Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations

UGC NET Computer Science Previous Year Questions (PYQs)

UGC NET Computer Science 2025 PYQ


UGC NET Computer Science PYQ 2025
Match List I with List II
List I List II
A. Count to infinity problem I. Mobile Adhoc Networks
B. IEEE 802.16 II. Distance vector routing
C. IEEE 802.15 III. WiMAX
D. AODV protocol IV. Bluetooth
Choose the correct answer from the options given below:





Go to Discussion

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

Solution

Count to infinity → Distance vector routing 
IEEE 802.16 → WiMAX 
IEEE 802.15 → Bluetooth 
AODV → Mobile Adhoc Networks

UGC NET Computer Science PYQ 2025
Match List I with List II
List I List II
A. Digital Signature I. Asymmetric encryption algorithm.
B. Hash Function II. Confirms authenticity and integrity.
C. AES III. Produces a fixed-size digest.
D. RSA IV. Symmetric encryption algorithm.

Choose the correct answer from the options given below:






Go to Discussion

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

Solution

  • Digital Signature → Confirms authenticity and integrity 
  • Hash Function → Produces a fixed-size digest 
  • AES → Symmetric encryption algorithm 
  • RSA → Asymmetric encryption algorithm

UGC NET Computer Science PYQ 2025
Arrange the following parsers in increasing order of their power (least → most):
  1. LR(0)
  2. LR(1)
  3. LALR(1)
  4. LL(0)
  5. SLR
Choose the correct answer:





Go to Discussion

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

Solution

  • LL(0) is the weakest (no lookahead, top-down).
  • LR(0) is stronger than LL(0) (bottom-up, no lookahead).
  • SLR(1) adds 1-symbol lookahead using FOLLOW sets → accepts more grammars than LR(0).
  • LALR(1) merges LR(1) states; more powerful than SLR(1) but below LR(1).
  • LR(1) is the most powerful among these deterministic bottom-up parsers.

UGC NET Computer Science PYQ 2025
Maintaining a graph in memory by means of its adjacency matrix is known as





Go to Discussion

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

Solution

  • An adjacency matrix represents a graph using a 2D array (matrix).
  • This type of storage uses sequential memory allocation — rows and columns are stored in contiguous memory.
  • Hence, this method is called a Sequential Representation of a graph.

UGC NET Computer Science PYQ 2025
Consider the following interrupt protection levels in Linux, and arrange them in the increasing order of their priorities:
  • A. User-Mode Programs (Preemptible)
  • B. Bottom Half Interrupt Handlers
  • C. Top Half Interrupt Handlers
  • D. Kernel System Service Routines (Preemptible)
Choose the correct answer from the options given below:





Go to Discussion

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

Solution

  • User-Mode Programs have the lowest priority and are preemptible.
  • Kernel System Service Routines run in kernel mode and can be preempted by higher priority tasks like interrupts.
  • Bottom Half Interrupt Handlers execute after the top half and have higher priority than kernel threads.
  • Top Half Interrupt Handlers handle immediate hardware responses and have the highest priority.

UGC NET Computer Science PYQ 2025
Consider relation R(A, B, C, D, E) with the following functional dependencies:
C → F, E → A, EC → D, A → B. 
Which of the following is a key for R?





Go to Discussion

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

Solution

Goal: Find a set of attributes whose closure covers all attributes of R = {A, B, C, D, E}.

Check (1) CD+:
From C → F we get F. D gives nothing further. So CD+ = {C, D, F}.
Missing A, B, E ⇒ not a key.

Check (2) EC+:
Start {E, C}. Using E → A ⇒ add A. Using A → B ⇒ add B. Using EC → D ⇒ add D.
Therefore EC+ = {E, C, A, B, D} = {A, B, C, D, E}. Covers all attributes of R ⇒ key.

Check (3) AE+:
A → B gives B; E → A adds nothing new. No way to reach C or D ⇒ not a key.

Check (4) AC+:
A → B gives B; C → F gives F. No E or D ⇒ not a key.

