$. The inputs $A$ and $B$ are fed into a $NAND$ gate. The output of the $NAND$ gate is $\overline{A \cdot B}$.
$2$. The input $C$ is fed into a $NOT$ gate. The output of the $NOT$ gate is $\overline{C}$.
$3$. These two outputs are then fed into an $OR$ gate. The $OR$ operation of two inputs $X$ and $Z$ is $X+Z$.
$4$. Therefore,the final output $Y$ is the $OR$ sum of the two previous outputs: $Y = \overline{A \cdot B} + \overline{C}$.

(D) The circuit consists of a $NAND$ gate followed by an $AND$ gate.
In the provided diagram,the output of the $NAND$ gate is connected to only one input terminal of the $AND$ gate.
An $AND$ gate is a multi-input logic gate that requires at least two input signals to perform its logical operation.
Since the $AND$ gate in this circuit is receiving only one input,it cannot function as a standard logic gate.
Therefore,the gate is not operational.

(C) The given circuit consists of an $OR$ gate and a $NAND$ gate whose outputs are fed into an $AND$ gate.
Let the inputs be $A$ and $B$.
The output of the $OR$ gate is $(A + B)$.
The output of the $NAND$ gate is $(\overline{A \cdot B})$.
These are the inputs to the final $AND$ gate.
Therefore,the final output $Y$ is given by:
$Y = (A + B) \cdot (\overline{A \cdot B})$
Using De Morgan's theorem,$\overline{A \cdot B} = \bar{A} + \bar{B}$.
$Y = (A + B) \cdot (\bar{A} + \bar{B})$
$Y = A \cdot \bar{A} + A \cdot \bar{B} + B \cdot \bar{A} + B \cdot \bar{B}$
Since $A \cdot \bar{A} = 0$ and $B \cdot \bar{B} = 0$,we have:
$Y = 0 + A \cdot \bar{B} + B \cdot \bar{A} + 0$
$Y = A \cdot \bar{B} + \bar{A} \cdot B$
This is the Boolean expression for an $XOR$ gate.

(D) The circuit consists of one $NAND$ gate,one $NOT$ gate,one $NOR$ gate,and one final $NOR$ gate.
Let the output of the top $NAND$ gate be $Y_1 = \overline{A \cdot B}$.
Let the output of the bottom branch ($NOT$ gate followed by $NOR$ gate) be $Y_2 = \overline{\overline{A} + B}$.
The final output $Y$ is the output of a $NOR$ gate with inputs $Y_1$ and $Y_2$,so $Y = \overline{Y_1 + Y_2} = \overline{(\overline{A \cdot B}) + (\overline{\overline{A} + B})}$.
Testing the options:
For $A=1, B=1$:
$Y_1 = \overline{1 \cdot 1} = 0$
$Y_2 = \overline{\overline{1} + 1} = \overline{0 + 1} = 0$
$Y = \overline{0 + 0} = 1$.
Thus,for $A=1, B=1$,the output $Y$ is $1$.

(C) $NAND$ gate has two inputs,say $A$ and $B$,and its output is given by $Y = \overline{A \cdot B}$.
If the two inputs are shorted,then $A = B$. Let this common input be $A$.
Substituting this into the $NAND$ expression,we get $Y = \overline{A \cdot A}$.
Since $A \cdot A = A$ in Boolean algebra,the expression becomes $Y = \overline{A}$.
The expression $Y = \overline{A}$ represents the operation of a $NOT$ gate.
Therefore,when the inputs of a $NAND$ gate are shorted,it functions as a $NOT$ gate.

(C) The given circuit consists of two $NOT$ gates,two $AND$ gates,and one $OR$ gate.
$1$. The input $A$ passes through a $NOT$ gate to become $\overline{A}$,and input $B$ goes directly to the first $AND$ gate. The output of this $AND$ gate is $\overline{A} \cdot B$.
$2$. The input $B$ passes through a $NOT$ gate to become $\overline{B}$,and input $A$ goes directly to the second $AND$ gate. The output of this $AND$ gate is $A \cdot \overline{B}$.
$3$. These two outputs are then fed into an $OR$ gate. The final output $Y$ is the sum of these two expressions: $Y = (\overline{A} \cdot B) + (A \cdot \overline{B})$.
This is the Boolean expression for an $XOR$ gate.

