Boolean Algebra and Logic Gates Questions in English

(D) In Boolean algebra,the following laws are fundamental:
$1$. Idempotent Law: $A + A = A$ and $A \cdot A = A$. Thus,option $A$ is correct.
$2$. $OR$ Law: $A + 1 = 1$ (since any input ORed with $1$ is always $1$). Thus,option $B$ is correct.
$3$. Complement Law: $A \cdot \overline{A} = 0$ (since one of the variables must be $0$). Thus,option $C$ is correct.
Since all the given relations are standard Boolean identities,the correct answer is $D$.

(D) logic gate whose output is low $(0)$ when any one of the inputs is high $(1)$ is a $NAND$ gate.
For a $NAND$ gate,the output $Y = \overline{A \cdot B}$.
If $A = 1$ or $B = 1$,then $A \cdot B = 1$,so $Y = \overline{1} = 0$.
Thus,the output is low when at least one input is high.

(C) $1$. The first gate is a $NOR$ gate with both inputs connected to $A$. The output of this $NOR$ gate is $\overline{A + A} = \bar{A}$.
$2$. The second gate is a $NOT$ gate. The input to this gate is $\bar{A}$,so its output is $\overline{\bar{A}} = A$.
$3$. The third gate is a $NAND$ gate with both inputs connected to the output of the $NOT$ gate (which is $A$). The output of this $NAND$ gate is $\overline{A \cdot A} = \bar{A}$.
$4$. Therefore,the final Boolean expression for the circuit is $Y = \bar{A}$.

(A) $NAND$ gate is defined by the Boolean expression $Y = \overline{A \cdot B}$.
If all inputs are connected together,then $A = B = X$.
Substituting this into the expression,we get $Y = \overline{X \cdot X}$.
Since $X \cdot X = X$,the expression simplifies to $Y = \overline{X}$.
This is the Boolean expression for a $NOT$ gate.
Therefore,connecting all inputs of a $NAND$ gate results in a $NOT$ gate.

(C) Logic gates are the fundamental building blocks of any digital circuit or system. They perform basic logical functions (such as $AND$,$OR$,$NOT$,$NAND$,$NOR$,$XOR$,and $XNOR$) on one or more binary inputs to produce a single binary output. Since digital systems operate on binary signals ($0$ and $1$),logic gates are essential for processing information in computers,calculators,and other electronic devices.

(D) universal gate is a logic gate that can be used to implement any other logic gate (such as $AND, OR, NOT, XOR, XNOR$) without the need for other types of gates.
$NAND$ and $NOR$ gates are known as universal gates.
In the given options,$NAND$ is a universal gate.
Therefore,the correct option is $D$.

(A) The given logic gate symbol consists of an $AND$ gate followed by a $NOT$ operation (indicated by the small circle at the output).
This combination is known as a $NAND$ gate.
The Boolean expression for a $NAND$ gate is $Y = \overline{A \cdot B}$.

(C) Let's analyze each logic gate:
$(a)$ This is a $NOR$ gate with inputs $1$ and $1$. The output of a $NOR$ gate is $Y = \overline{A+B}$. So,$Y = \overline{1+1} = \overline{1} = 0$.
$(b)$ This is a $NOR$ gate with inputs $0$ and $1$. The output is $Y = \overline{0+1} = \overline{1} = 0$.
$(c)$ This is a $NAND$ gate with inputs $0$ and $1$. The output of a $NAND$ gate is $Y = \overline{A \cdot B}$. So,$Y = \overline{0 \cdot 1} = \overline{0} = 1$.
$(d)$ This is a $NAND$ gate with inputs $0$ and $1$. The output is $Y = \overline{0 \cdot 1} = \overline{0} = 1$.
Thus,gates $(c)$ and $(d)$ produce an output of $1$.

