UGC NET Computer Science Digital Logic Previous Year Question Paper and Solution with PDF

UGC NET Computer Science Previous Year Questions (PYQs)

UGC NET Computer Science Digital Logic PYQ

Qus : 1 UGC NET Computer Science PYQ

The number of gate inputs, required to realize expression $ABC+A\overline{B}CD+E\overline{F}+AD$ is

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science Sep 2022 (Paper II) PYQ

Solution

Given expression:

$ABC+A\overline{B}CD+E\overline{F}+AD$

Number of inputs in product terms:

$ABC=3$

$A\overline{B}CD=4$

$E\overline{F}=2$

$AD=2$

Total inputs for AND gates:

$3+4+2+2=11$

There are $4$ product terms, so OR gate needs $4$ inputs.

Total gate inputs:

$11+4=15$

Qus : 2 UGC NET Computer Science PYQ

Consider a logic gate circuit, with 8 input lines $(D_0,D_1,\ldots,D_7)$ and 3 output lines $(A_0,A_1,A_2)$ specified by following operations

$A_2=D_4+D_5+D_6+D_7$

$A_1=D_2+D_3+D_6+D_7$

$A_0=D_1+D_3+D_5+D_0$

Where $+$ indicates logical OR operation. This circuit is

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science Sep 2022 (Paper II) PYQ

Solution

In an octal to binary encoder, there are $8$ input lines and $3$ output lines.

The outputs represent binary form of input lines $D_0$ to $D_7$.

Standard equations are:

$A_2=D_4+D_5+D_6+D_7$

$A_1=D_2+D_3+D_6+D_7$

$A_0=D_1+D_3+D_5+D_7$

So, the given circuit represents an Octal to Binary encoder.

Qus : 3 UGC NET Computer Science PYQ

What will be the number of states when a $MOD-2$ counter is followed by a $MOD-5$ counter?

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science June 2019 (Paper II) PYQ

Solution

When one counter is followed by another counter, the total number of states is the product of their mod values.

Here,

First counter is $MOD-2$

Second counter is $MOD-5$

So, total number of states is

$2 \times 5=10$

Qus : 4 UGC NET Computer Science PYQ

For line drawing, which of the following algorithms are used?

A. DDA algorithm

B. Bresenham algorithm

C. Mid-point algorithm

D. Curve algorithm

Choose the most appropriate answer from the options given below:

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science June 2024 (Paper II) PYQ

Solution

Line drawing algorithms used in computer graphics are:

DDA algorithm

Bresenham algorithm

Mid-point algorithm is generally used for circle drawing.

Qus : 5 UGC NET Computer Science PYQ

Consider a $3$-bit counter, designed using $T$ flip flops as shown below. Assume the initial state of the counter given by $PQR$ as $000$, what are the next three states?

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science Dec 2022 Shift I (Paper II) PYQ

Solution

From the circuit,

$T_P=R$

$T_Q=P'$

$T_R=Q'$

For $T$ flip-flop:

If $T=0$, output remains same.

If $T=1$, output toggles.

Initial state:

$PQR=000$

Step 1:

$T_P=R=0,\ T_Q=P'=1,\ T_R=Q'=1$

So,

$PQR=011$

Step 2:

For $PQR=011$,

$T_P=R=1,\ T_Q=P'=1,\ T_R=Q'=0$

So,

$PQR=101$

Step 3:

For $PQR=101$,

$T_P=R=1,\ T_Q=P'=0,\ T_R=Q'=1$

So,

$PQR=000$

Therefore, next three states are:

$011, 101, 000$

Qus : 6 UGC NET Computer Science PYQ

Match List-I with List-II.

List-I (computing systems)	List-II (working/output)
(A) Half Adder	(I) Has $n$ input and $2^n$ output
(B) Decoder	(II) CPU Storage unit
(C) Register	(III) Used to store program at runtime
(D) Main memory	(IV) 2-bit addition circuit

Choose the correct answer from the options given below :

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science August 2024 (Paper II) PYQ

Solution

Half Adder → performs addition of two bits → 2-bit addition circuit → (IV)

Decoder → has $n$ inputs and $2^n$ outputs → (I)

Main memory → stores program during execution → (III)

Thus

(A) → (IV)

(B) → (I)

(D) → (III)

Qus : 7 UGC NET Computer Science PYQ

Consider the following statements regarding combinational and sequential circuits.

(A) Output of combinational circuits depends on the only current input.
(B) Output of combinational circuit depends on the both current input and previous output.
(C) Output of sequential circuit depends on the current input.
(D) Output of sequential circuit depends on both current input and previous output.

Choose the correct answer from the options given below :

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science August 2024 (Paper II) PYQ

Solution

Combinational circuit → output depends only on current input → (A) True

Sequential circuit → output depends current input + previous state → (D) True

(B) False

Qus : 8 UGC NET Computer Science PYQ

Find the Boolean expression for the logic circuit shown below:

$(1-\text{NAND gate},\ 2-\text{NOR gate},\ 3-\text{NOR gate})$

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science Dec 2018 (Paper II) PYQ

Solution

Gate $1$ is a NAND gate with inputs $A$ and $B$.

So, output of gate $1$ is

$\overline{AB}$

Now, gate $2$ is a NOR gate.

Its inputs are $\overline{A}$ and $B$.