(D) The given circuit consists of two $NOR$ gates whose outputs are fed into an $OR$ gate. Let the inputs be $A$ and $B$. The output of the top $NOR$ gate is $C = \overline{A+B}$ and the output of the bottom $NOR$ gate is $D = \overline{A+B}$. These are fed into an $OR$ gate,so the final output is $Y = C + D = \overline{A+B} + \overline{A+B} = \overline{A+B}$.
The truth table for this combination is:

$A$	$B$	$C = \overline{A+B}$	$D = \overline{A+B}$	$Y = C + D$
$0$	$0$	$1$	$1$	$1$
$0$	$1$	$0$	$0$	$0$
$1$	$0$	$0$	$0$	$0$
$1$	$1$	$0$	$0$	$0$

Comparing this with the truth table of standard logic gates,the output $Y$ is $1$ only when both $A$ and $B$ are $0$,which is the characteristic behavior of a $NOR$ gate.

(A) The given circuit consists of a $NAND$ gate with inputs $A$ and $B$, a $NOT$ gate with input $C$, and an $OR$ gate that combines their outputs.
The Boolean expression for the output $Y$ is $Y = (\overline{A \cdot B}) + \overline{C}$.
Case $1$: When all inputs $A, B, C$ are low $(0, 0, 0)$:
$A = 0, B = 0 \implies A \cdot B = 0 \implies \overline{A \cdot B} = 1$.
$C = 0 \implies \overline{C} = 1$.
$Y = 1 + 1 = 1$.
Case $2$: When all inputs $A, B, C$ are high $(1, 1, 1)$:
$A = 1, B = 1 \implies A \cdot B = 1 \implies \overline{A \cdot B} = 0$.
$C = 1 \implies \overline{C} = 0$.
$Y = 0 + 0 = 0$.
Thus, the output sequence is $1, 0$.

(C) $NAND$ gate is a combination of an $AND$ gate followed by a $NOT$ gate.
The Boolean expression for a $NAND$ gate is $Y = \overline{A \cdot B}$.
Let us evaluate the output $Y$ for all possible input combinations of $A$ and $B$:
$1$. If $A = 0, B = 0$, then $A \cdot B = 0$, so $Y = \overline{0} = 1$.
$2$. If $A = 0, B = 1$, then $A \cdot B = 0$, so $Y = \overline{0} = 1$.
$3$. If $A = 1, B = 0$, then $A \cdot B = 0$, so $Y = \overline{0} = 1$.
$4$. If $A = 1, B = 1$, then $A \cdot B = 1$, so $Y = \overline{1} = 0$.
Comparing this with the given tables:
- Table $(P)$ represents an $OR$ gate.
- Table $(Q)$ represents an $XOR$ gate.
- Table $(R)$ represents an $AND$ gate.
- Table $(S)$ represents a $NAND$ gate.
Thus, the correct truth-table is $(S)$.

(D) The logic circuit consists of a $NOR$ gate followed by an $AND$ gate.
The inputs to the $NOR$ gate are $B$ and $C$,so its output is $\overline{B+C}$.
This output is then fed into an $AND$ gate along with input $A$.
Therefore,the final output $Y$ is given by the Boolean expression: $Y = A \cdot \overline{(B+C)}$.
For the output $Y$ to be ' $HIGH$ ' $(Y=1)$,both $A$ must be $1$ and $\overline{(B+C)}$ must be $1$.
$\overline{(B+C)} = 1$ implies that $(B+C) = 0$,which means both $B=0$ and $C=0$.
Thus,$Y=1$ when $A=1, B=0$,and $C=0$.

(A) Universal gate is a logic gate that can be used to implement any other logic gate or Boolean function without the need for any other type of gate.
Both $NAND$ and $NOR$ gates are classified as Universal gates.
Since $NOR$ is the only option provided that fits this definition,the correct answer is $NOR$.

(D) The given circuit consists of an $OR$ gate followed by a $NAND$ gate.
The output of the $OR$ gate is $Y = A + B$.
This output $Y$ is one of the inputs to the $NAND$ gate,and the other input is $C$.
The final output $D$ of the $NAND$ gate is $D = \overline{Y \cdot C} = \overline{(A + B) \cdot C}$.
Case $1$: $A=0, B=0, C=0$
$Y = 0 + 0 = 0$
$D = \overline{0 \cdot 0} = \overline{0} = 1$
Case $2$: $A=1, B=1, C=0$
$Y = 1 + 1 = 1$
$D = \overline{1 \cdot 0} = \overline{0} = 1$
Thus,the logic states of output $D$ are $1, 1$.