(A) We evaluate each expression for $A=0$ and $A=1$:
$(A)$ $A \cdot \overline{A}$:
For $A=0$: $0 \cdot \overline{0} = 0 \cdot 1 = 0$.
For $A=1$: $1 \cdot \overline{1} = 1 \cdot 0 = 0$.
Since the result is $0$ for both cases,this expression is always zero.
$(B)$ $A \cdot 0$:
For $A=0$: $0 \cdot 0 = 0$.
For $A=1$: $1 \cdot 0 = 0$.
This expression is also always zero.
$(C)$ $\overline{A + \overline{A}}$:
For $A=0$: $\overline{0 + \overline{0}} = \overline{0 + 1} = \overline{1} = 0$.
For $A=1$: $\overline{1 + \overline{1}} = \overline{1 + 0} = \overline{1} = 0$.
This expression is also always zero.
$(D)$ $\overline{\overline{A} \cdot 0}$:
For $A=0$: $\overline{\overline{0} \cdot 0} = \overline{1 \cdot 0} = \overline{0} = 1$.
For $A=1$: $\overline{\overline{1} \cdot 0} = \overline{0 \cdot 0} = \overline{0} = 1$.
This expression is always $1$.
Note: Options $A$,$B$,and $C$ are mathematically equivalent to $0$ in Boolean algebra. In standard multiple-choice contexts,$A \cdot \overline{A}$ is the fundamental identity for zero.

(D) In Boolean algebra,we evaluate the given expressions:
$1$. For $A(B + \overline{B}) = A$: Since $B + \overline{B} = 1$,the expression becomes $A(1) = A$. This is correct.
$2$. For $A + AB = A$: We can factor out $A$ to get $A(1 + B) = A$. Since $1 + B = 1$,the expression becomes $A(1) = A$. This is correct.
$3$. For $A + 0 = A$: This is the identity law of Boolean algebra,which is correct.
Since all three relations are correct,the correct option is $D$.

(A) The given circuit consists of two $NAND$ gates. The first $NAND$ gate has inputs $A$ and $B$,producing an output $X = \overline{A \cdot B}$. This output $X$ is fed into the second $NAND$ gate,which has both its inputs shorted together. $A$ $NAND$ gate with shorted inputs acts as a $NOT$ gate. Therefore,the final output $Y = \overline{X \cdot X} = \overline{X} = \overline{(\overline{A \cdot B})} = A \cdot B$. This is the Boolean expression for an $AND$ gate. Thus,the combination represents an $AND$ gate.

(B) According to De Morgan's first theorem,the complement of the sum of two variables is equal to the product of their individual complements.
Mathematically,this is expressed as: $\overline{A + B} = \overline{A} \cdot \overline{B}$.
Therefore,the expression $\overline{A} \cdot \overline{B}$ is equivalent to $\overline{A + B}$.

(A) The given circuit consists of two $NOR$ gates. The first $NOR$ gate has inputs $A$ and $B$,so its output is $Y_1 = \overline{A+B}$.
The second $NOR$ gate has both inputs connected to $Y_1$. $A$ $NOR$ gate with both inputs tied together acts as a $NOT$ gate.
Therefore,the final output $Y = \overline{Y_1 + Y_1} = \overline{Y_1} = \overline{(\overline{A+B})} = A+B$.
The Boolean expression $Y = A+B$ represents an $OR$ gate.
Thus,the circuit is equivalent to an $OR$ gate.

(B) By analyzing the given logic gate symbols:
$(a)$ represents a $NAND$ gate.
$(b)$ represents an $XOR$ gate.
$(c)$ represents a $NOR$ gate.
$(d)$ represents a $NOT$ gate.
Therefore,the $XOR$ gate is $(b)$ and the $NOR$ gate is $(c)$.

(B) $NOR$ gate is a combination of an $OR$ gate followed by a $NOT$ gate.
The Boolean expression for a $NOR$ gate is $Y = \overline{A + B}$.
If both inputs $A$ and $B$ are low $(0, 0)$,then $A + B = 0$,and $Y = \overline{0} = 1$ (High output).
If any input is high,the output becomes low.
Therefore,the $NOR$ gate produces a high output only when both inputs are low.

(C) The truth table provided is:
$A=0, B=0 \implies Y=0$
$A=0, B=1 \implies Y=1$
$A=1, B=0 \implies Y=1$
$A=1, B=1 \implies Y=1$
This truth table shows that the output $Y$ is $1$ if at least one of the inputs ($A$ or $B$) is $1$.
This is the characteristic behavior of an $OR$ gate,where the output is high $(1)$ if any input is high $(1)$.
Therefore,the correct option is $C$.

(B) The given Boolean expression is $Y = A \cdot \bar B$. To find the inputs for which $Y = 1$,we construct the truth table:

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

From the truth table,it is clear that the output $Y = 1$ is obtained only when $A = 1$ and $B = 0$.

