Boolean Algebra and Logic Gates Questions in English

(B) By observing the provided symbols:
$(1)$ represents a $NAND$ gate.
$(2)$ represents an $XOR$ gate.
$(3)$ represents a $NOR$ gate.
$(4)$ represents a $NOT$ gate.
Therefore,the $XOR$ gate is represented by symbol $2$ and the $NOR$ gate is represented by symbol $3$.
Thus,the correct option is $B$.

(C) By observing the given logic gate symbols:
$(1)$ represents an $OR$ gate.
$(2)$ represents an $AND$ gate.
$(3)$ represents a $NOR$ gate.
$(4)$ represents a $NAND$ gate.
Therefore,the symbols for $OR$,$NOR$,and $NAND$ gates are $1$,$3$,and $4$ respectively.
Thus,the correct option is $C$.

(B) The truth table provided is:
$A=0, B=0 \implies X=0$
$A=0, B=1 \implies X=1$
$A=1, B=0 \implies X=1$
$A=1, B=1 \implies X=1$
This behavior corresponds to the $OR$ gate,where the output $X$ is $1$ if at least one of the inputs $A$ or $B$ is $1$. The Boolean expression for an $OR$ gate is $X = A + B$.

(B) The given truth table is:
$A=0, B=0 \implies Y=1$
$A=1, B=0 \implies Y=0$
$A=0, B=1 \implies Y=0$
$A=1, B=1 \implies Y=0$
For a $NOR$ gate,the output is defined by the Boolean expression $Y = \overline{A + B}$.
Calculating for each case:
$1$. For $A=0, B=0$: $Y = \overline{0+0} = \overline{0} = 1$.
$2$. For $A=1, B=0$: $Y = \overline{1+0} = \overline{1} = 0$.
$3$. For $A=0, B=1$: $Y = \overline{0+1} = \overline{1} = 0$.
$4$. For $A=1, B=1$: $Y = \overline{1+1} = \overline{1} = 0$.
This matches the given truth table. Therefore,the correct option is $B$.

(C) The truth table represents the output $Y$ for inputs $A$ and $B$ as follows:
For $A=0, B=0$,$Y=1$.
For $A=0, B=1$,$Y=0$.
For $A=1, B=0$,$Y=0$.
For $A=1, B=1$,$Y=1$.
This truth table corresponds to the $XNOR$ gate,which outputs $1$ only when both inputs are the same.
The Boolean expression for an $XNOR$ gate is $Y = A \odot B = \overline{A \oplus B} = \bar{A}\bar{B} + AB$.
Checking the values:
$1$. For $(0, 0): Y = \bar{0}\cdot\bar{0} + 0\cdot 0 = 1\cdot 1 + 0 = 1$.
$2$. For $(0, 1): Y = \bar{0}\cdot\bar{1} + 0\cdot 1 = 1\cdot 0 + 0 = 0$.
$3$. For $(1, 0): Y = \bar{1}\cdot\bar{0} + 1\cdot 0 = 0\cdot 1 + 0 = 0$.
$4$. For $(1, 1): Y = \bar{1}\cdot\bar{1} + 1\cdot 1 = 0\cdot 0 + 1 = 1$.
Thus,the correct option is $C$.

(D) The given circuit consists of an $OR$ gate $(G_1)$ followed by a $NAND$ gate $(G_2)$.
Let the output of the $OR$ gate be $Y$. Then $Y = A + B$.
The final output $D$ is the output of the $NAND$ gate with inputs $Y$ and $C$, so $D = \overline{Y \cdot C} = \overline{(A + B) \cdot C}$.
Case $1$: $A = 0, B = 0, C = 0$.
$Y = A + B = 0 + 0 = 0$.
$D = \overline{Y \cdot C} = \overline{0 \cdot 0} = \overline{0} = 1$.
Case $2$: $A = 1, B = 1, C = 0$.
$Y = A + B = 1 + 1 = 1$.
$D = \overline{Y \cdot C} = \overline{1 \cdot 0} = \overline{0} = 1$.
Thus, the output states are $1, 1$.

(B) Boolean algebra is a branch of algebra in which the values of the variables are the truth values $true$ and $false$,usually denoted by $1$ and $0$ respectively.
It is essentially based on logic,as it deals with logical operations such as $AND$,$OR$,and $NOT$ to manipulate these truth values.
Therefore,the correct option is $B$.