(C) The $OR$ gate is a basic logic gate that performs logical addition.
For an $OR$ gate with inputs $A$ and $B$,the output $Y$ is given by the Boolean expression $Y = A + B$.
According to the truth table of an $OR$ gate,the output is $1$ if at least one of the inputs is $1$.
Specifically,if input $A$ is $1$ or input $B$ is $1$,or if both inputs $A$ and $B$ are $1$,the output $Y$ will be $1$.

(B) Given the Boolean expression: $\overline{(A+B) \cdot(A \cdot B)}=1$.
Taking the complement on both sides,we get: $(A+B) \cdot(A \cdot B) = 0$.
We test the given options:
For option $B$ $(A=0, B=0)$: $(0+0) \cdot (0 \cdot 0) = 0 \cdot 0 = 0$. Since this satisfies the equation,the input is $(0, 0)$.
For option $A$ $(A=1, B=0)$: $(1+0) \cdot (1 \cdot 0) = 1 \cdot 0 = 0$. This also results in $0$,but let's check the original expression: $\overline{0} = 1$.
For option $D$ $(A=1, B=1)$: $(1+1) \cdot (1 \cdot 1) = 1 \cdot 1 = 1$. Then $\overline{1} = 0 \neq 1$.
Since $(0, 0)$ is a standard option provided,it is the correct choice.

(B) Let the inputs $A, B$ and $C$ be given to the circuit.
Gate-$I$ is an $AND$ gate,and Gate-$II$ is a $NAND$ gate.
The output of the $AND$ gate is $X = A \cdot B$.
The final output $Y$ is the $NAND$ of $X$ and $C$,so $Y = \overline{X \cdot C} = \overline{(A \cdot B) \cdot C}$.
When $A=1, B=1, C=1$ (all high):
$Y = \overline{(1 \cdot 1) \cdot 1} = \overline{1 \cdot 1} = \overline{1} = 0$.
When $A=0, B=0, C=0$ (all low):
$Y = \overline{(0 \cdot 0) \cdot 0} = \overline{0 \cdot 0} = \overline{0} = 1$.

(B) The given logic circuit consists of an $OR$ gate followed by an $AND$ gate. Let the output of the $OR$ gate be $X$. Then $X = A + B$.
The final output $Y$ is obtained from the $AND$ gate,where $Y = X \cdot C = (A + B) \cdot C$.
For the output $Y$ to be $1$,both inputs to the $AND$ gate must be $1$. Therefore,$X = 1$ and $C = 1$.
Since $X = A + B = 1$,at least one of $A$ or $B$ must be $1$.
Checking the options:
$A) A=1, B=0, C=0 \implies Y = (1+0) \cdot 0 = 0$
$B) A=1, B=0, C=1 \implies Y = (1+0) \cdot 1 = 1$
$C) A=1, B=1, C=0 \implies Y = (1+1) \cdot 0 = 0$
$D) A=0, B=0, C=1 \implies Y = (0+0) \cdot 1 = 0$
Thus,option $B$ is correct.

(C) The truth table for an Ex-$OR$ gate is as follows:
| Input $A$ | Input $B$ | Output $Y = A \oplus B$ |
| :--- | :--- | :--- |
| $0$ | $0$ | $0$ |
| $0$ | $1$ | $1$ |
| $1$ | $0$ | $1$ |
| $1$ | $1$ | $0$ |
The output equation for an Ex-$OR$ gate is $Y = A \oplus B = (\bar{A} \cdot B) + (A \cdot \bar{B})$.
Key points to remember:
$(1)$ The output is low $(0)$ when both inputs are the same.
$(2)$ The output is high $(1)$ when both inputs are different.

