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[ Option A ]

A NOR gate is the complement of the OR gate. It gives output 1 only when all inputs are 0.
Boolean Expression : = (A+B)′
An XOR gate gives output 1 when the inputs are different or if the number of 1’s are odd and 0 when the inputs are the same.
Boolean Expression : = A⊕B
Now the given expression is (NOR)OR(XOR) i.e., (A+B)’ + (A⊕B)

Now, we construct the truth tables of NAND, OR, and AND gates and verify their behavior using the given equation.

The NAND gate behavior exactly matches with given expression.
Therefore, (NOR)OR(XOR) = NAND

[ Option B ]

A NAND gate produces the output as the negation of the AND operation, i.e., Y=(A⋅B)’. If one of its inputs is permanently connected to 0, then the AND operation becomes 0⋅B=0. So, Y=(0⋅B)′=0′=1. This means the output of the NAND gate will always be 1 regardless of the other input. Hence, the correct answer is 1.

[ Option B ]

The X-NOR gate, produces a high (1) output when both inputs are the same, either both 0 or both 1.
Truth—Table:

[ Option C ]

Given Boolean expression:
=AB+AB’C+AB’C’
=AB+AB’(C+C’)
=AB+AB’.1
=AB+AB’
=A(B+B’)
=A.1
=A

To get output A, there is two NAND gate required.

[ Option D ]

The EX-NOR (Exclusive-NOR) gate is the complement of the EX-OR (Exclusive-OR) gate. While an EX-OR gate gives output 1 when the inputs are different, the EX-NOR gate gives output 1 when the inputs are the same that means, both inputs are 0 or both are 1.

The Boolean expression for an EX-NOR gate is:
Y=A⊙B = (A⊕B)’ =  A.B+A’B’

[ Option D ]

Digital electronics are devices that operate using binary signals. They process information in discrete steps rather than continuous signals.
Examples of digital electronic devices include computers, which use digital circuits for all processing tasks, mobile phones, which contain digital processors and memory and digital cameras, which convert images into digital data.

[ Option D ]

The Exclusive-NOR (XNOR) function gives output 1 when both inputs are equal. The standard Boolean expression of XNOR is:
x⊙y = xy+x′y′
The option (A) is xy+x′y′. This is the standard form of XNOR.
The option (B) is x⊕y’. Complementing one input of an XOR gate inverts the output, so it becomes XNOR.
In other word, as we know x⊕y is equivalent to x’y+xy’. Similarly, x⊕y’ is equivalent to x’y’+xy’’, which further evaluated to x’y’+xy. This is the standard form of XNOR.
In other words, as we know that x⊕y is equivalent to x′y+xy′. Similarly, x⊕y′ is equivalent to x′y′+xy′′ (replacing y by y’). Since y′′=y, this expression further simplifies to x′y′+xy, which is the standard form of the XNOR function.
The option (C) is x’⊕y. Same logic as option (B).
The option (D) is x’⊕y’. Complementing both inputs of XOR does not change the operation.
The expression x′⊕y′ = x⊕y. This is XOR, not XNOR.

[ Option D ]

First simplify the given boolean expression.
=AB+AB’C+AB’C’
=AB+AB’(C+C’)
=AB+AB’.1
=AB+AB’
=A(B+B’)
=A.1
=A
The given expression simplifies to A, so the output can be directly taken from input A. Therefore, the minimum number of NAND gates required is zero.

[ Option C ]

The XOR (Exclusive OR) is a logical operation in which the output is 1 when the inputs are different, and the output is 0 when the inputs are the same.
In other words, the output of an XOR gate is 1 when the number of 1’s at the input is odd, otherwise, the output is 0.

XOR operation between (4AC0)₁₆ and (B53F)₁₆:

So, (4AC0)₁₆ XOR (B53F)₁₆ = (FFFF)₁₆

[ Option D ]

In digital electronics, a Universal Gate is a gate that can be used to build any digital circuit. Both NAND and NOR gates are considered universal gates. This means that by repeatedly using only NAND gates, we can construct AND, OR, NOT, XOR, XNOR and any other logic circuit. Similarly, by repeatedly using only NOR gates, we can implement all other logic functions as well.

[ Option C ]

The gate through this we can design or construct any type of digital circuit without using any other gate is called universal gate. Basically, these are of two types:

• NAND (AND+NOT) gate.
• NOR (OR+NOT) gate.

[ Option B ]

A NOR gate performs the operation : (A+B)′
To design a NOR gate using only two-input NAND gates, a total of four NAND gates are required. Consider the below given circuit diagram.