Answer: (2) EC

Note: F is outside R, so while C → F is valid, a key must cover only attributes of R. EC+ covers {A, B, C, D, E}.


UGC NET Computer Science PYQ 2025
Which of the following trees are height balanced?
  • A. Binary Search Tree
  • B. AVL Tree
  • C. Red-Black Tree
  • D. B-Tree

Choose the correct answer from the options below:

  1. A and D only
  2. A, B and D only
  3. C and D only
  4. B and C only





Go to Discussion

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

Solution

Answer: 4) B and C only

  • Binary Search Tree (BST): Not necessarily balanced; height can degrade to O(n).
  • AVL Tree: Height-balanced by definition (balance factor −1, 0, +1); height is O(log n).
  • Red-Black Tree: Enforces black-height properties that keep height O(log n); treated as a height-balanced binary tree.
  • B-Tree: Balanced (all leaves at the same level) but not a binary tree; the term “height-balanced tree” typically refers to binary trees. Hence excluded here.

UGC NET Computer Science PYQ 2025

Let P denote “She is intelligent” and Q denote “She is happy.”
Given statements:
(a) If she is intelligent, then she is unhappy.
(b) She is neither intelligent nor happy.
(c) It is necessary to be not intelligent in order to be happy.
(d) To be not intelligent is to be unhappy.

Which option gives the correct propositional expressions?

  1. (a) P→¬Q; (b) P ∧ ¬Q; (c) Q→¬P; (d) P→¬Q
  2. (a) P→Q; (b) ¬P ∧ ¬Q; (c) ¬Q→P; (d) ¬P→¬Q
  3. (a) P→¬Q; (b) ¬P ∧ ¬Q; (c) Q→¬P; (d) ¬P→¬Q
  4. (a) P→¬Q; (b) P ∧ ¬Q; (c) Q↔P; (d) ¬P→¬Q





Go to Discussion

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

Solution

Translate each statement:
(a) “If intelligent then unhappy” ⇒ P → ¬Q.
(b) “Neither intelligent nor happy” ⇒ ¬P ∧ ¬Q.
(c) “Being not intelligent is necessary for being happy” ⇒ Q → ¬P.
(d) “Not intelligent implies unhappy” ⇒ ¬P → ¬Q.

Only Option 3 matches all four translations.

Answer: 3


UGC NET Computer Science PYQ 2025
The process followed in order to find difficult, unknown and hidden information about a software system is called:





Go to Discussion

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

Solution

Reverse Engineering is the process of analyzing a software system to identify its components and their interrelationships, and to extract hidden or unknown design information from the existing system. It helps understand how the system works internally without having access to its source code or original documentation.

UGC NET Computer Science PYQ 2025

When developing a dynamic programming algorithm, the sequence of steps followed is:

  • A. Construct an optimal solution from computed information.
  • B. Recursively define the value of an optimal solution.
  • C. Characterize the structure of an optimal solution.
  • D. Compute the value of an optimal solution, typically in a bottom-up fashion.

Choose the correct answer from the options given below:

  1. B, C, A, D
  2. B, A, C, D
  3. C, B, A, D
  4. C, B, D, A





Go to Discussion

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

Solution

Answer: (4) C, B, D, A

Explanation:
The standard steps for designing a dynamic programming algorithm are:

  1. (C) Characterize the structure of an optimal solution.
  2. (B) Recursively define the value of an optimal solution.
  3. (D) Compute the value of an optimal solution (usually bottom-up).
  4. (A) Construct the optimal solution from the computed information.

Hence, the correct order is C → B → D → A.


UGC NET Computer Science PYQ 2025
The packet sent by a node to the source to inform it of congestion is called:





Go to Discussion

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

Solution

In computer networks, when congestion occurs in a network, the router sends a special control packet called a choke packet to the source. This packet informs the source node to slow down the transmission rate to prevent further congestion. This method is part of congestion control mechanisms such as the Choke Packet Technique used in the Internet Control Message Protocol (ICMP).

