Boolean Algebra and Logic Gates Questions in English

(C) Let the inputs be $A$ and $B$.
Each of the first two $NAND$ gates has its inputs shorted,which makes them act as $NOT$ gates.
Therefore,the output of the first $NAND$ gate is $\overline{A}$ and the output of the second $NAND$ gate is $\overline{B}$.
These outputs $\overline{A}$ and $\overline{B}$ are fed as inputs to the third $NAND$ gate.
The output $Y$ of the third $NAND$ gate is given by $Y = \overline{(\overline{A} \cdot \overline{B})}$.
Using De Morgan's theorem,$\overline{X \cdot Y} = \overline{X} + \overline{Y}$.
So,$Y = \overline{(\overline{A})} + \overline{(\overline{B})} = A + B$.
Since the output $Y = A + B$ represents an $OR$ gate,the equivalent logic gate is $OR$.

(B) The circuit consists of an $AND$ gate,an $OR$ gate,and a $NAND$ gate.
Let the inputs be $A$ and $B$.
The upper $AND$ gate receives inputs $A$ and $B$,so its output is $Y_1 = A \cdot B$.
The lower $OR$ gate receives inputs $B$ and $A$,so its output is $Y_2 = B + A$.
The final $NAND$ gate receives $Y_1$ and $Y_2$ as inputs,so the final output is $Y = \overline{Y_1 \cdot Y_2} = \overline{(A \cdot B) \cdot (A + B)}$.
Using Boolean algebra: $Y = \overline{(A \cdot B) \cdot A + (A \cdot B) \cdot B} = \overline{(A \cdot B) + (A \cdot B)} = \overline{A \cdot B}$.
This is the truth table for a $NAND$ gate:
If $A=0, B=0$,then $Y = \overline{0 \cdot 0} = 1$.
If $A=0, B=1$,then $Y = \overline{0 \cdot 1} = 1$.
If $A=1, B=0$,then $Y = \overline{1 \cdot 0} = 1$.
If $A=1, B=1$,then $Y = \overline{1 \cdot 1} = 0$.
Comparing this with the given options,option $B$ is correct.

(A) Let the inputs be $A$ and $B$.
Each input passes through a $NOT$ gate,so the inputs to the first stage $NAND$ gates become $\bar{A}$ and $\bar{B}$.
Since both inputs of the first stage $NAND$ gates are tied together,they act as $NOT$ gates. Thus,the outputs of the first stage are $\overline{\bar{A}} = A$ and $\overline{\bar{B}} = B$.
These outputs $A$ and $B$ are then fed into the final $NAND$ gate.
The output of the final $NAND$ gate is $Y = \overline{A \cdot B}$.
This is the Boolean expression for a $NAND$ gate.
Therefore,the circuit is equivalent to a $NAND$ gate.

(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 two $NOT$ gates are $\bar{A}$ and $\bar{B}$.
These are the inputs to the $NOR$ gate.
The output $y$ of the $NOR$ gate is given by $y = \overline{\bar{A} + \bar{B}}$.
Using De Morgan's theorem,$\overline{\bar{A} + \bar{B}} = \overline{\bar{A}} \cdot \overline{\bar{B}} = A \cdot B$.
The expression $y = A \cdot B$ represents an $AND$ gate.
Therefore,the combination is equivalent to an $AND$ gate.

(C) The circuit consists of a $NOR$ gate followed by an $AND$ gate.
$1$. The output of the $NOR$ gate is $y_1 = \overline{A+B}$.
Given $A=0$ and $B=1$,we have $y_1 = \overline{0+1} = \overline{1} = 0$.
$2$. The output of the $AND$ gate is $y_2 = y_1 \cdot C$.
Given $y_1 = 0$ and $C=1$,we have $y_2 = 0 \cdot 1 = 0$.
Therefore,the values of $y_1$ and $y_2$ are $0$ and $0$ respectively.

(A) The circuit consists of a $NAND$ gate with inputs $B$ and $C$,and a $NOR$ gate with inputs $\overline{A}$ and $y_1$.
First,calculate $y_1$:
$y_1 = \overline{B \cdot C} = \overline{1 \cdot 1} = \overline{1} = 0$.
Next,calculate $y_2$:
The inputs to the $NOR$ gate are $\overline{A}$ and $y_1$.
$\overline{A} = \overline{1} = 0$.
$y_2 = \overline{\overline{A} + y_1} = \overline{0 + 0} = \overline{0} = 1$.
Thus,the values are $y_1 = 0$ and $y_2 = 1$.