(A) logic gate is an electronic circuit that makes logic decisions.
$A$ digital logic gate is an electronic device that makes logical decisions based on the different combinations of digital signals present on its inputs.
Digital logic gates may have more than one input ($A, B, C$,etc.) but generally have only one digital output $(Q)$.
Individual logic gates can be connected together to form combinational or sequential circuits or larger logic gate functions.
$A$ logic gate is an elementary building block of a digital circuit.
Most logic gates have two inputs and one output.
At any given moment,every terminal is in one of the two binary conditions: low $(0)$ or high $(1)$,represented by different voltage levels.
Therefore,the correct answer is $A$.

(D) The Boolean expression for an $AND$ gate is $R = P \cdot Q$.
Checking the values from the truth table:
For $P=1, Q=1$,$R = 1 \cdot 1 = 1$.
For $P=1, Q=0$,$R = 1 \cdot 0 = 0$.
For $P=0, Q=1$,$R = 0 \cdot 1 = 0$.
For $P=0, Q=0$,$R = 0 \cdot 0 = 0$.
Since the calculated values match the given truth table,the gate is an $AND$ gate.

(B) To form an $AND$ gate using $NAND$ gates,we need two $NAND$ gates.
First,the inputs $A$ and $B$ are fed into the first $NAND$ gate,which produces an output of $\overline{A \cdot B}$.
This output is then fed into the second $NAND$ gate,which acts as a $NOT$ gate (since its inputs are shorted together).
The final output $Y$ is given by:
$Y = \overline{(\overline{A \cdot B}) \cdot (\overline{A \cdot B})} = \overline{\overline{A \cdot B}} = A \cdot B$
Thus,two $NAND$ gates are required to realize an $AND$ gate.

(C) Let us analyze the output for each gate with the given inputs:
$(A)$ For an $AND$ gate,the output $Y = A \cdot B$. With inputs $1$ and $0$,$Y = 1 \cdot 0 = 0$.
$(B)$ For a $NOR$ gate,the output $Y = \overline{A + B}$. With inputs $0$ and $1$,$Y = \overline{0 + 1} = \overline{1} = 0$.
$(C)$ For a $NAND$ gate,the output $Y = \overline{A \cdot B}$. With inputs $0$ and $1$,$Y = \overline{0 \cdot 1} = \overline{0} = 1$.
$(D)$ For an $XNOR$ gate,the output $Y = A \odot B$ (or $\overline{A \oplus B}$). With inputs $0$ and $1$,$Y = \overline{0 \oplus 1} = \overline{1} = 0$.
Thus,the $NAND$ gate produces an output of $1$.

(A) $NAND$ gate is defined as the logical combination of an $AND$ gate followed by a $NOT$ gate.
Mathematically,if the inputs are $A$ and $B$,the output of the $AND$ gate is $Y' = A \cdot B$.
Passing this through a $NOT$ gate results in the final output $Y = \overline{A \cdot B}$.
Therefore,the correct representation is an $AND$ gate followed by a $NOT$ gate.

(C) $NOT$ gate is a logic gate that implements logical negation. It inverts the input signal.
If the input $A = 0$,the output $X = 1$.
If the input $A = 1$,the output $X = 0$.
This behavior is represented by the Boolean expression $X = \bar{A}$.
Comparing this with the given truth table,it matches the operation of a $NOT$ gate.

(B) For an $AND$ gate,the output $Y$ is defined by the Boolean expression $Y = A \cdot B$.
This means the output is $1$ if and only if both inputs $A$ and $B$ are $1$.
If any input is $0$,the output will be $0$.
Therefore,the output is $1$ only when $A = 1$ and $B = 1$.

(B) $NOR$ gate is a combination of an $OR$ gate followed by a $NOT$ gate.
$1$. The output of an $OR$ gate for inputs $A$ and $B$ is given by $Y = A + B$.
$2$. The $NOT$ gate inverts this output,resulting in $C = \overline{Y}$.
$3$. Substituting the $OR$ gate output into the $NOT$ gate expression,we get $C = \overline{A + B}$.
Therefore,the correct option is $B$.

$(A)$ The given symbol shows a $NAND$ gate where the two inputs $A$ and $B$ are shorted together to form a single input.
For a $NAND$ gate, the output is $Y = \overline{A \cdot B}$.
Since the inputs are shorted, $A = B = X$.
Therefore, the output becomes $Y = \overline{X \cdot X} = \overline{X}$.
This is the Boolean expression for a $NOT$ gate.
Thus, the combination acts as a $NOT$ gate.

