CS Free MCQ Bank

[ 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 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 ]

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 ]

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 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 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 ]

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 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.