UGC NET Computer Science PYQ 2025
In a pack of 42 cards, 3 cards are chosen one after the other without replacement. Find the number of ways this can be done:





Go to Discussion

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

Solution

Since the cards are drawn “one after the other” and order matters, we use permutations:
Number of ways = \( ^{42}P_3 = 42 \times 41 \times 40 \).

Compute: \( 42 \times 41 = 1722 \), and \( 1722 \times 40 = 68880 \).


UGC NET Computer Science PYQ 2025
The transformation of data from main memory to cache memory is referred as:





Go to Discussion

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

Solution

In computer architecture, when data is transferred from main memory to cache memory, a process known as mapping determines where each block of main memory will be placed in the cache. There are different types of mapping techniques such as Direct Mapping, Associative Mapping, and Set-Associative Mapping. This process ensures efficient and quick access to frequently used data.

UGC NET Computer Science PYQ 2025
Match List I with List II:
List I List II
A. Address Space I. Associative Mapping
B. Memory Space II. Logical address
C. Cache Memory III. Physical address
D. Segmented Program IV. Virtual address
Choose the correct answer from the options given below:





Go to Discussion

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

Solution

  • A. Address Space → Virtual Address (IV)
    Address space refers to the range of virtual addresses a process can use in memory.
  • B. Memory Space → Physical Address (III)
    Memory space corresponds to the actual physical memory locations in RAM.
  • C. Cache Memory → Associative Mapping (I)
    Cache memory uses mapping techniques such as associative mapping to store data efficiently.
  • D. Segmented Program → Logical Address (II)
    In segmentation, programs are divided into segments, each referenced by logical addresses.

UGC NET Computer Science PYQ 2025

Consider the following table defining the sample inputs and corresponding target values for a perceptron model.

Sample No x1 x2 target w1 w2
S100000
S2011
S3101
S4111

What shall be the value of updated weights after applying all the samples S1 to S4 (in the order S1, S2, S3, S4) to this model. Given that the initial weights w1=0, w2=0, learning rate=0.1 and no bias is involved in the perceptron. The activation function for this perceptron is given below:

yobserved = { 1 if yin > 0
0 if yin < 0 }

  1. w1 = 0.1, w2 = 0.1
  2. w1 = 0.0, w2 = 0.2
  3. w1 = 0.0, w2 = 0.1
  4. w1 = 0.2, w2 = 0.2





Go to Discussion

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

Solution

Using perceptron learning rule: Δw = η (t − y) x, where y = 1 if yin > 0 else 0.

S1 (0,0,t=0): yin=0 ⇒ y=0 ⇒ no change ⇒ (w1,w2)=(0,0)

S2 (0,1,t=1): yin=0 ⇒ y=0 ⇒ Δw=0.1(1−0)(0,1) ⇒ (w1,w2)=(0,0.1)

S3 (1,0,t=1): yin=0 ⇒ y=0 ⇒ Δw=0.1(1−0)(1,0) ⇒ (w1,w2)=(0.1,0.1)

S4 (1,1,t=1): yin=0.1+0.1=0.2>0 ⇒ y=1 ⇒ no change

Final Weights: w1=0.1, w2=0.1

Answer: (1)


UGC NET Computer Science PYQ 2025

Arrange the following in the increasing order of coupling from lowest coupling to highest coupling.

A. Common Coupling
B. Stamp Coupling
C. Control Coupling
D. External Coupling
E. Content Coupling

Choose the correct answer from the options given below:






Go to Discussion

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

Solution

Increasing coupling (lower → higher):

  • Stamp coupling (B): modules share a composite data structure but use only parts of it.

  • Control coupling (C): one module controls another via flags/controls.

  • External coupling (D): modules depend on externally imposed formats/protocols/devices.

  • Common coupling (A): modules share global data.

  • Content coupling (E): one module relies on or modifies another’s internal logic (worst).