(A) The provided diagram shows the standard symbol for an $AND$ gate.
An $AND$ gate is a digital logic gate that implements logical conjunction; it behaves according to the truth table where the output is high $(1)$ only if all its inputs are high $(1)$.

(B) The provided symbol consists of an $OR$ gate followed by a small circle (inversion bubble) at the output.
This combination of an $OR$ gate and a $NOT$ gate is known as a $NOR$ gate.
Therefore,the correct option is $B$.

(C) The given circuit consists of an $OR$ gate followed by an $AND$ gate. The output of the $OR$ gate is $(A + B)$. This output is then fed into an $AND$ gate along with input $C$. Thus,the final Boolean expression for the output $Y$ is $Y = (A + B) \cdot C$.
To get an output $Y = 1$,both inputs to the $AND$ gate must be $1$. This means $(A + B) = 1$ and $C = 1$.
For $(A + B) = 1$,at least one of $A$ or $B$ must be $1$.
Checking the options:
- For option $A$: $A=0, B=1, C=0 \implies Y = (0+1) \cdot 0 = 0$.
- For option $B$: $A=1, B=0, C=0 \implies Y = (1+0) \cdot 0 = 0$.
- For option $C$: $A=1, B=0, C=1 \implies Y = (1+0) \cdot 1 = 1$.
- For option $D$: $A=1, B=1, C=0 \implies Y = (1+1) \cdot 0 = 0$.
Thus,the correct input is $A = 1, B = 0, C = 1$.

(C) The output $Y$ of a $NAND$ gate for inputs $A$ and $B$ is given by the Boolean expression $Y = \overline{A \cdot B}$.
For the output to be $0$,we must have $\overline{A \cdot B} = 0$,which implies $A \cdot B = 1$.
This condition is satisfied only when both inputs $A$ and $B$ are $1$.
Therefore,if $A = 1$ and $B = 1$,then $Y = \overline{1 \cdot 1} = \overline{1} = 0$.

(C) For a $NOR$ gate, the output is high $(1)$ only when all inputs are low $(0)$.
If inputs are $A$ and $B$, the output $Y = \overline{A+B}$.
If $A=0$ and $B=0$, then $Y = \overline{0+0} = \overline{0} = 1$.
Thus, the correct gate is a $NOR$ gate.
Therefore, the correct option is $C$.

(B) The output $Y$ of an $OR$ gate is given by the Boolean expression $Y = A + B$.
According to the logic of the $OR$ gate,the output is $1$ if at least one of the inputs ($A$ or $B$) is $1$.
Therefore,if either input is $1$ or both inputs are $1$,the output is $1$.

(A) The given figure shows an $AND$ gate followed by a small circle at the output,which represents an inversion ($NOT$ operation).
An $AND$ gate combined with a $NOT$ gate is known as a $NAND$ gate.
Therefore,the correct option is $A$.

(A) First,convert the binary numbers to decimal:
$(100010)_2 = (1 \times 2^5) + (0 \times 2^4) + (0 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (0 \times 2^0) = 32 + 0 + 0 + 0 + 2 + 0 = (34)_{10}$
$(11011)_2 = (1 \times 2^4) + (1 \times 2^3) + (0 \times 2^2) + (1 \times 2^1) + (1 \times 2^0) = 16 + 8 + 0 + 2 + 1 = (27)_{10}$
Sum in decimal: $(34)_{10} + (27)_{10} = (61)_{10}$
Now,convert $(61)_{10}$ back to binary:

$61 \div 2 = 30$	Remainder $1$ $(LSD)$
$30 \div 2 = 15$	Remainder $0$
$15 \div 2 = 7$	Remainder $1$
$7 \div 2 = 3$	Remainder $1$
$3 \div 2 = 1$	Remainder $1$
$1 \div 2 = 0$	Remainder $1$ $(MSD)$

Reading remainders from bottom to top,we get $(111101)_2$.

(D) For a $NAND$ gate,the output $C$ is given by the Boolean expression $C = \overline{A \cdot B}$.
Checking the values:
$1$. For $A = 0, B = 0$: $C = \overline{0 \cdot 0} = \overline{0} = 1$.
$2$. For $A = 0, B = 1$: $C = \overline{0 \cdot 1} = \overline{0} = 1$.
$3$. For $A = 1, B = 0$: $C = \overline{1 \cdot 0} = \overline{0} = 1$.
$4$. For $A = 1, B = 1$: $C = \overline{1 \cdot 1} = \overline{1} = 0$.
Comparing these results with the given table,the output matches the $NAND$ gate logic.