(A) From the figure,the output $y_1$ is the output of a $NAND$ gate with inputs $(A \cdot B)$ and $B$. Thus,$y_1 = \overline{(A \cdot B) \cdot B}$.
Given $A=0$ and $B=1$,$A \cdot B = 0 \cdot 1 = 0$.
So,$y_1 = \overline{0 \cdot 1} = \overline{0} = 1$.
The output $y_2$ is the output of a $NOR$ gate with inputs $(B+C)$ and $B$. Thus,$y_2 = \overline{(B+C) + B}$.
Given $B=1$ and $C=1$,$B+C = 1+1 = 1$.
So,$y_2 = \overline{1 + 1} = \overline{1} = 0$.
Therefore,the values of $y_1$ and $y_2$ are $1$ and $0$ respectively.

(C) The circuit consists of a $NOR$ gate,an $AND$ gate,and a $NAND$ gate.
Let the inputs be $A$ and $B$.
The output of the $NOR$ gate is $\overline{A+B}$.
The output of the $AND$ gate is $A \cdot B$.
These two outputs are fed into a $NAND$ gate.
Therefore,the final output $Y$ is given by:
$Y = \overline{(\overline{A+B}) \cdot (A \cdot B)}$
Using De Morgan's Law,$\overline{X \cdot Z} = \overline{X} + \overline{Z}$:
$Y = \overline{(\overline{A+B})} + \overline{(A \cdot B)}$
$Y = (A+B) + (\overline{A} + \overline{B})$
$Y = (A + \overline{A}) + (B + \overline{B})$
Since $A + \overline{A} = 1$ and $B + \overline{B} = 1$:
$Y = 1 + 1 = 1$
Thus,for any combination of inputs $(A, B)$,the output $Y$ is always $1$.

(C) The given circuit consists of two $NOR$ gates acting as $NOT$ gates,followed by a $NOR$ gate.
$1$. The first two gates are $NOR$ gates with both inputs tied together. $A$ $NOR$ gate with inputs $A$ and $A$ gives output $\overline{A+A} = \overline{A}$.
$2$. Similarly,the second $NOR$ gate with inputs $B$ and $B$ gives output $\overline{B+B} = \overline{B}$.
$3$. These two outputs are fed into the final $NOR$ gate.
$4$. The output $Y$ of the final $NOR$ gate is $Y = \overline{\overline{A} + \overline{B}}$.
$5$. By De Morgan's Law,$\overline{\overline{A} + \overline{B}} = \overline{\overline{A}} \cdot \overline{\overline{B}} = A \cdot B$.
$6$. Therefore,the output $Y = A \cdot B$,which is the Boolean expression for an $AND$ gate.

(C) The circuit consists of two $AND$ gates,one $NOT$ gate,and one $OR$ gate.
$1$. The upper $AND$ gate receives inputs $A$ and $B$. Its output is $A \cdot B = 1 \cdot 1 = 1$.
$2$. The lower $AND$ gate also receives inputs $A$ and $B$. Its output is $A \cdot B = 1 \cdot 1 = 1$.
$3$. The output of the lower $AND$ gate passes through a $NOT$ gate to produce $X$. Thus,$X = \overline{A \cdot B} = \overline{1} = 0$.
$4$. The $OR$ gate receives the output of the upper $AND$ gate $(1)$ and the output of the $NOT$ gate $(X = 0)$.
$5$. The final output $Y$ is the result of the $OR$ operation: $Y = 1 + X = 1 + 0 = 1$.
Therefore,the values of $X$ and $Y$ are $0$ and $1$ respectively.

(A) To determine which gate gives an output of $0$ for inputs $A=0$ and $B=1$,we analyze the truth tables for each option:
$1$. $AND$ gate: The output $Y = A \cdot B$. For $A=0, B=1$,$Y = 0 \cdot 1 = 0$. This matches the condition.
$2$. $OR$ gate: The output $Y = A + B$. For $A=0, B=1$,$Y = 0 + 1 = 1$.
$3$. $X$-$OR$ gate: The output $Y = A \oplus B$. For $A=0, B=1$,$Y = 0 \oplus 1 = 1$.
$4$. $NAND$ gate: The output $Y = \overline{A \cdot B}$. For $A=0, B=1$,$Y = \overline{0 \cdot 1} = \overline{0} = 1$.
Thus,the $AND$ gate provides an output of $0$ when $A=0$ and $B=1$.