(B) For figure $(a)$: The gate is an $OR$ gate with inverted inputs. The output $Y = \bar{A} + \bar{B}$. According to De Morgan's theorem,$\bar{A} + \bar{B} = \overline{A \cdot B}$. This is the Boolean expression for a $NAND$ gate.
For figure $(b)$: The gate is an $AND$ gate with inverted inputs. The output $Y = \bar{A} \cdot \bar{B}$. According to De Morgan's theorem,$\bar{A} \cdot \bar{B} = \overline{A + B}$. This is the Boolean expression for a $NOR$ gate.
Therefore,$(a)$ is equivalent to a $NAND$ gate and $(b)$ is equivalent to a $NOR$ gate.

(C) The circuit consists of three $NAND$ gates and one $NOR$ gate.
Let the inputs be $A$ and $B$.
The first two $NAND$ gates act as $NOT$ gates because their inputs are shorted.
So,the outputs of the first two gates are $\bar{A}$ and $\bar{B}$.
These are fed into the third $NAND$ gate.
The output of the third $NAND$ gate is $Y' = \overline{\bar{A} \cdot \bar{B}} = A + B$.
This $Y'$ is then passed through a $NOR$ gate.
The final output is $Y = \overline{Y' + 0} = \overline{A + B}$.
This is the Boolean expression for a $NOR$ gate.

(A) The truth table provided is:
$A=0, B=0 \implies Y=1$
$A=0, B=1 \implies Y=1$
$A=1, B=0 \implies Y=1$
$A=1, B=1 \implies Y=0$
This truth table shows that the output $Y$ is $0$ only when both inputs $A$ and $B$ are $1$. In all other cases,the output is $1$.
This is the characteristic behavior of a $NAND$ gate,which is the inverse of an $AND$ gate.
The Boolean expression for a $NAND$ gate is $Y = \overline{A \cdot B}$.
Thus,the correct option is $A$.

(D) To identify the logic gate,we analyze the truth table from the given waveforms:
$1$. At the start,$A=0, B=0$,and the output $C=0$.
$2$. Next,$A=1, B=0$,and the output $C=1$.
$3$. Then,$A=1, B=1$,and the output $C=0$.
$4$. Finally,$A=0, B=1$,and the output $C=1$.
Summarizing the truth table:

$A$	$B$	$C$
$0$	$0$	$0$
$1$	$0$	$1$
$1$	$1$	$0$
$0$	$1$	$1$

Comparing this with standard logic gates:
- For an $OR$ gate,$C = A + B$. Here,$1+1=1$,but the output is $0$. So,it is not $OR$.
- For an $AND$ gate,$C = A \cdot B$. Here,$1 \cdot 0 = 0$,but the output is $1$. So,it is not $AND$.
- For a $NOR$ gate,$C = \overline{A+B}$. Here,$0+0=1$,but the output is $0$. So,it is not $NOR$.
- For an $XOR$ gate,$C = A \oplus B$. The truth table matches the $XOR$ gate,as the output is $1$ only when the inputs are different.
Therefore,the correct logic gate is the $XOR$ gate.

(D) The truth table provided is:
$A=1, B=1 \implies Y=0$
$A=0, B=1 \implies Y=1$
$A=1, B=0 \implies Y=1$
$A=0, B=0 \implies Y=1$
Analyzing the output $Y$,we see that the output is $0$ only when both inputs $A$ and $B$ are $1$. In all other cases,the output is $1$.
This is the characteristic behavior of a $NAND$ gate,where $Y = \overline{A \cdot B}$.
Therefore,the correct option is $D$.

(A) The given circuit consists of two $NOT$ gates followed by a $NOR$ gate.
Let the inputs be $A$ and $B$. The outputs of the $NOT$ gates are $\bar{A}$ and $\bar{B}$.
These are fed into a $NOR$ gate,so the final output $Y$ is given by the Boolean expression:
$Y = \overline{\bar{A} + \bar{B}}$
Using De Morgan's Law,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
Thus,the circuit acts as an $AND$ gate.
By analyzing the input waveforms:
$1$. When both $A$ and $B$ are $1$ (high),the output $Y$ is $1$.
$2$. In all other cases,the output $Y$ is $0$.
Comparing this with the given options,the waveform that represents the $AND$ operation of inputs $A$ and $B$ is correct.