(A) Given inputs are $P = 0$,$Q = 1$,$R = 0$,and $S = 1$.
$1$. The output $X$ is obtained from an $AND$ gate with inputs $P$ and $Q$. Thus,$X = P \cdot Q = 0 \cdot 1 = 0$.
$2$. The input to the $NOT$ gate is the output of an $AND$ gate with inputs $R$ and $S$. Let this be $W$. So,$W = R \cdot S = 0 \cdot 1 = 0$. The output $Y$ is the $NOT$ of $W$,so $Y = \overline{W} = \overline{0} = 1$.
$3$. The output $Z$ is obtained from a $NOR$ gate with inputs $X$ and $Y$. Thus,$Z = \overline{X + Y} = \overline{0 + 1} = \overline{1} = 0$.
Therefore,the outputs are $X = 0$,$Y = 1$,and $Z = 0$.

(D) The Boolean expression for an Exclusive-$OR$ (Ex-$OR$) gate is given by:
$Y = A \oplus B = \bar{A} \cdot B + A \cdot \bar{B}$
The output of an Ex-$OR$ gate is $HIGH$ $(1)$ only when the inputs are at different logic levels. If both inputs are the same ($0,0$ or $1,1$),the output is $LOW$ $(0)$.
Evaluating the truth table:
$1$. For $A=0, B=0$: $Y = 0 \oplus 0 = 0$
$2$. For $A=0, B=1$: $Y = 0 \oplus 1 = 1$
$3$. For $A=1, B=0$: $Y = 1 \oplus 0 = 1$
$4$. For $A=1, B=1$: $Y = 1 \oplus 1 = 0$
Comparing this with the given tables,Table $(S)$ matches this truth table.

(B) The Boolean expression for a $NAND$ gate is $Y = \overline{A \cdot B}$.
$1$. For $A = 0, B = 1$: $Y = \overline{0 \cdot 1} = \overline{0} = 1$. Thus, $C = 1$.
$2$. For $A = 0, B = 0$: $Y = \overline{0 \cdot 0} = \overline{0} = 1$. Thus, $D = 1$.
$3$. For $A = 1, B = 0$: $Y = \overline{1 \cdot 0} = \overline{0} = 1$. Thus, $E = 1$.
$4$. For $A = 1, B = 1$: $Y = \overline{1 \cdot 1} = \overline{1} = 0$. Thus, $F = 0$.
Therefore, the values are $C = 1, D = 1, E = 1, F = 0$.

(D) The given circuit consists of two $NAND$ gates used as $NOT$ gates (since their inputs are shorted) followed by a $NAND$ gate.
Let the inputs be $A$ and $B$.
The output of the first $NAND$ gate (acting as $NOT$) is $\bar{A}$.
The output of the second $NAND$ gate (acting as $NOT$) is $\bar{B}$.
These two outputs are fed into a third $NAND$ gate.
The final output $Y$ is given by $Y = \overline{(\bar{A} \cdot \bar{B})}$.
Using De Morgan's theorem,$\overline{(\bar{A} \cdot \bar{B})} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$.
This is the Boolean expression for an $OR$ gate.
Therefore,the combination acts as an $OR$ gate.

(D) An $OR$ gate performs the logical addition operation. The Boolean expression for an $OR$ gate with inputs $A$ and $B$ is $Y = A + B$.
According to the truth table of an $OR$ gate:
- If $A = 0$ and $B = 0$,then $Y = 0$.
- If $A = 0$ and $B = 1$,then $Y = 1$.
- If $A = 1$ and $B = 0$,then $Y = 1$.
- If $A = 1$ and $B = 1$,then $Y = 1$.
Therefore,the output is $1$ if either input $A$ or input $B$ (or both) is $1$.

(D) For an $AND$ gate,the output is $1$ only if both inputs are $1$. For inputs $(0,1)$ and $(1,0)$,the output is $0$.
For an $OR$ gate,the output is $0$ only if both inputs are $0$. For inputs $(0,1)$ and $(1,0)$,the output is $1$.
For a $NAND$ gate,the output is $0$ only if both inputs are $1$. For inputs $(0,1)$ and $(1,0)$,the output is $1$.
For a $NOR$ gate,the output is $1$ only if both inputs are $0$. For inputs $(0,1)$ and $(1,0)$,the output is $0$.
Thus,both $AND$ and $NOR$ gates provide an output of $0$ for the given input combinations $(0,1)$ and $(1,0)$.