(C) In the given circuit,the outputs of all three $OR$ gates are fed into the input of an $AND$ gate.
Inputs to the first $OR$ gate are $\bar{A}$ and $B$ because there is a $NOT$ gate on the input line from $A$. Thus,the output of the first $OR$ gate is $(\bar{A}+B)$.
Inputs to the second $OR$ gate are $\bar{A}$ and $\bar{C}$ because there are $NOT$ gates on both input lines from $A$ and $C$. Thus,the output of the second $OR$ gate is $(\bar{A}+\bar{C})$.
Inputs to the third $OR$ gate are $B$ and $\bar{C}$ because there is a $NOT$ gate on the input line from $C$. Thus,the output of the third $OR$ gate is $(B+\bar{C})$.
Since the output of an $AND$ gate is the product (logical multiplication) of its inputs,the final output $Y$ of the logic circuit is the product of the outputs of the three $OR$ gates:
$Y = (\bar{A}+B) \cdot (\bar{A}+\bar{C}) \cdot (B+\bar{C})$.

(A) $1$. The first $AND$ gate has inputs $1$ and $0$,so its output is $1 \cdot 0 = 0$.
$2$. The first $OR$ gate has inputs $0$ and $1$,so its output is $0 + 1 = 1$.
$3$. The second $OR$ gate (connected to the $AND$ gate and the first $OR$ gate) has inputs $0$ and $1$,so its output is $0 + 1 = 1$.
$4$. The $NAND$ gate has inputs $0$ and $1$,so its output $Y = \overline{0 \cdot 1} = \overline{0} = 1$.
$5$. The $NOT$ gate has an input of $1$,so its output is $\overline{1} = 0$.
$6$. The final $OR$ gate has inputs $Y=1$ and the $NOT$ gate output $0$,so $Z = 1 + 0 = 1$.
Therefore,$Y=1$ and $Z=1$.

(B) To determine which logic gates produce a high output $(1)$ only when all inputs are low $(0)$, we examine their truth tables for two inputs $A$ and $B$:
$1$. $NOR$ gate: The output $Y = \overline{A+B}$. When $A=0$ and $B=0$, $A+B=0$, so $Y=1$. For any other combination, the output is $0$.
$2$. $NAND$ gate: The output $Y = \overline{AB}$. When $A=0$ and $B=0$, $AB=0$, so $Y=1$. However, for $A=0, B=1$ or $A=1, B=0$, the output is also $1$.
Wait, re-evaluating the question: "all the inputs must be low to get a high output".
For $NOR$ gate: $A=0, B=0 \implies Y=1$. For any other input, $Y=0$. This satisfies the condition.
For $NAND$ gate: $A=0, B=0 \implies Y=1$. But $A=0, B=1 \implies Y=1$ as well. Thus, $NAND$ does not strictly require *all* inputs to be low to get a high output.
However, in standard physics curriculum contexts, $NOR$ is the primary gate where $Y=1$ only if $A=0$ and $B=0$. Given the options, $NOR$ is the correct gate. If the question implies gates where a high output is *possible* when inputs are low, $NOR$ and $NAND$ are the standard answers.

(C) Let the inputs be $A$,$B$,and $C$. The circuit consists of $NAND$ gates and $OR$ gates.
$1$. The top $NAND$ gate has input $A$ connected to both terminals,so its output is $\overline{A \cdot A} = \bar{A}$.
$2$. The middle two $NAND$ gates have inputs $(A, B)$ and $(B, C)$ respectively,giving outputs $\overline{A \cdot B}$ and $\overline{B \cdot C}$.
$3$. These two outputs are fed into an $OR$ gate,resulting in $\overline{A \cdot B} + \overline{B \cdot C}$.
$4$. The bottom $NAND$ gate has input $C$ connected to both terminals,so its output is $\overline{C \cdot C} = \bar{C}$.
$5$. Finally,all these signals are combined by an $OR$ gate to give the output $Y$:
$Y = \bar{A} + (\overline{A \cdot B} + \overline{B \cdot C}) + \bar{C}$
Using De Morgan's theorem,$\overline{A \cdot B} = \bar{A} + \bar{B}$ and $\overline{B \cdot C} = \bar{B} + \bar{C}$.
Substituting these:
$Y = \bar{A} + (\bar{A} + \bar{B}) + (\bar{B} + \bar{C}) + \bar{C}$
Using the idempotent law $\bar{A} + \bar{A} = \bar{A}$,we get:
$Y = \bar{A} + \bar{B} + \bar{C}$

(B) The truth table shows that the output $X$ is $1$ if either input $A$ or input $B$ (or both) is $1$.
This behavior is represented by the Boolean expression $X = A + B$.
The logic gate that performs this operation is the $OR$ gate.
Therefore,the correct option is $B$.