(A) The given 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 the output of the $AND$ gate,which takes $X$ and $C$ as inputs. Thus,$Y = X \cdot C = (A + B) \cdot C$.
For the output $Y = 1$,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:
$A) 101: A=1, B=0, C=1 \implies Y = (1+0) \cdot 1 = 1 \cdot 1 = 1$.
$B) 100: A=1, B=0, C=0 \implies Y = (1+0) \cdot 0 = 1 \cdot 0 = 0$.
$C) 110: A=1, B=1, C=0 \implies Y = (1+1) \cdot 0 = 1 \cdot 0 = 0$.
$D) 010: A=0, B=1, C=0 \implies Y = (0+1) \cdot 0 = 1 \cdot 0 = 0$.
Thus,the correct input is $101$.

(D) To convert a binary number to decimal,we multiply each digit by $2$ raised to the power of its position relative to the binary point.
For the binary number $1011.01$:
$1 \times 2^3 + 0 \times 2^2 + 1 \times 2^1 + 1 \times 2^0 + 0 \times 2^{-1} + 1 \times 2^{-2}$
$= 8 + 0 + 2 + 1 + 0 + 0.25$
$= 11.25$
Thus,the decimal equivalent is $11.25$.

(B) The given circuit consists of two $AND$ gates,two $NOT$ gates,and one $OR$ gate.
Let the inputs be $A$ and $B$.
The upper $AND$ gate receives inputs $A$ and $\overline{B}$,so its output is $A\overline{B}$.
The lower $AND$ gate receives inputs $\overline{A}$ and $B$,so its output is $\overline{A}B$.
The $OR$ gate combines these two outputs.
Therefore,the final output $Y$ is given by the Boolean expression: $Y = A\overline{B} + \overline{A}B$.
This expression corresponds to the $XOR$ (Exclusive-$OR$) logic gate.

(C) The given circuit consists of two $NOT$ gates followed by a $NOR$ gate.
Let the inputs be $A$ and $B$.
The outputs of the two $NOT$ gates are $\overline{A}$ and $\overline{B}$.
These are the inputs to the $NOR$ gate.
The output $Y$ of the $NOR$ gate is given by:
$Y = \overline{\overline{A} + \overline{B}}$
Using De Morgan's theorem,$\overline{X + Y} = \overline{X} \cdot \overline{Y}$.
Therefore,$Y = \overline{\overline{A}} \cdot \overline{\overline{B}}$
Since $\overline{\overline{A}} = A$ and $\overline{\overline{B}} = B$,we get:
$Y = A \cdot B$
This is the Boolean expression for an $AND$ gate.

(A) In Boolean algebra,the variable $A$ can only take two values: $0$ or $1$.
If $A = 0$,then $\bar{A} = 1$. Therefore,$A \cdot \bar{A} = 0 \cdot 1 = 0$.
If $A = 1$,then $\bar{A} = 0$. Therefore,$A \cdot \bar{A} = 1 \cdot 0 = 0$.
In both possible cases,the result is $0$. Thus,$A \cdot \bar{A} = 0$.

(C) The 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 $(\overline{A \cdot B})$.
The final output $Y$ of the $AND$ gate is:
$Y = (A + B) \cdot (\overline{A \cdot B})$
Using De Morgan's theorem,$\overline{A \cdot B} = \overline{A} + \overline{B}$.
So,$Y = (A + B) \cdot (\overline{A} + \overline{B})$
$Y = A \cdot \overline{A} + A \cdot \overline{B} + B \cdot \overline{A} + B \cdot \overline{B}$
Since $A \cdot \overline{A} = 0$ and $B \cdot \overline{B} = 0$,we get:
$Y = A \cdot \overline{B} + B \cdot \overline{A}$
This is the Boolean expression for an $XOR$ gate.

(D) The logic gate shown in the image is a $3$-input $AND$ gate.
For an $AND$ gate,the output $Y$ is $1$ if and only if all the inputs are $1$.
Therefore,for $Y = 1$,the inputs must be $A = 1, B = 1,$ and $C = 1$.
Thus,the correct option is $D$.