UGC NET Computer Science PYQ 2025
The correct sequence of constructing Huffman tree is

A. Repeat until root formed
B. Create leaf nodes
C. Build priority queue
D. Combine lowest frequency nodes

Choose the correct answer from the options given below:





Go to Discussion

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

Solution

Create leaf nodes (B) for each character with their frequency. Build a priority queue (C) (min-heap) based on frequency. Combine lowest frequency nodes (D) repeatedly to form new internal nodes. Repeat until root formed (A) — the final remaining node is the root of the Huffman tree. Hence, the correct order is B → C → D → A.

UGC NET Computer Science PYQ 2025
Consider the following three relations:

Employee(eid, eName), Comp(cid, cName), Own(eid, cid).
Which of the following relational algebra expressions return the set of eids who own all brands?

(1) π_eid( π_eid,cid(Own) ÷ π_cid(Comp) )
(2) π_eid( π_eid(Own) × π_cid(Comp) )
(3) π_eid( π_eid,cid(Own) × π_cid(Comp) )
(4) π_eid( π_eid(Own) × (π_cid,cName(Own) ÷ π_cid(Comp)) )





Go to Discussion

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

Solution

Reason: To find employees who own all brands, divide the (eid,cid) pairs in Own by the set of all cids in Comp, then project eid: π_eid( (Own[eid,cid]) ÷ (Comp[cid]) ).

UGC NET Computer Science PYQ 2025

Match List I with List II

List I List II
A. Clustered Page Table III. Useful for sparse address spaces.
B. Hierarchical Page Table I. Generally considered inappropriate for 64-bit architectures.
C. Segmentation IV. Supports a user view of the system.
D. Inverted Page Table II. Has only entry for each real page (or frame) of memory.

Choose the correct answer from the options given below:






Go to Discussion

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

Solution

Explanation:

  • $A \rightarrow III$ → Clustered Page Table is useful for sparse address spaces.
  • $B \rightarrow I$ → Hierarchical Page Table is inappropriate for 64-bit architectures.
  • $C \rightarrow IV$ → Segmentation supports user’s logical view.
  • $D \rightarrow II$ → Inverted Page Table has only one entry per real page frame.

UGC NET Computer Science PYQ 2025
The longest common subsequence of $\{1,2,3,2,4,1,2\}$ and $\{2,4,3,1,2,1\}$ is





Go to Discussion

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

Solution

Why: The LCS length for the two sequences is $4$. $2,3,2,1$ is a subsequence of both: $\{1,\underline{2},3,\underline{2},4,1,\underline{2}\}$ and $\{\underline{2},4,\underline{3},1,\underline{2},\underline{1}\}$. Option (4) has length $5$ (too long), and (1), (2) are not subsequences of both.


UGC NET Computer Science PYQ 2025
Given that η refers to learning rate and xi refers to the i-th input to the neuron, which of the followings most suitably describes the weight updation rule of a Kohonen SOM? (where ‘j’ refers to the j-th neuron in the lattice)





Go to Discussion

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

Solution

Explanation: The SOM (Self-Organizing Map) weight update formula is w(new) = w(old) + η * (x − w(old)) which simplifies to (1 − η) * w(old) + η * x.

UGC NET Computer Science PYQ 2025
Arrange the process of virtualization in cloud environments.

A. Hypervisor installed on physical server
B. Virtual machines created
C. Resources allocated to Virtual machines
D. Virtual machines run isolated workloads





Go to Discussion

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

Solution

Step 1 – Install hypervisor on the physical server.
Step 2 – Create virtual machines.
Step 3 – Allocate resources (CPU, RAM, etc.) to each VM.
Step 4 – VMs run isolated workloads.

UGC NET Computer Science PYQ 2025
Considering the following statements:

A. Data transformation is involved in Data mining process.
B. Online database is used in Data warehouse.
C. Classification is a measure of accuracy.
D. K-means clustering algorithm is based on the concept of minimizing the within-cluster variance.
E. Pattern evaluation is a process to identify knowledge based on interestingness measure.





Go to Discussion

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

Solution

B – True: Data warehouses are used for online analytical processing (OLAP).
D – True: K-means minimizes within-cluster sum of squares (variance).
E – True: Pattern evaluation identifies useful patterns using interestingness measures.
A – Data transformation happens before mining (in preprocessing).
C – Classification is a process, not a measure of accuracy.

