XOR is a derived logic gate made from combination of basic gates (AND, OR, NOT). The three basic gates are AND, OR, and NOT.
Q.2Easy
The output of a NAND gate is 0 when:
Answer: B
NAND gate output is 0 only when all inputs are 1. NAND is NOT of AND operation. For 2-input: output is 0 only when both inputs are 1.
Q.3Easy
How many 2-input NAND gates are required to implement a 2-input AND gate?
Answer: B
AND gate can be implemented using two NAND gates. First NAND gate as AND, second NAND gate as NOT (both inputs same) to negate the output.
Q.4Medium
The truth table of which gate has odd number of 1s in output column for 2-input gate?
Answer: C
XOR gate has 2 ones in output (when inputs are different). OR has 3 ones. AND has 1 one. These are odd. XNOR has 2 ones (even). XOR is most commonly cited.
Q.5Medium
A combinational circuit with n inputs and m outputs is completely specified by:
Answer: D
A combinational circuit can be completely defined using truth table, Boolean expressions, or logic diagram. All three are equivalent representations.
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Q.6Medium
What is the simplified form of the Boolean expression A + AB + ABC?
Answer: A
Using Boolean algebra: A + AB + ABC = A(1 + B + BC) = A(1) = A. The dominant term A absorbs all other terms.
Q.7Easy
The decimal equivalent of binary number 10110.101 is:
Gray code (Reflected Binary Code) is designed such that successive values differ by only one bit. This is useful in error detection and mechanical encoders.
Q.9Medium
What is the Hamming distance between binary numbers 1011101 and 1001001?
Answer: B
Hamming distance = number of positions where bits differ. 1011101 vs 1001001: positions 3, 5, 6 differ. Hamming distance = 3.
Q.10Easy
A multiplexer with 8 inputs requires how many select lines?
Answer: B
For n select lines, 2^n inputs are possible. For 8 inputs: 2^n = 8, so n = 3. An 8:1 multiplexer needs 3 select lines (A, B, C).
Q.11Easy
The BCD equivalent of decimal 47 is:
Answer: A
In BCD, each decimal digit is converted separately to 4-bit binary. 4 = 0100, 7 = 0111. So 47 in BCD = 0100 0111.
Q.12Medium
Which of the following Boolean expressions represents the majority function for 3 inputs?
Answer: A
Majority function outputs 1 when majority (2 or more) inputs are 1. Expression: AB + BC + AC covers all cases where at least 2 inputs are 1.
Q.13Medium
In a 4-bit binary counter, after state 1111, the next state is:
Answer: B
In a 4-bit binary counter, states cycle from 0000 to 1111 (0 to 15). After 1111, it wraps around to 0000 (overflow occurs).
Q.14Easy
A full adder has how many inputs and outputs?
Answer: B
Full adder has 3 inputs (A, B, Cin) and 2 outputs (Sum, Cout). It differs from half adder which has 2 inputs and 2 outputs.
Q.15Medium
For a Karnaugh map of 4 variables, what is the minimum number of cells needed to cover all minterms?
Answer: C
A 4-variable K-map has 2^4 = 16 cells, each representing one minterm. To cover all minterms, minimum 16 cells are needed.
Q.16Medium
What is the maximum frequency of a synchronous counter if each flip-flop has a propagation delay of 10 ns?
Answer: B
In synchronous counters, all flip-flops change simultaneously, so maximum frequency is determined by one flip-flop delay. f_max = 1/t_pd = 101ns = 100 MHz.
Q.17Medium
An asynchronous counter with n flip-flops counts from 0 to:
Answer: B
An n-bit counter has 2^n possible states (from 0 to 2^n - 1). For 3-bit counter: states are 0 to 7 (000 to 111).
Q.18Medium
In Mealy machine, the output depends on:
Answer: C
Mealy machine output depends on both present state and present input. Moore machine output depends only on present state. Mealy machines are more efficient.
Q.19Easy
A JK flip-flop with J=1, K=1 performs:
Answer: D
JK flip-flop truth table: J=0,K=0 (No change), J=1,K=0 (Set), J=0,K=1 (Reset), J=1,K=1 (Toggle). When both are 1, output toggles.
Q.20Hard
For a 3-bit Johnson counter, the maximum count is:
Answer: B
Johnson counter (twisted ring counter) has 2n states for n flip-flops. For 3-bit: 2×3 = 6 states. It's more efficient than ripple counter but less than binary counter.