(C) The two $NAND$ gates whose inputs are joined together act as $NOT$ gates. Let the inputs be $A$ and $B$. The outputs of the first two $NAND$ gates are $y_1 = \overline{A}$ and $y_2 = \overline{B}$. These are fed into the third $NAND$ gate. The final output is $y = \overline{y_1 \cdot y_2} = \overline{\overline{A} \cdot \overline{B}}$. By De Morgan's theorem, $y = \overline{\overline{A}} + \overline{\overline{B}} = A + B$. This is the Boolean expression for an $OR$ gate. The truth table is as follows:
| $A$ | $B$ | $y_1$ | $y_2$ | $y$ |
|---|---|---|---|---|
| $0$ | $0$ | $1$ | $1$ | $0$ |
| $0$ | $1$ | $1$ | $0$ | $1$ |
| $1$ | $0$ | $0$ | $1$ | $1$ |
| $1$ | $1$ | $0$ | $0$ | $1$ |
Thus, the combination is equivalent to an $OR$ gate.

(D) The $X-OR$ (Exclusive-$OR$) gate is a digital logic gate that implements exclusive disjunction.
Its output is '$HIGH$' $(1)$ if and only if the inputs are different (i.e.,one input is $0$ and the other is $1$).
If both inputs are the same ($0,0$ or $1,1$),the output is '$LOW$' $(0)$.
Therefore,the $X-OR$ gate satisfies the condition of giving a '$HIGH$' output only when its two input terminals are at different logic levels.

(A) Let us analyze each gate based on its truth table:
$(I)$ The gate is a $NAND$ gate with inputs $1, 1$. The output is $Y = \overline{A \cdot B} = \overline{1 \cdot 1} = \overline{1} = 0$.
$(II)$ The gate is a $NOR$ gate with inputs $0, 0$. The output is $Y = \overline{A + B} = \overline{0 + 0} = \overline{0} = 1$.
$(III)$ The gate is a $NAND$ gate with inputs $0, 1$. The output is $Y = \overline{A \cdot B} = \overline{0 \cdot 1} = \overline{0} = 1$.
$(IV)$ The gate is an $EX$-$NOR$ gate with inputs $1, 0$. The output is $Y = A \odot B = 1 \odot 0 = 0$.
Thus,gates $II$ and $III$ provide an output of ' $1$ '.

(B) The given circuit consists of an $OR$ gate followed by a $NOT$ gate, which together form a $NOR$ gate.
Let the output of the $OR$ gate be $Y'$. The output of the $NOR$ gate is $Y = \overline{A + B}$.
For the final output $Y$ to be '$1$', the input to the $NOT$ gate must be '$0$'.
This means the output of the $OR$ gate $Y' = A + B$ must be '$0$'.
An $OR$ gate gives an output of '$0$' only when both its inputs are '$0$'.
Therefore, $A = 0$ and $B = 0$.

(C) In the given circuit,we have a combination of a $NOT$ gate and an $OR$ gate.
For the $NOT$ gate,the input is $A$,so the output is $X = \bar{A}$.
This output $X$ acts as one of the inputs to the $OR$ gate,while $B$ is the other input.
For the $OR$ gate,the output $Y$ is the sum of its inputs:
$Y = X + B$
Substituting the value of $X$ from the $NOT$ gate:
$Y = \bar{A} + B$
Thus,the Boolean expression for the given circuit is $Y = \bar{A} + B$.

(B) The logic operation for an $AND$ gate is defined by the Boolean expression $Y = A \cdot B$.
This means the output $Y$ is $1$ only if both inputs $A$ and $B$ are $1$. Otherwise,the output $Y$ is $0$.
Let us evaluate the given entries:
Entry $1$: $A=0, B=1$. $Y = 0 \cdot 1 = 0$. This is correct.
Entry $2$: $A=1, B=0$. $Y = 1 \cdot 0 = 0$. This is correct.
Entry $3$: $A=1, B=1$. $Y = 1 \cdot 1 = 1$. This is correct.
Entry $4$: $A=0, B=0$. $Y = 0 \cdot 0 = 0$. The table shows $Y=1$,which is incorrect.
Therefore,entries $1, 2,$ and $3$ are correct.