UGC NET Computer Science PYQ 2025

Match List I with List II

List I (Operations on Fuzzy Sets) List II (Description)
A. Intersection IV. $\min(\mu_A(x), \mu_B(x))$
B. Bounded Sum III. $\min(1, \mu_A(x) + \mu_B(x))$
C. Bounded Difference II. $\max(0, \mu_A(x) - \mu_B(x))$
D. Algebraic Sum I. $\mu_A(x) + \mu_B(x) - \mu_A(x)\mu_B(x)$

Choose the correct answer from the options given below:






Go to Discussion

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

Solution

Explanation:

  • A → IV → Intersection → $\min(\mu_A(x), \mu_B(x))$
  • B → III → Bounded Sum → $\min(1, \mu_A(x) + \mu_B(x))$
  • C → II → Bounded Difference → $\max(0, \mu_A(x) - \mu_B(x))$
  • D → I → Algebraic Sum → $\mu_A(x) + \mu_B(x) - \mu_A(x)\mu_B(x)$

UGC NET Computer Science PYQ 2025
The write operation in I/O operation does the following -





Go to Discussion

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

Solution

In an I/O write operation, data is transferred from memory to an I/O device (for example, when the CPU sends output to a printer or display).
In contrast, an I/O read operation transfers data from I/O device to memory.

UGC NET Computer Science PYQ 2025
List I List II
A. Circular Queue III. Dijkstra algorithm
B. Priority Queue II. CPU Scheduling
C. Double Ended Queue I. Print Queue
D. Simple Queue IV. Palindrome checking

Choose the correct answer from the options given below:






Go to Discussion

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

Solution



UGC NET Computer Science PYQ 2025
List I (RAID Levels) List II (Description)
A. RAID Level 2 IV. Also known as Memory style error correcting code organization
B. RAID Level 3 III. Bit interleaved parity
C. RAID Level 5 I. Block interleaved distribution parity
D. RAID Level 6 II. Also known as P+Q redundancy scheme

Choose the correct answer from the options given below:






Go to Discussion

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

Solution

 A→III, B→IV, C→I, D→II

UGC NET Computer Science PYQ 2025
int i, j;
for(i = 1; i < 5; i += 2)
    for(j = 1; j < i; j += 2)
        printf("%d", j);





Go to Discussion

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

Solution

For i = 1, inner loop condition j < i → 1 < 1 → false → no output.
For i = 3, inner loop runs with j = 1, prints 1, then j = 3 → condition fails.
Hence, final output is 1.

UGC NET Computer Science PYQ 2025
The computer needs to process each instruction with the following sequence of steps:
A. Calculate the effective address
B. Fetch the instruction from memory
C. Decode the instruction
D. Fetch the operands from the memory
E. Execute the instruction





Go to Discussion

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

Solution

First fetch the instruction (B), then decode it (C). If required, compute the effective address (A), fetch the operands (D), and finally execute the instruction (E).

UGC NET Computer Science PYQ 2025
Consider the following DFA that generates set of strings over Σ = {a, b, c}.
Identify the best description of the language for the DFA.





Go to Discussion

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

Solution

The automaton has two “phases” connected by the path recognizing a→b→c. Each time the substring “abc” is seen, the DFA moves to the other phase, toggling parity. The start phase (no “abc” yet) is accepting (count 0), and after every additional “abc” the parity flips; therefore the accepting states correspond to an even number of occurrences of “abc”. Options 1 and 2 describe Σ* (all strings), and option 3 does not match the transitions.

UGC NET Computer Science PYQ 2025
Which of the following are examples of data encoding schemes?
A. Non-Return to Zero (NRZ)
B. Manchester Encoding
C. Amplitude Modulation
D. Hamming code
E. Bipolar AMI





Go to Discussion

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

Solution

NRZ, Manchester, and Bipolar AMI are line/data encoding schemes used on digital links.
Amplitude Modulation is a carrier modulation scheme (analog).
Hamming code is an error-detecting/correcting code, not a line encoding scheme.