(B) The output of the $OR$ gate $(G_1)$ is $(A + B)$.
The output of the $NAND$ gate $(G_2)$ is $\overline{A \cdot B}$.
These two outputs are fed into the $AND$ gate $(G_3)$,so the final output $Y$ is:
$Y = (A + B) \cdot (\overline{A \cdot B})$
Using De Morgan's theorem,$\overline{A \cdot B} = \bar{A} + \bar{B}$.
Substituting this into the equation:
$Y = (A + B) \cdot (\bar{A} + \bar{B})$
$Y = A \cdot \bar{A} + A \cdot \bar{B} + B \cdot \bar{A} + B \cdot \bar{B}$
Since $A \cdot \bar{A} = 0$ and $B \cdot \bar{B} = 0$,the expression simplifies to:
$Y = 0 + A \cdot \bar{B} + \bar{A} \cdot B + 0$
$Y = A \cdot \bar{B} + \bar{A} \cdot B$
This is the Boolean expression for an $XOR$ gate.

(D) The given circuit consists of two $NOT$ gates ($G_1$ and $G_2$) at the inputs $A$ and $B$,followed by a $NOR$ gate $(G_3)$ and another $NOT$ gate $(G_4)$.
$1$. The outputs of $G_1$ and $G_2$ are $\bar{A}$ and $\bar{B}$ respectively.
$2$. These are fed into the $NOR$ gate $G_3$,which produces the output: $Y' = \overline{\bar{A} + \bar{B}}$.
$3$. Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
$4$. This output $Y'$ is then passed through the final $NOT$ gate $G_4$,resulting in $Y = \overline{Y'} = \overline{A \cdot B}$.
$5$. The expression $\overline{A \cdot B}$ represents the output of a $NAND$ gate.
Therefore,option $(d)$ is correct.

(B) The circuit consists of two $NAND$ gates whose outputs $X$ and $Y$ are fed into a $NOR$ gate to produce output $Z$.
Inputs are $A, B, C$. The outputs of the $NAND$ gates are $X = \overline{AB}$ and $Y = \overline{BC}$.
The output of the $NOR$ gate is $Z = \overline{X + Y}$.
Substituting $X$ and $Y$: $Z = \overline{\overline{AB} + \overline{BC}}$.
Using De Morgan's Law: $Z = \overline{\overline{AB}} \cdot \overline{\overline{BC}} = AB \cdot BC = ABC$.
Since the output $Z = ABC$ corresponds to a $3$-input $AND$ gate,the system is equivalent to an $AND$ gate.

$A$	$B$	$C$	$X = \overline{AB}$	$Y = \overline{BC}$	$Z = \overline{X+Y}$
$0$	$0$	$0$	$1$	$1$	$0$
$1$	$0$	$0$	$1$	$1$	$0$
$0$	$0$	$1$	$1$	$1$	$0$
$1$	$0$	$1$	$1$	$1$	$0$
$0$	$1$	$0$	$1$	$1$	$0$
$1$	$1$	$0$	$0$	$1$	$0$
$0$	$1$	$1$	$1$	$0$	$0$
$1$	$1$	$1$	$0$	$0$	$1$

(A) From the given waveforms,the following truth table can be constructed:

Time interval	Input $A$	Input $B$	Output $Y$
$0 \to T_1$	$0$	$0$	$0$
$T_1 \to T_2$	$0$	$1$	$0$
$T_2 \to T_3$	$1$	$0$	$0$
$T_3 \to T_4$	$1$	$1$	$1$

Comparing this truth table with the standard logic gates,we observe that the output $Y$ is $1$ only when both inputs $A$ and $B$ are $1$. This behavior corresponds to an $AND$ gate.