(D) The given circuit consists of two $NOT$ gates ($G_1$ and $G_2$) and one $NOR$ gate $(G_3)$.
$1$. The input $A$ passes through the $NOT$ gate $G_1$,so the output at $C$ is $C = \overline{A}$.
$2$. The input $B$ passes through the $NOT$ gate $G_2$,so the output at $D$ is $D = \overline{B}$.
$3$. These outputs $C$ and $D$ are fed into the $NOR$ gate $G_3$. The output $Y$ of a $NOR$ gate is the complement of the $OR$ of its inputs.
$4$. Therefore,$Y = \overline{C + D}$.
$5$. Substituting the values of $C$ and $D$,we get $Y = \overline{\overline{A} + \overline{B}}$.
$6$. According to De Morgan's theorem,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$.
$7$. The Boolean expression $Y = A \cdot B$ represents an $AND$ gate.
Thus,the resultant gate is an $AND$ gate with the expression $A \cdot B$.

(A) The logic gate for which the output is high $(1)$ if and only if all inputs are high $(1)$ is the $AND$ gate.
For an $AND$ gate with inputs $A$ and $B$,the output $Y$ is given by $Y = A \cdot B$.
If $A = 1$ and $B = 1$,then $Y = 1 \cdot 1 = 1$.
For any other combination of inputs,the output is low $(0)$.

(B) To form an $AND$ gate using $NAND$ gates,we first pass the inputs $A$ and $B$ through a $NAND$ gate to get $\overline{A \cdot B}$.
Then,we pass this output through another $NAND$ gate configured as a $NOT$ gate (by shorting its inputs together).
Let the output of the first $NAND$ gate be $X = \overline{A \cdot B}$.
The second $NAND$ gate acts as a $NOT$ gate,so its output $Y = \overline{X \cdot X} = \overline{X} = \overline{\overline{A \cdot B}} = A \cdot B$.
Thus,two $NAND$ gates are required to form an $AND$ gate.

(C) $NAND$ gate is defined as an $AND$ gate followed by a $NOT$ gate. The Boolean expression for a $NAND$ gate with inputs $A$ and $B$ is $Y = \overline{A \cdot B}$.
When the two inputs are shorted,$A = B$. Substituting this into the expression,we get $Y = \overline{A \cdot A} = \overline{A}$.
This is the Boolean expression for a $NOT$ gate.
The truth table for this configuration is:

$A$	$Y = \overline{A \cdot A}$
$0$	$1$
$1$	$0$

Since the output is the inverse of the input,the resulting gate is a $NOT$ gate.

(C) truth table defines the output of a logic gate for all possible input combinations.
For the given table:
- When $A=0, B=0$,output $Y=1$.
- When $A=0, B=1$,output $Y=1$.
- When $A=1, B=0$,output $Y=1$.
- When $A=1, B=1$,output $Y=0$.
This behavior corresponds to the $NAND$ gate,which is the inverse of the $AND$ gate. The Boolean expression for a $NAND$ gate is $Y = \overline{A \cdot B}$.

(D) The given circuit consists of two $NOR$ gates acting as $NOT$ gates, followed by a $NOR$ gate. Let the inputs be $A$ and $B$.
The first two gates are $NOR$ gates with both inputs tied together. The output of the first gate is $\overline{A+A} = \bar{A}$.
The output of the second gate is $\overline{B+B} = \bar{B}$.
These outputs are fed into the final $NOR$ gate. The output $Y$ is given by:
$Y = \overline{\bar{A} + \bar{B}}$
Using De Morgan's theorem, $\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
Thus, the circuit represents an $AND$ gate.

$A$	$B$	$Y = A \cdot B$
$0$	$0$	$0$
$0$	$1$	$0$
$1$	$0$	$0$
$1$	$1$	$1$

(D) The given logic circuit consists of a $NOR$ gate and an $AND$ gate whose output is fed into a $NAND$ gate. Let the inputs be $A$ and $B$.
$1$. The output of the $NOR$ gate is $\overline{A+B}$.
$2$. The output of the $AND$ gate is $A \cdot B$.
$3$. These two outputs are fed into a $NAND$ gate. The final output $Y$ is given by:
$Y = \overline{(\overline{A+B}) \cdot (A \cdot B)}$
Using De Morgan's law,$\overline{X \cdot Y} = \overline{X} + \overline{Y}$:
$Y = \overline{(\overline{A+B})} + \overline{(A \cdot B)}$
$Y = (A+B) + (\overline{A} + \overline{B})$
$Y = (A + \overline{A}) + (B + \overline{B})$
Since $A + \overline{A} = 1$ and $B + \overline{B} = 1$:
$Y = 1 + 1 = 1$
Thus,the output $Y$ is always $1$ for all combinations of inputs $A$ and $B$. Therefore,the truth table will have $1$ in all rows for output $Y$. Comparing this with the given options,option $(D)$ is correct.