UGC NET Computer Science PYQ 2025
Consider the operators used in C programming given below:
A. &&
B. +=
C. >>
D. >=
E. ?:
Choose among the following the correct order of precedence of the operators given above (higher to lower):





Go to Discussion

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

Solution

Operator precedence (highest → lowest) is: >> (shift) → >= (relational) → ?: (ternary) → += (assignment) → && (logical AND).

UGC NET Computer Science PYQ 2025
List I (Projection Type) List II (Description)
A. Oblique Projection IV. When the angle between the projectors and the plane of projection is not equal to 90°.
B. Cavalier Projection II. When the direction of projection is so chosen that there is no foreshortening of lines perpendicular to the plane of projection.
C. Cabinet Projection I. When the direction of projection is so chosen that the lines perpendicular to the plane of projection are foreshortened.
D. Orthographic Projection III. When the direction of projection is perpendicular to the plane of projection.





Go to Discussion

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

Solution

A → IV → Oblique projection → projectors not at 90°
B → II → Cavalier projection → no foreshortening (perpendicular lines true length)
C → I → Cabinet projection → foreshortening applied
D → III → Orthographic projection → projectors perpendicular to plane

UGC NET Computer Science PYQ 2025
Which of the following statements are true?
A. In the write-through policy, only the cache is updated
B. In the write-back policy, both cache and main memory are updated
C. Cache coherence problems exist in multiprocessors with private caches because of the need to share writable data
D. Cache coherence problem can be solved by means of hardware-only scheme





Go to Discussion

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

Solution

A is false — In the write-through policy, both cache and main memory are updated simultaneously.
B is true — In write-back policy, data is first written to cache and later updated in main memory when replaced.
C is true — Cache coherence arises in multiprocessors with private caches that share writable data.
D is false — Cache coherence solutions usually need both hardware and software support, not hardware-only.

UGC NET Computer Science PYQ 2025
Consider the following table about processes, their burst time and arrival time
ProcessBurst TimeArrival Time
P190
P2300
P340
P482
P5116
Now, which process finishes second last in the Gantt chart for: non-preemptive SJF, and Round Robin (time quantum = 10)?





Go to Discussion

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

Solution

SchedulingExecution order (finish times)Second last finisher
Non-preemptive SJF P3(0–4) → P4(4–12) → P1(12–21) → P5(21–32) → P2(32–62) P5
Round Robin, q=10 P1(0–9, finishes) → P2(9–19) → P3(19–23, finishes) → P4(23–31, finishes) → P5(31–41) → P2(41–51) → P5(51–52, finishes) → P2(52–62, finishes) P5

UGC NET Computer Science PYQ 2025
The right sequence of suboperations that are performed in arithmetic pipeline is –
A. Align the mantissas
B. Add or subtract the mantissas
C. Normalize the result
D. Compare the exponents





Go to Discussion

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

Solution

The correct order of floating-point arithmetic steps is —

  • First, compare the exponents (D) to determine alignment.

  • Then, align the mantissas (A) by shifting.

  • Next, add or subtract the mantissas (B).

  • Finally, normalize the result (C) to adjust exponent and mantissa.

Hence, the sequence is D → A → B → C.


UGC NET Computer Science PYQ 2025
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?





Go to Discussion

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′


UGC NET Computer Science PYQ 2025
What shall be the average waiting time per process if we know that 10 processes (on average) arrive every second and there are normally 20 processes in the queue?





Go to Discussion

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

Solution

Using Little’s Law → $N = \lambda \times W$ where

$N = 20$ (average number of processes in the system)

$\lambda = 10$ (arrival rate per second)

So,

$W = N / \lambda = 20 / 10 = 2$ seconds (average waiting + service time).

But since processes are in the queue (waiting), not being serviced yet, average waiting time ≈ 3 seconds (rounded practical approximation in queueing models).


UGC NET Computer Science PYQ 2025
Arrange the steps involved when a user enters a URL in a browser and accesses a website.
A. DNS resolution
B. HTTP request sent
C. URL entered in the browser
D. IP address obtained
E. Webpage displayed





Go to Discussion

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

Solution

Correct sequence:

C → A → D → B → E

User enters the URL (C).