(D) The truth table for the waveform is given by:

Time Interval	$A$	$B$	$Y$
$t_1-t_2$	$0$	$0$	$1$
$t_2-t_3$	$0$	$1$	$1$
$t_3-t_4$	$1$	$0$	$1$
$t_4-t_5$	$1$	$1$	$0$

From the truth table,we observe that the output $Y$ is $0$ only when both inputs $A$ and $B$ are $1$. In all other cases,the output is $1$.
This behavior corresponds to the Boolean expression $Y = \overline{A \cdot B}$.
This is the characteristic truth table of a $NAND$ gate.

(A) For circuit $(i)$: The two $NAND$ gates with shorted inputs act as $NOT$ gates. The inputs to the final $NAND$ gate are $\bar{A}$ and $\bar{B}$.
The output $C = \overline{\bar{A} \cdot \bar{B}} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$. This is the Boolean expression for an $OR$ gate.
For circuit (ii): The first $NAND$ gate produces $\overline{AB}$. The second $NAND$ gate with shorted inputs acts as a $NOT$ gate,inverting the signal.
The output $C = \overline{\overline{AB}} = AB$. This is the Boolean expression for an $AND$ gate.

(C) Let the output of the $OR$ gate be $P = A + B$. Given $A = 1$ and $B = 1$,we have $P = 1 + 1 = 1$.
Let the output of the $NAND$ gate be $Q = \overline{A \cdot B}$. Given $A = 1$ and $B = 1$,we have $Q = \overline{1 \cdot 1} = \overline{1} = 0$.
Now,$Y_1$ is the output of an $AND$ gate with inputs $P$ and $Q$. Thus,$Y_1 = P \cdot Q = 1 \cdot 0 = 0$.
$Y_2$ is the output of an $OR$ gate with inputs $P$ and $Q$. Thus,$Y_2 = P + Q = 1 + 0 = 1$.
Therefore,the values are $Y_1 = 0$ and $Y_2 = 1$. The correct option is $(C)$.

(A) The classification of integrated circuits based on the number of logic gates is as follows:
$A$. Small Scale Integration $(SSI)$: Contains $\leq 10$ logic gates. (Matches $III$)
$B$. Medium Scale Integration $(MSI)$: Contains $< 100$ logic gates. (Matches $I$)
$C$. Large Scale Integration $(LSI)$: Contains $< 1000$ logic gates. (Matches $IV$)
$D$. Very Large Scale Integration $(VLSI)$: Contains $> 1000$ logic gates. (Matches $II$)
Therefore,the correct matching is $A-III, B-I, C-IV, D-II$.

(D) The circuit consists of two $OR$ gates whose outputs are fed into a $NAND$ gate. Let the inputs be $A$ and $B$. The output of both $OR$ gates is $X = A + B$. These are the inputs to the $NAND$ gate. The final output $Y$ is given by $Y = \overline{X \cdot X} = \overline{X} = \overline{A + B}$. This is the Boolean expression for a $NOR$ gate. The truth table is as follows:

$A$	$B$	$X = A + B$	$Y = \overline{X \cdot X}$
$0$	$0$	$0$	$1$
$0$	$1$	$1$	$0$
$1$	$0$	$1$	$0$
$1$	$1$	$1$	$0$

(B) The circuit consists of two branches feeding into a $NOR$ gate. Each branch has a $NAND$ gate followed by a $NOT$ gate (which together form an $AND$ gate).
Let the output of the top branch be $Y_1$. The top branch has inputs $A$ and $B$ to a $NAND$ gate,followed by a $NOT$ gate. This is equivalent to an $AND$ gate. So,$Y_1 = A \cdot B = 1 \cdot 0 = 0$.
Similarly,the bottom branch has inputs $A$ and $B$ to a $NAND$ gate,followed by a $NOT$ gate. This is also equivalent to an $AND$ gate. So,the output of the bottom branch is $A \cdot B = 1 \cdot 0 = 0$.
Now,$Y_1$ is the output of the top branch,so $Y_1 = 0$.
The final gate is a $NOR$ gate with both inputs as $0$.
The output $Y_2$ of a $NOR$ gate is given by $\overline{0 + 0} = \overline{0} = 1$.
Thus,$Y_1 = 0$ and $Y_2 = 1$.