(B) The given circuit consists of two $NAND$ gates connected in series,where the second $NAND$ gate acts as a $NOT$ gate because its inputs are shorted together.
Let the inputs to the first $NAND$ gate be $A$ and $B$. The output of the first $NAND$ gate is $\overline{A \cdot B}$.
This output is fed as the input to the second $NAND$ gate. Since both inputs of the second $NAND$ gate are connected to the same signal,its output $Y$ is given by:
$Y = \overline{(\overline{A \cdot B}) \cdot (\overline{A \cdot B})}$
Using the property of Boolean algebra $\overline{X \cdot X} = \overline{X}$,we get:
$Y = \overline{(\overline{A \cdot B})} = A \cdot B$
Thus,the circuit performs the operation of an $AND$ gate.

(B) The circuit consists of two cross-coupled $NAND$ gates, forming an $S-R$ latch.
Let the inputs be $S=1$ and $R=0$.
The output $P$ is given by $P = \overline{1 \cdot Q} = \overline{Q}$.
The output $Q$ is given by $Q = \overline{0 \cdot P} = \overline{0} = 1$.
Since $Q=1$, substituting this into the equation for $P$ gives $P = \overline{1} = 0$.
Thus, the stable state of the circuit is $P=0$ and $Q=1$.

(A) The circuit consists of two cross-coupled $NOR$ gates forming an $SR$ latch.
For a $NOR$ gate,the output $Y = \overline{A+B}$.
Let the top gate be Gate $1$ and the bottom gate be Gate $2$.
The inputs to Gate $1$ are $1$ and $Q$. Thus,$P = \overline{1+Q} = 0$.
The inputs to Gate $2$ are $0$ and $P$. Thus,$Q = \overline{0+P}$.
Substituting $P=0$ into the equation for $Q$,we get $Q = \overline{0+0} = \overline{0} = 1$.
Therefore,$P=0$ and $Q=1$.

(A) Given $ A=1 $ and $ B=0 $.
In Boolean algebra,the complement of $ A $ is $ \bar{A} = \bar{1} = 0 $.
Now,substitute the values into the expression $ \bar{A}+B $:
$ \bar{A}+B = 0 + 0 = 0 $.
Since $ B=0 $,we can see that $ \bar{A}+B = B $.
Therefore,the correct option is $ A $.

(D) Let the inputs of the $OR$ gate be $A$ and $B$. The output of the $OR$ gate is $Y = A + B$.
This output $Y$ is connected to both inputs of a $NAND$ gate. Let the inputs of the $NAND$ gate be $X_1$ and $X_2$,where $X_1 = X_2 = Y = A + B$.
The output of a $NAND$ gate with inputs $X_1$ and $X_2$ is given by $\overline{X_1 \cdot X_2}$.
Substituting $X_1 = X_2 = A + B$,the final output $Y^{\prime}$ is $\overline{(A + B) \cdot (A + B)}$.
Using the Boolean identity $X \cdot X = X$,we get $Y^{\prime} = \overline{A + B}$.
The expression $\overline{A + B}$ represents the Boolean operation of a $NOR$ gate.
Therefore,the combination acts as a $NOR$ gate.

(D) Let us analyze the truth table provided:
$1$. When $A=1, B=0$,Output = $1$.
$2$. When $A=1, B=1$,Output = $0$.
$3$. When $A=0, B=1$,Output = $1$.
$4$. When $A=0, B=0$,Output = $0$.
Comparing this with standard logic gates:
- For an $XOR$ gate,the output is $1$ only when the inputs are different $(A \neq B)$.
- In this table,the output is $1$ when $(A=1, B=0)$ or $(A=0, B=1)$,and $0$ when $(A=1, B=1)$ or $(A=0, B=0)$.
- This behavior perfectly matches the $XOR$ gate (Exclusive-$OR$ gate),which follows the Boolean expression $Y = A \oplus B$.