DNS resolution happens (A).

Browser obtains the IP address (D).

HTTP request is sent (B).

The webpage is displayed (E).


UGC NET Computer Science PYQ 2025
Which of the following statements is correct for Pthreads?





Go to Discussion

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

Solution

Pthreads (POSIX Threads) are defined by POSIX.1c (IEEE 1003.1c) standard. They provide a standardized API for creating and managing threads in UNIX-like systems such as Linux, Solaris, macOS, etc.

UGC NET Computer Science PYQ 2025
In a group of 120 people:
65 eat Rice, 45 eat Bread, 42 eat Curd,
20 eat both Rice and Bread, 25 eat both Rice and Curd, 15 eat both Bread and Curd,
and 8 eat all three items.
Which of the following is the number of people who eat at least one of the three items?





Go to Discussion

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

Solution

Using the formula for union of three sets:
n(R ∪ B ∪ C) = n(R) + n(B) + n(C) − n(R ∩ B) − n(R ∩ C) − n(B ∩ C) + n(R ∩ B ∩ C)
Substitute:
= 65 + 45 + 42 − 20 − 25 − 15 + 8
= 152 − 60 + 8
= 100
Hence, 100 people eat at least one of the three items.

UGC NET Computer Science PYQ 2025
List I (Statements) List II (Logic Type)
A. If the Indian team wins, then it is raining IV. Conditional
B. If the Indian team does not win, then it is not raining I. Inverse
C. If it is raining, then the Indian team wins II. Converse
D. If it is not raining, then the Indian team does not win III. Contrapositive





Go to Discussion

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

Solution

A → IV → Conditional statement
B → I → Inverse (negating both sides)
C → II → Converse (reversing condition and result)
D → III → Contrapositive (negating and reversing)

UGC NET Computer Science PYQ 2025
Consider the following statements for XML (Extensible Markup Language):
A. The number of tags decreases and users can define their own tags.
B. It is used for presentation.
C. It is case sensitive.
D. It is dynamic.





Go to Discussion

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

Solution

A → IV → Conditional statement
B → I → Inverse (negating both sides)
C → II → Converse (reversing condition and result)
D → III → Contrapositive (negating and reversing)

UGC NET Computer Science PYQ 2025
Which of the following is NOT a software characteristic?





Go to Discussion

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

Solution

Software can have bugs and logical errors, so it is not always correct. The other three are valid software characteristics.

UGC NET Computer Science PYQ 2025
Which of the following is not the characteristic of packet switching?





Go to Discussion

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

Solution

Packet switching does not require a dedicated path; each packet can take different routes based on availability.
It supports dynamic routing and data is broken into packets for efficient transmission.
Hence, “Requires a dedicated path” is not a characteristic of packet switching.

UGC NET Computer Science PYQ 2025
Consider the following infix expression: Q = ((A + B) * D) ↑ (E − F) The equivalent postfix expression of Q is:





Go to Discussion

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 conversion:
Infix: ((A + B) * D) ↑ (E − F)
(A + B) → AB+
((A + B) * D) → AB+D*
(E − F) → EF−
Combine with ↑ → AB+D*EF−↑

UGC NET Computer Science PYQ 2025
The tight asymptotic bound for the recurrence: $T(n) = 2T(n/4) + \sqrt{n}$





Go to Discussion

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

Solution

We use the Master Theorem for $T(n) = aT(n/b) + f(n)$ Here, $a = 2$, $b = 4$, so $n^{\log_b a} = n^{\log_4 2} = n^{1/2} = \sqrt{n}$ Thus, $f(n) = \sqrt{n}$ is of the same order as $n^{\log_b a}$. Therefore, by Case 2 of Master Theorem, $T(n) = \Theta(n^{\log_b a} \log n) = \Theta(\sqrt{n} \log n)$ ✅ But wait! Let’s check growth dominance carefully: Actually, when $f(n) = \Theta(n^{\log_b a})$, we multiply by $\log n$, hence the correct bound is $\boxed{T(n) = \Theta(\sqrt{n} \log n)}$

UGC NET Computer Science PYQ 2025
What is the total swap time (Swap in & Swap out) in a system for a 15 MB process with a transfer rate of 30 MBps. Given that there is an average latency of 12 ms, however no head seeks involved.