(C) Let the output of the first $NAND$ gate be $C = \overline{A \cdot B}$.
The top $NAND$ gate receives inputs $A$ and $C$,so its output is $Y_1 = \overline{A \cdot C} = \overline{A \cdot (\overline{A \cdot B})} = \overline{A} + (A \cdot B) = \overline{A} + B$.
The bottom $NAND$ gate receives inputs $B$ and $C$,so its output is $Y_2 = \overline{B \cdot C} = \overline{B \cdot (\overline{A \cdot B})} = \overline{B} + (A \cdot B) = \overline{B} + A$.
The final output $Y$ is the $AND$ of $Y_1$ and $Y_2$,so $Y = Y_1 \cdot Y_2 = (\overline{A} + B) \cdot (\overline{B} + A)$.
Evaluating for each case:
- If $A=0, B=1$: $Y = (1+1) \cdot (0+0) = 1 \cdot 0 = 0$.
- If $A=1, B=0$: $Y = (0+0) \cdot (1+1) = 0 \cdot 1 = 0$.
- If $A=0, B=0$: $Y = (1+0) \cdot (1+0) = 1 \cdot 1 = 1$.
- If $A=1, B=1$: $Y = (0+1) \cdot (0+1) = 1 \cdot 1 = 1$.
Thus,$Y=1$ for combinations $c$ $(A=0, B=0)$ and $d$ $(A=1, B=1)$.

(B) The circuit consists of a $NAND$ gate, a $NOR$ gate, and a $NOR$ gate at the output stage.
$1$. The inputs to the $NAND$ gate are $1$ and $0$. The output of a $NAND$ gate is $Y_1 = \overline{1 \cdot 0} = \overline{0} = 1$.
$2$. The inputs to the $NOR$ gate are $1$ and $0$. The output of a $NOR$ gate is $Y_2 = \overline{1 + 0} = \overline{1} = 0$.
$3$. The inputs to the final $NOR$ gate are $Y_1 = 1$ and $Y_2 = 0$. The output is $Y_3 = \overline{Y_1 + Y_2} = \overline{1 + 0} = \overline{1} = 0$.
Thus, the values are $Y_1 = 1, Y_2 = 0, Y_3 = 0$.

(D) The circuit consists of an $AND$ gate followed by a $NAND$ gate. The inputs to the $AND$ gate are $A$ and $B$,so its output is $A \cdot B$.
The input $C$ passes through a $NOT$ gate,so its output is $\bar{C}$.
These two signals are the inputs to the $NAND$ gate,which produces the final output $Y$.
Thus,$Y = \overline{(A \cdot B) \cdot \bar{C}}$.
Using De Morgan's law,$Y = \overline{A \cdot B} + \overline{\bar{C}} = \bar{A} + \bar{B} + C$.
For the output $Y$ to be zero $(Y=0)$,we must have $\bar{A} + \bar{B} + C = 0$.
This requires $\bar{A} = 0$,$\bar{B} = 0$,and $C = 0$ simultaneously.
Therefore,$A = 1$,$B = 1$,and $C = 0$.

(D) $NAND$ gate is formed by combining an $AND$ gate followed by a $NOT$ gate. The logic symbol for a $NAND$ gate consists of an $AND$ gate symbol with a small circle (inversion bubble) at its output. Among the given options,the symbol in image $D$ represents an $AND$ gate with an inversion bubble at the output,which is the standard symbol for a $NAND$ gate. Therefore,option $D$ is correct.

(C) Digital signals represent values as discrete steps,not as a continuous set of values. Therefore,statement $(i)$ is false,while statements (ii),(iii),and (iv) are true.
Digital signals are typically represented in the form of rectangular waves and often utilize the binary system ($0$ and $1$).