(A) The circuit consists of an $OR$ gate followed by an $AND$ gate.
$1$. The inputs to the $OR$ gate are $A$ and $B$. Therefore,the output of the $OR$ gate is $(A + B)$.
$2$. The inputs to the $AND$ gate are $A$ and the output of the $OR$ gate $(A + B)$.
$3$. The final output $Y$ of the $AND$ gate is given by the Boolean expression: $Y = A \cdot (A + B)$.
$4$. Using the distributive law of Boolean algebra: $Y = A \cdot A + A \cdot B$.
$5$. Since $A \cdot A = A$,we have $Y = A + A \cdot B$.
$6$. Factoring out $A$,we get $Y = A \cdot (1 + B)$.
$7$. Since $(1 + B) = 1$,the final expression is $Y = A \cdot 1 = A$.

(B) To construct an $AND$ gate using $NAND$ gates,we first pass the inputs $A$ and $B$ through a $NAND$ gate. The output of this first gate is $Y_1 = \overline{A \cdot B}$.
Next,we pass this output $Y_1$ through a second $NAND$ gate whose inputs are shorted together. $A$ $NAND$ gate with shorted inputs acts as a $NOT$ gate. Therefore,the final output is $Y = \overline{Y_1} = \overline{(\overline{A \cdot B})} = A \cdot B$.
Thus,$2$ $NAND$ gates are required to construct an $AND$ gate.

(A) In the first circuit,two $NAND$ gates are connected such that the second gate acts as a $NOT$ gate (since its inputs are shorted). The output of the first $NAND$ gate is $\overline{A \cdot B}$. Passing this through a $NOT$ gate gives $Y = \overline{\overline{A \cdot B}} = A \cdot B$,which is the $AND$ operation.
In the second circuit,there are three $NAND$ gates. The inputs $A$ and $B$ are passed through $NAND$ gates acting as $NOT$ gates,resulting in $\overline{A}$ and $\overline{B}$. These are then fed into a third $NAND$ gate. The output is $Y = \overline{\overline{A} \cdot \overline{B}}$. By De Morgan's Law,$Y = \overline{\overline{A}} + \overline{\overline{B}} = A + B$,which is the $OR$ operation.
Therefore,the circuits represent $AND$ and $OR$ operations respectively.

(C) The given Boolean expression is $Y = A + \bar B$.
To find the inputs for which $Y = 0$,we construct the truth table:

$A$	$B$	$\bar B$	$Y = A + \bar B$
$1$	$1$	$0$	$1 + 0 = 1$
$1$	$0$	$1$	$1 + 1 = 1$
$0$	$1$	$0$	$0 + 0 = 0$
$0$	$0$	$1$	$0 + 1 = 1$

From the truth table,it is clear that the output $Y = 0$ only when $A = 0$ and $B = 1$.

(B) The 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 $\overline{A}$.
The output of the second $NAND$ gate (acting as $NOT$) is $\overline{B}$.
These are fed into the final $NAND$ gate.
The output $X = \overline{(\overline{A} \cdot \overline{B})}$.
Using De Morgan's Law,$X = \overline{\overline{A}} + \overline{\overline{B}} = A + B$.
Thus,the combination acts as an $OR$ gate.

(C) The given logic circuit consists of a $NOR$ gate followed by a $NOT$ gate (represented by the $NOR$ gate with shorted inputs).
Let the output of the first $NOR$ gate be $Y' = \overline{A + B}$.
This output $Y'$ is fed into a second $NOR$ gate whose inputs are shorted together,which acts as a $NOT$ gate.
Thus,the final output is $Y = \overline{Y'} = \overline{\overline{A + B}} = A + B$.
This is the Boolean expression for an $OR$ gate.
The truth table for an $OR$ gate is given in option $C$.

(D) The inputs to the $NOR$ gate are the inverted inputs $A$ and $B$. Let the inputs to the $NOR$ gate be $A' = \bar{A}$ and $B' = \bar{B}$.
The output $Y$ of the $NOR$ gate is given by $Y = \overline{A' + B'}$.
Substituting the values of $A'$ and $B'$, we get $Y = \overline{\bar{A} + \bar{B}}$.
According to De Morgan's theorem, $\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
Therefore, the output $Y = A \cdot B$, which is the Boolean expression for an $AND$ gate.