[ Option B ]

Principle of Duality says, every algebraic expression remains valid if you interchange:

• AND (·) with OR (+) 
• 0 with 1

In simple words, for every Boolean law or expression, there exists a dual expression that is also valid. E.g.: 

• A + 0 = A, its Dual is A . 1 = A
• A + A =A, its Dual is A . A = A 
• A . 0 = 0, its Dual is A + 1 =1

[ Option A ]

A functionally complete set of gates is a set using which any Boolean function can be implemented.
In digital circuit, the Universal Gates are functionally complete set of gates because a universal gate is a logic gate using which any Boolean function can be implemented without using any other type of gate.
The NAND and NOR gates are called universal gates because each of them alone is sufficient to construct all basic logic gates such as AND, OR and NOT.

• The NOT gate alone is not functionally complete because it can only perform inversion.

[ Option D ]

Digital circuits work with discrete (binary) voltage levels, usually represented as 0 (Low) and 1 (High). They do not use analog or continuous signals for processing, instead, they rely on step-like voltage changes.
The digital circuits are actually highly suitable for high-speed operations, that’s is why they are used in modern computers, microprocessors, and communication systems.

[ Option B ]

An inhibit gate is a logic gate in which the output of the circuit is controlled by a specific input called the inhibit (or control) input. When this inhibit input is applied with a particular logic level, it prevents or blocks the signal applied at the other input from affecting the output.
For a NAND gate, if any input is 0, the output becomes 1, regardless of the other input. Thus, when a low (0) is applied to one input, it inhibits the effect of the other input and forces the output to be high (1).

NAND ACTS AS AN INHIBIT GATE:

When the control input is 0, the output is forced to 1, so the signal input has no effect on the output.

[ Option C ]

A truth table shows all possible input combinations for a logic gate. If a gate has n inputs, the number of possible combinations is given by the formula 2ⁿ. For an OR gate with 4 inputs, the number of rows in the truth table will be 2⁴=16. This means there are 16 different input combinations, each producing an output.

[ Option D ]

The given output expression is Y=A′+B′+C′. By applying De Morgan’s law, this can be rewritten as Y=(A⋅B⋅C)′. This form directly represents the operation of a NAND gate, since a NAND gate produces the complement of the AND operation.

[ Option A ]

In a staircase wiring system, toggling either switch changes the state of the bulb. If the bulb is ON, it turns OFF and if it is OFF, it turns ON, regardless of the position of the other switch. This behavior matches the XOR (Exclusive OR) gate, where the output is 1 when the inputs are different and 0 when the inputs are the same.

[ Option A ]

The AND operation compares each corresponding bit of two binary numbers. The output is 1 only when both bits are 1, otherwise it is 0. 1010 AND 1100 = 1000

[ Option C ]

To perform XOR between two hexadecimal numbers, each hex digit is first converted into its 4-bit binary equivalent. For (FE35)₁₆ and (CB15)₁₆, the binary forms are 1111111000110101 and 1100101100010101, respectively. Performing bitwise XOR gives 0011 0101 0010 0000, which in hexadecimal is (3520)₁₆.

(FE35)₁₆ ⊕ (CB15)₁₆

First of all, write both numbers in binary form, so,

So,
FE35 = 1111 1110 0011 0101
CB15 = 1100 1011 0001 0101
Perform bitwise XOR:

Finally, (FE35)₁₆⊕(CB15)₁₆ = (3520)₁₆.

[ Option A ]

As we know, XOR gives output 1 when inputs are different, while XNOR gives output 1 when inputs are the same.
Suppose A and B are two boolean variable then:
A XOR B = A⊕B = A’B+AB’
A XNOR B = A⊙B = A’B’+AB
Option (A):
P’⊕Q’ = P⊙Q
P’⊕Q’ = P’’Q’+P’Q’’
P’⊕Q’ = PQ’+P’Q
Therefore, P’⊕Q’ = P⊕Q not P⊙Q
Option (B):
P’⊕Q =P⊙Q
P’⊕Q = P’’Q+P’Q’
P’⊕Q = PQ+P’Q’
Therefore, P’⊕Q = P⊙Q
Option (C):
P’⊕Q’ = P⊕Q
As we already discussed in Option (A) P’⊕Q’ = P⊕Q
Option (D):
(P⊕P’)⊕Q = (P⊙P’)⊙Q

[ Option D ]

In the ASCII coding system, uppercase and lowercase letters are represented by different binary patterns. The important point is that the difference between an uppercase letter and its corresponding lowercase letter is only one bit. This bit is known as the case bit and has a decimal value of 32, which in binary is 0100000. For example, the ASCII code of uppercase A is 1000001, while lowercase a is 1100001. The difference between them is the bit 0100000.
To change an uppercase letter into a lowercase letter, this case bit must be set to 1. In digital logic, the OR operation is used when we want to force a particular bit to become 1.
Therefore, performing an OR operation between the ASCII code of an uppercase letter and the mask 0100000 will convert it into the corresponding lowercase letter.

[ Option B ]

The given Boolean function f=x’y+xy’ represents the XOR (Exclusive-OR) operation between variables x and y. It is well known that an XOR function can be implemented using 4 two-input NAND gates.