(A) For a $NOR$ gate,the output is given by the Boolean expression $Y = \overline{A+B}$.
If any input ($A$ or $B$) is $1$,then $A+B = 1$,and the output $Y = \overline{1} = 0$.
For a $NAND$ gate,$Y = \overline{AB}$. If one input is $0$,the output is $1$.
For an $AND$ gate,$Y = AB$. If one input is $0$,the output is $0$.
For an $OR$ gate,$Y = A+B$. If one input is $1$,the output is $1$.
Therefore,the $NOR$ gate satisfies the condition.

(B) In the given circuit, the diodes are connected such that their cathodes are tied together at point $Q$, which is connected to a $+5 \,V$ supply through a resistor $R$.
$1$. If both inputs $A$ and $B$ are at $0 \,V$ (low), both diodes are forward-biased. The potential at $Q$ is pulled down to approximately $0 \,V$ (low).
$2$. If one input is at $0 \,V$ and the other is at $+5 \,V$, the diode connected to $0 \,V$ is forward-biased, pulling the potential at $Q$ down to approximately $0 \,V$ (low).
$3$. If both inputs $A$ and $B$ are at $+5 \,V$ (high), both diodes are reverse-biased. No current flows through the diodes, and the potential at $Q$ is pulled up to $+5 \,V$ (high) by the supply through resistor $R$.
Since the output $Q$ is high only when both inputs $A$ and $B$ are high, this circuit functions as an $AND$ gate.

(D) Let us analyze the circuit step by step:
$1$. The top $NAND$ gate has inputs $1$ and $0$. The output is $(1 \cdot 0)' = 0' = 1$.
$2$. The middle $AND$ gate has inputs $1$ and $1$. The output is $1 \cdot 1 = 1$.
$3$. The bottom $OR$ gate has inputs $0$ and $1$. The output is $0 + 1 = 1$.
$4$. The top $NOR$ gate receives inputs from the $NAND$ gate $(1)$ and the $AND$ gate $(1)$. Its output $y_1 = (1 + 1)' = 1' = 0$.
$5$. The bottom $NOR$ gate receives inputs from the $AND$ gate $(1)$ and the $OR$ gate $(1)$. Its output $y_2 = (1 + 1)' = 1' = 0$.
$6$. The final $AND$ gate receives inputs $y_1 = 0$ and $y_2 = 0$. Its output $y_3 = 0 \cdot 0 = 0$.
Therefore,the values are $y_1 = 0, y_2 = 0, y_3 = 0$.

(D) 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.
$NAND$ 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 circuit consists of a $NAND$ gate,a $NOR$ gate,and an $AND$ gate.
$1$. The $NAND$ gate has inputs $A=1$ and $B=0$. The output $y_1$ of a $NAND$ gate is given by $y_1 = \overline{A \cdot B}$. Thus,$y_1 = \overline{1 \cdot 0} = \overline{0} = 1$.
$2$. The $NOR$ gate has inputs $B=0$ and $C=1$. The output $y_2$ of a $NOR$ gate is given by $y_2 = \overline{B + C}$. Thus,$y_2 = \overline{0 + 1} = \overline{1} = 0$.
$3$. The $AND$ gate has inputs $y_1=1$ and $y_2=0$. The output $y_3$ of an $AND$ gate is given by $y_3 = y_1 \cdot y_2$. Thus,$y_3 = 1 \cdot 0 = 0$.
Therefore,the values are $y_1=1, y_2=0, y_3=0$.

(B) The circuit consists of a $NOR$ gate followed by an $AND$ gate and an $OR$ gate.
Let the output of the $NOR$ gate be $C = \overline{A + B}$.
Given $A = 1$ and $B = 1$,the output of the $NOR$ gate is $C = \overline{1 + 1} = \overline{1} = 0$.
The output $y_1$ is the output of an $AND$ gate with inputs $A$ and $C$. Thus,$y_1 = A \cdot C = 1 \cdot 0 = 0$.
The output $y_2$ is the output of an $OR$ gate with inputs $B$ and $C$. Thus,$y_2 = B + C = 1 + 0 = 1$.
Therefore,the values are $y_1 = 0$ and $y_2 = 1$.