(C) The given circuit consists of a $NOR$ gate followed by a $NOT$ gate (represented by the second gate with shorted inputs).
$1$. The output of the first $NOR$ gate is $C = \overline{A+B}$.
$2$. The second gate is a $NOT$ gate (since both inputs are connected together),so its output is $Y = \overline{C}$.
$3$. Substituting $C$,we get $Y = \overline{(\overline{A+B})} = A+B$.
$4$. This is the Boolean expression for an $OR$ gate.
$5$. The truth table for an $OR$ 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 $Y=1$.
This matches option $C$.

(C) To convert the binary number $1000101.101$ to decimal,we expand it using powers of $2$:
$1000101.101 = (1 \times 2^6) + (0 \times 2^5) + (0 \times 2^4) + (0 \times 2^3) + (1 \times 2^2) + (0 \times 2^1) + (1 \times 2^0) + (1 \times 2^{-1}) + (0 \times 2^{-2}) + (1 \times 2^{-3})$
$= 64 + 0 + 0 + 0 + 4 + 0 + 1 + 0.5 + 0 + 0.125$
$= 69 + 0.625$
$= 69.625$
Thus,the decimal equivalent is $69.625$.

(C) The given circuit consists of two $NOT$ gates, two $AND$ gates, and one $OR$ gate.
Let the inputs be $A$ and $B$.
The output of the first $AND$ gate is $Y_1 = (\text{NOT } A) \text{ AND } B = \bar{A} \cdot B$.
The output of the second $AND$ gate is $Y_2 = A \text{ AND } (\text{NOT } B) = A \cdot \bar{B}$.
The final output $Y$ is the $OR$ combination of $Y_1$ and $Y_2$: $Y = Y_1 + Y_2 = \bar{A}B + A\bar{B}$.
This is the Boolean expression for an $XOR$ gate.
The truth table for an $XOR$ gate is:
- If $A=0, B=0$, then $Y = \bar{0} \cdot 0 + 0 \cdot \bar{0} = 1 \cdot 0 + 0 \cdot 1 = 0 + 0 = 0$.
- If $A=0, B=1$, then $Y = \bar{0} \cdot 1 + 0 \cdot \bar{1} = 1 \cdot 1 + 0 \cdot 0 = 1 + 0 = 1$.
- If $A=1, B=0$, then $Y = \bar{1} \cdot 0 + 1 \cdot \bar{0} = 0 \cdot 0 + 1 \cdot 1 = 0 + 1 = 1$.
- If $A=1, B=1$, then $Y = \bar{1} \cdot 1 + 1 \cdot \bar{1} = 0 \cdot 1 + 1 \cdot 0 = 0 + 0 = 0$.
Thus, the correct truth table is option $C$.

(B) $NAND$ gate performs the operation $Y = \overline{A \cdot B}$.
Based on the input waveforms:
For $t = 0-2 \ s$: $A=1, B=0 \implies A \cdot B = 0 \implies Y = \overline{0} = 1$.
For $t = 2-4 \ s$: $A=0, B=0 \implies A \cdot B = 0 \implies Y = \overline{0} = 1$.
For $t = 4-6 \ s$: $A=1, B=0 \implies A \cdot B = 0 \implies Y = \overline{0} = 1$.
For $t = 6-8 \ s$: $A=1, B=1 \implies A \cdot B = 1 \implies Y = \overline{1} = 0$.
For $t > 8 \ s$: $A=0, B=0 \implies A \cdot B = 0 \implies Y = \overline{0} = 1$.
Thus,the output $Y$ is $1$ for $0-6 \ s$,$0$ for $6-8 \ s$,and $1$ for $t > 8 \ s$. This matches the waveform shown in option $B$.

(D) The truth table shows that the output $Y$ is $1$ only when both inputs $A$ and $B$ are $0$.
For all other combinations of inputs ($1,1$; $1,0$; $0,1$),the output $Y$ is $0$.
This behavior corresponds to the Boolean expression $Y = \overline{A + B}$.
This is the definition of a $NOR$ gate,which is an $OR$ gate followed by a $NOT$ gate.
Therefore,the correct option is $D$.