(D) In a negative logic system,the lower voltage level $(5 \text{ V})$ is assigned the logic state '$0$' and the higher voltage level $(10 \text{ V})$ is assigned the logic state '$1$'.
However,in negative logic,the interpretation is inverted: a low signal $(5 \text{ V})$ represents logic '$1$' and a high signal $(10 \text{ V})$ represents logic '$0$'.
Let us divide the time axis into intervals of $5 \text{ } \mu\text{s}$ each.
The voltage levels for each interval are:
$0-5 \text{ } \mu\text{s}: 5 \text{ V} \rightarrow \text{Logic } 1$
$5-10 \text{ } \mu\text{s}: 10 \text{ V} \rightarrow \text{Logic } 0$
$10-15 \text{ } \mu\text{s}: 5 \text{ V} \rightarrow \text{Logic } 1$
$15-20 \text{ } \mu\text{s}: 10 \text{ V} \rightarrow \text{Logic } 0$
$20-25 \text{ } \mu\text{s}: 10 \text{ V} \rightarrow \text{Logic } 0$
$25-30 \text{ } \mu\text{s}: 5 \text{ V} \rightarrow \text{Logic } 1$
$30-35 \text{ } \mu\text{s}: 10 \text{ V} \rightarrow \text{Logic } 0$
$35-40 \text{ } \mu\text{s}: 5 \text{ V} \rightarrow \text{Logic } 1$
$40-45 \text{ } \mu\text{s}: 5 \text{ V} \rightarrow \text{Logic } 1$
$45-50 \text{ } \mu\text{s}: 5 \text{ V} \rightarrow \text{Logic } 1$
Combining these,we get the sequence: $1010010111$.

(B) Digital signals are signals that represent data as a sequence of discrete values at any given time.
$(i)$ Unlike analog signals,digital signals do not provide continuous values; they are discontinuous.
$(ii)$ They represent information in discrete steps (e.g.,$0$ and $1$).
$(iii)$ Digital signals are typically represented using a binary system ($0$ and $1$).
$(iv)$ While they can be used to encode decimal numbers,the fundamental nature of a digital signal is binary.
Therefore,statements $(i), (ii),$ and $(iii)$ are correct descriptions of digital signals.

(C) Let us analyze each logic gate:
$(a)$ This is a $NOR$ gate with inputs $1$ and $1$. The output of a $NOR$ gate is $Y = \overline{A+B}$. For inputs $1, 1$,$Y = \overline{1+1} = \overline{1} = 0$.
$(b)$ This is a $NOR$ gate with inputs $0$ and $1$. For inputs $0, 1$,$Y = \overline{0+1} = \overline{1} = 0$.
$(c)$ This is a $NAND$ gate with inputs $0$ and $1$. The output of a $NAND$ gate is $Y = \overline{A \cdot B}$. For inputs $0, 1$,$Y = \overline{0 \cdot 1} = \overline{0} = 1$.
$(d)$ This is an $XOR$ gate with inputs $0$ and $0$. The output of an $XOR$ gate is $Y = A \oplus B$. For inputs $0, 0$,$Y = 0 \oplus 0 = 0$.
Therefore,only gate $(c)$ has an output of $1$.

(B) In Boolean algebra,the $OR$ operation between a variable and its complement always results in $1$.
Case $1$: If $A = 1$,then $\bar{A} = 0$. Therefore,$A + \bar{A} = 1 + 0 = 1$.
Case $2$: If $A = 0$,then $\bar{A} = 1$. Therefore,$A + \bar{A} = 0 + 1 = 1$.
Thus,for any value of $A$,$A + \bar{A} = 1$.

(C) Let the inputs of the $OR$ gate be $A$ and $B$. The output of the $OR$ gate is $Y_1 = A + B$.
This output $Y_1$ is connected to both inputs of a $NAND$ gate. Let the inputs of the $NAND$ gate be $X_1$ and $X_2$,where $X_1 = X_2 = Y_1 = A + B$.
The output of the $NAND$ gate is $Y = \overline{X_1 \cdot X_2} = \overline{(A + B) \cdot (A + B)} = \overline{A + B}$.
The expression $\overline{A + B}$ represents the Boolean operation of a $NOR$ gate.
Therefore,the combination is equivalent to a $NOR$ gate.

(D) universal gate is a logic gate that can be used to implement any other logic gate or Boolean function.
$NAND$ and $NOR$ gates are known as universal gates because any basic logic gate (like $AND, OR, NOT$) can be constructed using only $NAND$ gates or only $NOR$ gates.
Therefore,the correct pair is $NAND$ and $NOR$.

(A) The given circuit consists of a $NOT$ gate followed by a $NOR$ gate where both inputs of the $NOR$ gate are connected to the output of the $NOT$ gate.
$1$. The input $A$ passes through the $NOT$ gate,producing an output of $\bar{A}$.
$2$. This signal $\bar{A}$ is fed into both inputs of the $NOR$ gate.
$3$. The output of a $NOR$ gate with inputs $X$ and $Y$ is $\overline{X + Y}$.
$4$. Here,both inputs are $\bar{A}$,so the output $Y = \overline{\bar{A} + \bar{A}}$.
$5$. Using Boolean algebra,$\bar{A} + \bar{A} = \bar{A}$.
$6$. Therefore,$Y = \overline{\bar{A}} = A$.