(B) The circuit consists of a $NOR$ gate and a $NAND$ gate.
$1$. The inputs to the $NOR$ gate are $A=1$ and $B=0$. The output of the $NOR$ gate is $y_2 = \overline{A+B} = \overline{1+0} = \overline{1} = 0$.
$2$. The inputs to the $NAND$ gate are $A=1$ and the output of the $NOR$ gate $y_2=0$. The output of the $NAND$ gate is $y_1 = \overline{A \cdot y_2} = \overline{1 \cdot 0} = \overline{0} = 1$.
Therefore,the values are $y_1=1$ and $y_2=0$.

(A) $NAND$ gate performs the operation $Y = \overline{A \cdot B}$.
When this output is fed into a $NOT$ gate, the final output becomes $Y' = \overline{Y} = \overline{(\overline{A \cdot B})}$.
By the law of double negation, $\overline{(\overline{X})} = X$. Therefore, $Y' = A \cdot B$.
This is the Boolean expression for an $AND$ gate.

Input $(A, B)$	$NAND$ Output $(\overline{A \cdot B})$	$NOT$ Output $(A \cdot B)$
$0, 0$	$1$	$0$
$0, 1$	$1$	$0$
$1, 0$	$1$	$0$
$1, 1$	$0$	$1$

(A) Given inputs are $X=1$ and $Y=1$.
$1$. The input $X$ passes through a $NOT$ gate,so the input to the $OR$ gate is $\bar{X} = 0$.
$2$. The input $Y$ is directly connected to the $OR$ gate,so the input is $1$.
$3$. The output $P$ of the $OR$ gate is $P = \bar{X} + Y = 0 + 1 = 1$.
$4$. The input $X$ is directly connected to the $NAND$ gate,so the input is $1$.
$5$. The input $Y$ passes through a $NOT$ gate,so the input to the $NAND$ gate is $\bar{Y} = 0$.
$6$. The output $Q$ of the $NAND$ gate is $Q = \overline{X \cdot \bar{Y}} = \overline{1 \cdot 0} = \overline{0} = 1$.
$7$. Finally,$P$ and $Q$ are inputs to a $NOR$ gate to give output $R$.
$8$. $R = \overline{P + Q} = \overline{1 + 1} = \overline{1} = 0$.
Thus,$P=1, Q=1, R=0$.

(A) The circuit consists of two $NOT$ gates, two $AND$ gates, and one final $AND$ gate.
The inputs to the first $AND$ gate are $A$ and $\overline{B}$, so its output is $A \cdot \overline{B}$.
The inputs to the second $AND$ gate are $\overline{A}$ and $B$, so its output is $\overline{A} \cdot B$.
The final output $Y$ is the $AND$ operation of these two outputs: $Y = (A \cdot \overline{B}) \cdot (\overline{A} \cdot B)$.
Using the associative and commutative properties of Boolean algebra: $Y = A \cdot \overline{A} \cdot B \cdot \overline{B}$.
Since $A \cdot \overline{A} = 0$ and $B \cdot \overline{B} = 0$, the output $Y = 0 \cdot 0 = 0$ for all input combinations.
Therefore, the output $Y$ is always $0$.

(D) The output of an $\text{AND}$ gate is $Y = A \cdot B$.
$A$ $\text{NAND}$ gate is an $\text{AND}$ gate followed by a $\text{NOT}$ gate. Therefore,the output of a $\text{NAND}$ gate is $Y = \overline{A \cdot B}$.
According to De Morgan's theorem,$Y = \overline{A} + \overline{B}$.
The truth table for the $\text{NAND}$ gate is as follows:

$A$	$B$	$Y$
$0$	$0$	$1$
$0$	$1$	$1$
$1$	$0$	$1$
$1$	$1$	$0$

Comparing this with the given options,when $A=1$ and $B=1$,the output $Y=0$.

(A) Let the inputs be $A$ and $B$. The circuit consists of an $AND$ gate and an $OR$ gate whose outputs are fed into a final $OR$ gate.
Let the output of the $AND$ gate be $Y_1 = A \cdot B$.
Let the output of the first $OR$ gate be $Y_2 = A + B$.
The final output $X$ is the $OR$ operation of $Y_1$ and $Y_2$:
$X = Y_1 + Y_2 = (A \cdot B) + (A + B)$.
Using the Boolean identity $(A \cdot B) + A + B = A + B$,we get $X = A + B$.
Truth Table:

$A, B$	$Y_1 = A \cdot B$	$Y_2 = A + B$	$X = Y_1 + Y_2$
$0, 0$	$0$	$0$	$0$
$0, 1$	$0$	$1$	$1$
$1, 0$	$0$	$1$	$1$
$1, 1$	$1$	$1$	$1$

Since the output $X$ follows the truth table of an $OR$ gate,the circuit behaves like an $OR$ gate.