(B) The truth table provided is:
$A=0, B=0 \implies Y=0$
$A=0, B=1 \implies Y=1$
$A=1, B=0 \implies Y=1$
$A=1, B=1 \implies Y=0$
This truth table corresponds to the Exclusive-$OR$ $(XOR)$ gate.
In an $XOR$ gate,the output is $1$ if the inputs are different,and the output is $0$ if the inputs are the same.
Therefore,the correct option is $B$.

(B) In Boolean algebra,the expression $Y = A + B$ represents the logical $OR$ operation.
In this operation,the output $Y$ is $1$ (high) if at least one of the inputs $A$ or $B$ is $1$ (high).
This means $Y$ is $1$ if $A=1$,or $B=1$,or both $A=1$ and $B=1$.
Therefore,the output $Y$ exists if $A$ or $B$ or both $A$ and $B$ exist.

(C) By observing the given logic gate symbols:
$(a)$ represents an $OR$ gate.
$(b)$ represents an $AND$ gate.
$(c)$ represents a $NOR$ gate.
$(d)$ represents a $NAND$ gate.
Therefore,the symbols for $OR$,$NOR$,and $NAND$ gates are $a$,$c$,and $d$ respectively.

(D) The circuit consists of two $OR$ gates followed by an $AND$ gate.
Let the inputs to the first $OR$ gate be $W$ and $Y$. Its output is $(W + Y)$.
Let the inputs to the second $OR$ gate be $W$ and $X$. Its output is $(W + X)$.
These two outputs are fed into an $AND$ gate.
Therefore, the final output $Z = (W + Y) . (W + X)$.
Using the distributive law of Boolean algebra: $Z = W . W + W . X + Y . W + Y . X$.
Since $W . W = W$, we have $Z = W + W . X + W . Y + X . Y$.
Factoring out $W$: $Z = W . (1 + X + Y) + X . Y$.
Since $(1 + X + Y) = 1$, we get $Z = W . (1) + X . Y = W + X . Y$.

(A) In the given circuit,when input $A$ or $B$ or both $A$ and $B$ are at logic level $'1'$,at least one diode becomes forward-biased,and the output $C$ becomes $'1'$.
When both inputs $A$ and $B$ are at logic level $'0'$,both diodes are reverse-biased,and the output $C$ remains at logic level $'0'$.
This behavior corresponds to the truth table of an $OR$ gate.

(D) In the given circuit,two switches $A$ and $B$ are connected in parallel.
If switch $A$ is closed (represented as $1$) or switch $B$ is closed (represented as $1$),the current will flow through the circuit and the lamp will glow (output $Y = 1$).
If both switches $A$ and $B$ are open (represented as $0$),the current will not flow and the lamp will not glow (output $Y = 0$).
The truth table for this circuit is:
$A=0, B=0 \implies Y=0$
$A=0, B=1 \implies Y=1$
$A=1, B=0 \implies Y=1$
$A=1, B=1 \implies Y=1$
This truth table corresponds to the $OR$ gate.

(C) The given truth table is:
$A=0, B=0 \implies Y=1$
$A=0, B=1 \implies Y=0$
$A=1, B=0 \implies Y=0$
$A=1, B=1 \implies Y=1$
This truth table corresponds to the $XNOR$ gate (Exclusive $NOR$ gate).
The Boolean expression for an $XNOR$ gate is $Y = A \odot B = \overline{A \oplus B} = AB + \overline{A}\overline{B}$.
Checking the values:
For $(0,0): Y = (0)(0) + (1)(1) = 0 + 1 = 1$.
For $(0,1): Y = (0)(1) + (1)(0) = 0 + 0 = 0$.
For $(1,0): Y = (1)(0) + (0)(1) = 0 + 0 = 0$.
For $(1,1): Y = (1)(1) + (0)(0) = 1 + 0 = 1$.
Thus,the correct option is $C$.

(A) The circuit consists of a $NOR$ gate,a $NAND$ gate with both inputs tied together (acting as a $NOT$ gate),and a $NOT$ gate.
$1$. The output of the $NOR$ gate is $X = \overline{A+B}$.
$2$. The $NAND$ gate with both inputs tied together acts as a $NOT$ gate. Its output is $Z = \overline{X \cdot X} = \overline{X} = \overline{\overline{A+B}} = A+B$.
$3$. The final $NOT$ gate inverts this output: $Y = \overline{Z} = \overline{A+B}$.
Thus,the circuit is equivalent to a $NOR$ gate.