(D) $NOR$ gate is the combination of an $OR$ gate followed by a $NOT$ gate.
The Boolean expression for an $OR$ gate is $Y = A + B$.
By applying the $NOT$ operation (inversion) to the output of the $OR$ gate,we get the expression for the $NOR$ gate:
$Y = \overline{A + B}$

(D) $NOR$ gate can be constructed using $NAND$ gates by following these steps:
$1$. Connect the inputs of two $NAND$ gates together to act as $NOT$ gates (inverters).
$2$. Feed the inputs $A$ and $B$ into these two $NOT$ gates to get $\bar{A}$ and $\bar{B}$.
$3$. Connect $\bar{A}$ and $\bar{B}$ to the inputs of a third $NAND$ gate.
$4$. The output of this third $NAND$ gate will be $\overline{\bar{A} \cdot \bar{B}}$.
$5$. By De Morgan's Law,$\overline{\bar{A} \cdot \bar{B}} = A + B$,which is the output of an $OR$ gate.
$6$. To get a $NOR$ gate,we need to invert this output,requiring a fourth $NAND$ gate used as a $NOT$ gate.
$7$. However,a more efficient configuration exists using $4$ $NAND$ gates to realize a $NOR$ gate. Thus,the minimum number of $NAND$ gates required is $4$.

(D) In digital electronics,$NAND$ and $NOR$ gates are known as universal gates. This is because any basic logic gate (such as $AND$,$OR$,or $NOT$) can be implemented using only $NAND$ gates or only $NOR$ gates. Among the given options,the $NAND$ gate is a universal gate,making it the fundamental building block for complex digital circuits.

(B) The $XOR$ (Exclusive $OR$) gate is a digital logic gate that implements exclusive disjunction.
Its Boolean expression is $Y = A \oplus B$,which is equivalent to $Y = A\bar{B} + \bar{A}B$.
The truth table for a two-input $XOR$ gate is:
If $A=0, B=0$,then $Y=0$.
If $A=0, B=1$,then $Y=1$.
If $A=1, B=0$,then $Y=1$.
If $A=1, B=1$,then $Y=0$.
From the truth table,it is clear that the output is $1$ (High) only when the inputs $A$ and $B$ are different.

(C) In Boolean algebra,the variable $A$ can only take two values: $0$ or $1$.
If $A = 0$,then $\overline{A} = 1$. Thus,$A + \overline{A} = 0 + 1 = 1$.
If $A = 1$,then $\overline{A} = 0$. Thus,$A + \overline{A} = 1 + 0 = 1$.
Therefore,for any value of $A$,the relation $A + \overline{A} = 1$ always holds true.

(D) According to De Morgan's Laws:
$1$. $\overline{A.B} = \overline{A} + \overline{B}$
$2$. $\overline{A+B} = \overline{A}.\overline{B}$
Let's evaluate each option:
Option $A$: $\overline{\overline{A}.\overline{B}} = \overline{\overline{A}} + \overline{\overline{B}} = A + B$. (Correct)
Option $B$: $\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} . \overline{\overline{B}} = A.B$. (Correct)
Option $C$: $\overline{(\overline{A.B}).(\overline{A.B})} = \overline{\overline{A.B}} = A.B$. (Correct)
Option $D$: $\overline{A} + \overline{A} = \overline{A}$. The equation states $\overline{A} + \overline{A} = 1$,which is incorrect because it is only true if $A=0$. Therefore,option $D$ is the incorrect equation.

(A) The given circuit consists of a $NOR$ gate followed by another $NOR$ gate where both inputs are tied together.
$1$. The first gate is a $NOR$ gate with inputs $A$ and $B$. Its output is $Y_1 = \overline{A+B}$.
$2$. The second gate is also a $NOR$ gate with both inputs connected to $Y_1$. $A$ $NOR$ gate with both inputs tied together acts as a $NOT$ gate.
$3$. Therefore,the final output is $Y = \overline{Y_1} = \overline{(\overline{A+B})} = A+B$.
$4$. The Boolean expression $Y = A+B$ represents an $OR$ gate.