(C) The given circuit consists of two $OR$ gates followed by an $AND$ gate.
$1$. The inputs to the first $OR$ gate are $A$ and $B$. Therefore,its output is $(A+B)$.
$2$. The inputs to the second $OR$ gate are $A$ and $C$. Therefore,its output is $(A+C)$.
$3$. These two outputs $(A+B)$ and $(A+C)$ are fed as inputs to the final $AND$ gate.
$4$. The output $Y$ of the $AND$ gate is the product of its inputs: $Y = (A+B) \cdot (A+C)$.

(A) The given circuit consists of a $NOR$ gate followed by a $NOR$ gate with its inputs shorted together,which acts as a $NOT$ gate.
Let the inputs to the first $NOR$ gate be $A$ and $B$. The output of the first $NOR$ gate is $Y' = \overline{A+B}$.
This output $Y'$ is fed into both inputs of the second $NOR$ gate. The output of the second $NOR$ gate is $Y = \overline{Y' + Y'} = \overline{Y'} = \overline{\overline{A+B}}$.
Using the law of double negation,$\overline{\overline{X}} = X$,we get $Y = A+B$.
This is the Boolean expression for an $OR$ gate.

(A) For circuit $A$: The inputs to the $NAND$ gate are $1$ and $1$,so its output is $0$. This $0$ passes through a $NOT$ gate,becoming $1$. The $OR$ gate receives inputs $1$ and $0$,resulting in an output of $1$.
For circuit $B$: The inputs $0$ and $1$ pass through $NOT$ gates,becoming $1$ and $0$ respectively. These are inputs to a $NAND$ gate. Since $1 \text{ AND } 0 = 0$,the $NAND$ gate (which is $NOT$-$AND$) outputs $1$.
For circuit $C$: The $OR$ gate receives inputs $1$ and $1$,resulting in an output of $1$. However,the diagram shows the $OR$ gate is followed by a $NOT$ bubble (making it a $NOR$ gate),so the output is $0$. This $0$ and the input $1$ are fed into an $AND$ gate,resulting in an output of $0$.
Thus,the outputs are $1, 1, 0$ respectively.

(C) The given circuit consists of an $OR$ gate followed by an $AND$ gate. The Boolean expression for the output $Y$ is given by $Y = (A + B) \cdot C$.
For the output to be $Y = 1$,both inputs to the $AND$ gate must be $1$. Therefore,$(A + B) = 1$ and $C = 1$.
Checking the options:
For option $C$: $A = 1, B = 0, C = 1$.
Substituting these values: $Y = (1 + 0) \cdot 1 = 1 \cdot 1 = 1$.
Thus,the inputs $A = 1, B = 0, C = 1$ result in an output $Y = 1$.

(D) The given truth table shows that the output $Y$ is $1$ when either $A$ or $B$ or both are $0$,and the output $Y$ is $0$ only when both inputs $A$ and $B$ are $1$.
This behavior corresponds to the Boolean expression $Y = \overline{A \cdot B}$.
This is the characteristic truth table of a $NAND$ gate,which is an $AND$ gate followed by a $NOT$ gate.

(D) universal gate is a logic gate that can be used to implement any other logic gate without the need for any other type of gate. The $NAND$ gate and the $NOR$ gate are known as universal gates. In the given options,the $NAND$ gate is a universal gate because any basic logic gate ($AND$,$OR$,$NOT$) can be constructed using only $NAND$ gates.

(A) Let the inputs be $A=0, B=1, C=0$.
$1$. The top $NAND$ gate has inputs $A=0$ and $B=1$. Its output is $\overline{0 \cdot 1} = \overline{0} = 1$.
$2$. The second $NAND$ gate has inputs $A=0$ and the output of the first $NAND$ gate $(1)$. Its output $y_1 = \overline{0 \cdot 1} = \overline{0} = 1$.
$3$. The bottom $NOR$ gate has inputs $B=1$ and $C=0$. Its output is $\overline{1 + 0} = \overline{1} = 0$.
$4$. The second $NOR$ gate has inputs $C=0$ and the output of the first $NOR$ gate $(0)$. Its output $y_2 = \overline{0 + 0} = \overline{0} = 1$.
$5$. The final $OR$ gate has inputs $y_1=1$ and $y_2=1$. Its output $y_3 = 1 + 1 = 1$.
Thus,the values are $y_1=1, y_2=1, y_3=1$.