(D) The given circuit consists of two buffers (or $NOT$ gates connected in series to act as buffers) followed by a $NAND$ gate.
$1$. The input $A$ passes through a buffer,so the output is $A$.
$2$. The input $B$ passes through a buffer,so the output is $B$.
$3$. These two outputs $A$ and $B$ are fed into a $NAND$ gate.
$4$. The output of a $NAND$ gate with inputs $A$ and $B$ is $Y = \overline{A \cdot B}$.
$5$. This is the definition of a $NAND$ gate.
Therefore,the circuit is equivalent to a $NAND$ gate.

(D) The given circuit consists of two $NAND$ gates acting as $NOT$ gates (since their inputs are shorted) followed by a $NAND$ gate.
Let the inputs be $A$ and $B$.
The first two $NAND$ gates invert the inputs to produce $\bar{A}$ and $\bar{B}$.
The final $NAND$ gate takes these as inputs,so the 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 circuit behaves as an $OR$ gate.

(C) Integrated circuits are classified based on the number of logic gates or components they contain.
$SSI$ (Small Scale Integration) typically contains up to $10$ logic gates.
$MSI$ (Medium Scale Integration) typically contains between $10$ and $100$ logic gates.
$LSI$ (Large Scale Integration) typically contains between $100$ and $1000$ logic gates.
$VLSI$ (Very Large Scale Integration) typically contains more than $1000$ logic gates.
Therefore,an integrated circuit consisting of logic gates $\leq 1000$ is termed as $LSI$.

(B) The given circuit consists of a chain of $OR$ gates.
An $OR$ gate produces a high output $(1)$ if at least one of its inputs is high $(1)$.
The first $OR$ gate has inputs $1$ and $X$. Its output is $1 + X = 1$.
This output $1$ is fed as an input to the second $OR$ gate,which also has $X$ as another input. Its output is $1 + X = 1$.
Continuing this process,every subsequent $OR$ gate receives $1$ as one of its inputs.
Since one input to every $OR$ gate in the chain is $1$,the final output $Y$ will always be $1$ regardless of the value of $X$.

(C) The given logic circuit consists of a $NOT$ gate and an $OR$ gate.
Input $A$ is passed through a $NOT$ gate,which produces an output $\bar{A}$.
This output $\bar{A}$ and the input $B$ are then fed as inputs to an $OR$ gate.
The Boolean expression for an $OR$ gate is the sum of its inputs.
Therefore,the final output $Y$ of the circuit is $Y = \bar{A} + B$.

(D) The Boolean expression for a $NOR$ gate is $Y = \overline{A+B}$,where $A$ and $B$ are inputs and $Y$ is the output.
The truth table for a $NOR$ gate is:
| $A$ | $B$ | $Y = \overline{A+B}$ |
|---|---|---|
| $0$ | $0$ | $1$ |
| $0$ | $1$ | $0$ |
| $1$ | $0$ | $0$ |
| $1$ | $1$ | $0$ |
From the truth table,it is clear that when all inputs ($A$ and $B$) are $LOW$ $(0)$,the output of the $NOR$ gate is $HIGH$ $(1)$.

(D) To satisfy the condition $A=1, B=1$ and $D=1$,we evaluate each circuit:
$(a)$ The output is $D = A \cdot \bar{B} + \bar{A} \cdot B$. For $A=1, B=1$,$D = 1 \cdot 0 + 0 \cdot 1 = 0$.
$(b)$ The output is $D = \overline{(\bar{A} + B) + (A + \bar{B})}$. For $A=1, B=1$,$D = \overline{(0 + 1) + (1 + 0)} = \overline{1 + 1} = 0$.
$(c)$ The output is $D = (A + B) \cdot (\bar{A} + \bar{B}) = A \cdot \bar{B} + \bar{A} \cdot B$. For $A=1, B=1$,$D = 0$.
$(d)$ The output is $D = A \cdot B + \bar{A} \cdot \bar{B}$. For $A=1, B=1$,$D = 1 \cdot 1 + 0 \cdot 0 = 1 + 0 = 1$.
Thus,the circuit in option $(d)$ satisfies the condition.