So, output of gate $2$ is

$\overline{\overline{A}+B}$

Using De Morgan's law,

$\overline{\overline{A}+B}=A\overline{B}$

Now, gate $3$ is also a NOR gate.

Its inputs are $\overline{AB}$ and $A\overline{B}$.

So,

$Y=\overline{\overline{AB}+A\overline{B}}$

Now,

$\overline{AB}=\overline{A}+\overline{B}$

Therefore,

$Y=\overline{\overline{A}+\overline{B}+A\overline{B}}$

Since $\overline{B}+A\overline{B}=\overline{B}$,

$Y=\overline{\overline{A}+\overline{B}}$

Using De Morgan's law,

$Y=AB$

Qus : 9 UGC NET Computer Science PYQ

Consider the Boolean expression A(x, y, z) = x (y′ + z′). Which of the following is the complete sum-of-products form of the given Boolean expression?

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science 26 June 2025 (Paper II) PYQ

Solution

Given,

A = x (y′ + z′)

= xy′ + xz′

To express as a complete sum-of-products, expand using all variable combinations:

= xy′(z + z′) + xz′(y + y′)

= xy′z + xy′z′ + xyz′ + xy′z′

After simplification, duplicate terms are removed:

A = xyz′ + xy′z + xy′z′

Qus : 10 UGC NET Computer Science PYQ

Out of following steps in the proper sequence for simplifying a Boolean function using a Karnaugh map (K-map).

(A) Identify and group the largest possible cluster of $1$'s

(B) Draw the K-map for the given Boolean function

(D) Transfer the truth table values to the K-map

Choose the correct answer from the options given below :

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science August 2024 (Paper II) PYQ

Solution

Correct K-map procedure:

Step 1 → Draw the K-map

Step 2 → Fill truth table values

Step 3 → Group largest clusters of $1$'s

Step 4 → Write simplified expression

Sequence:

(B) → (D) → (A) → (C)

Qus : 11 UGC NET Computer Science PYQ

Binary equivalent to $(A0F)_{16}$ is:

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science 26 June 2025 (Paper II) PYQ

Solution

A = 1010, 0 = 0000, F = 1111 So, $(A0F){16} = 101000001111{2}$.

Qus : 12 UGC NET Computer Science PYQ

Simplify the following using K-Map

$F(A,B,C,D)=\sum(0,2,5,7,8,10,13,15)$

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science Dec 2022 Shift II (Paper II) PYQ

Solution

Given minterms are:

$0,2,5,7,8,10,13,15$

In binary form:

$0=0000,\ 2=0010,\ 8=1000,\ 10=1010$

Here, $B=0$ and $D=0$

So, group gives:

$B'D'$

Also,

$5=0101,\ 7=0111,\ 13=1101,\ 15=1111$

Here, $B=1$ and $D=1$

So, group gives:

$BD$

Therefore,

$F=BD+B'D'$

Qus : 13 UGC NET Computer Science PYQ

Which of the following gates do not give output 1 when both inputs are 0?

A. NAND gate

B. NOR gate

C. XOR gate

D. XNOR gate

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science 26 June 2025 (Paper II) PYQ

Solution

Step-by-step explanation:

a = 12, b = 2

a = a + b → a = 14

b = a / b → b = 7

a = a * b → a = 98

b = a - b → b = 91

→ c = 91

Hence, -(*i) = -91

Output: 91 (negative of c printed as -(-91) → 91)

Qus : 14 UGC NET Computer Science PYQ

Gray code equivalent to decimal number 8 is:

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science 26 June 2025 (Paper II) PYQ

Solution

Binary of 8 = 1000 Gray code is obtained by XORing each bit with the next bit. → MSB remains same. → Remaining bits: perform XOR with the bit to its left. So, $1000 \rightarrow 1100$

Qus : 15 UGC NET Computer Science PYQ

. Match List I with List II

List-I	List-II
A. Odd Function	I. NAND Gate
B. Universal Gate	II. XOR Gate
C. 2421 Code	III. Amplification
D. Buffer	IV. Self-complementing

Choose the correct answer from the options given below:

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science Nov 2021 (Paper II) PYQ

Solution

Odd function means output is $1$ when odd number of inputs are $1$. This is the property of XOR gate. So, $A-II$.

NAND gate is a universal gate because any logic circuit can be made using only NAND gates. So, $B-I$.

$2421$ code is a self-complementing code. So, $C-IV$.

Buffer is used for amplification or increasing driving capability. So, $D-III$.

Final matching is $A-II,\ B-I,\ C-IV,\ D-III$.

Qus : 16 UGC NET Computer Science PYQ

The Boolean expression for the following would be:

$F(A,B,C) = \Sigma(0,2,6)$

$d(A,B,C) = \Sigma(1,3,5)$

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UGC NET Computer Science UGC NET PYQ UGC NET Computer Science UGC NET Computer Science June 2024 (Paper II) PYQ

Solution

Minterms:

$0 = 000$

$2 = 010$

$6 = 110$

Don't care terms:

$1 = 001$

$3 = 011$

$5 = 101$

Using K-map simplification:

Group $(0,2)$ → $A'C'$

Group $(2,6)$ → $BC'$

Thus,

$F = A'C' + BC'$

Factorizing:

$F = C'(A' + B)$

Which simplifies to:

$F = A' + BC'$

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