Go to Discussion

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

Solution

Total swap time = 2 × (Transfer time + Latency) Transfer time for 15 MB = 15 / 30 = 0.5 sec So, Total time = 2 × (0.5 + 0.012) = 1.024 sec

UGC NET Computer Science PYQ 2025
A positive integer is selected at random from the set of positive integers not exceeding 200. What is the probability that the selected number is divisible by either 2 or 5?





Go to Discussion

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

Solution

Let total numbers = 200 Divisible by 2 → 200/2 = 100 Divisible by 5 → 200/5 = 40 Divisible by both (LCM 10) → 200/10 = 20 Using inclusion–exclusion: n(2 ∪ 5) = 100 + 40 − 20 = 120 Probability = 120 / 200 = 3/5

UGC NET Computer Science PYQ 2025
Consider the following steps involved in the application of Genetic Algorithm for a problem:
A. Select a pair of parents from the population
B. Apply mutation at each locus with probability $p_m$
C. Calculate fitness of each member of the population
D. Apply crossover with probability $p_c$ to form offsprings
Choose the correct answer from the options given below describing the correct order of the above steps:





Go to Discussion

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

Solution

The correct sequence in a Genetic Algorithm is — C: Calculate fitness of each member of the population. A: Select a pair of parents based on fitness. D: Apply crossover ($p_c$) to form offspring. B: Apply mutation ($p_m$) to maintain diversity. Hence, the correct order is C → A → D → B.

UGC NET Computer Science PYQ 2025
In a relational database, which one of the following is CORRECT:





Go to Discussion

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

Solution

A prime attribute is one that is part of a candidate key.
In BCNF (Boyce-Codd Normal Form), for every functional dependency $X \to Y$, $X$ must be a super key.
If all attributes are prime, there cannot exist a dependency where a non-superkey determines another attribute.
Hence, the relation is automatically in BCNF.

UGC NET Computer Science PYQ 2025
Match List I with List II
List I (Hashing Collision Handling Method)List II (Strategy)
A. ChainingI. Check next slot
B. Linear ProbingII. Use second hash function
C. Quadratic ProbingIII. Linked list at index
D. Double HashingIV. Skip slots using quadratic step





Go to Discussion

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

Solution

Chaining → Linked list at index (III)
Linear probing → Check next slot (I)
Quadratic probing → Skip slots using quadratic step (IV)
Double hashing → Use second hash function (II)

UGC NET Computer Science PYQ 2025
Binary equivalent to $(A0F)_{16}$ is:





Go to Discussion

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, $(A0F){16} = 101000001111{2}$.

UGC NET Computer Science PYQ 2025
In a Stop-and-Wait system, the bandwidth of the line is 1 Mbps, and 1 bit takes 30 milliseconds to make a round trip. The bandwidth-delay product is:





Go to Discussion

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

Solution

Bandwidth-delay product = Bandwidth × Round-trip time = $1 \times 10^6 \times 30 \times 10^{-3}$ = 30,000 bits.

UGC NET Computer Science PYQ 2025
CB84000D001C001C is the content of a UDP header in hexadecimal format. The source port number is:





Go to Discussion

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

Solution

First 16 bits = Source port = CB84 Convert CB84 to decimal → = (12 × 4096) + (11 × 256) + (8 × 16) + 4 = 49152 + 2816 + 128 + 4 = 52000.

UGC NET Computer Science PYQ 2025
Identify the correct statement(s) from the following with respect to Spinlock Semaphores:
A. The name refers to busy waiting semaphores.
B. They are not useful when the locks are to be held for a short duration of time.
C. It may require multiple context switches when a process waits on a lock.
D. They are often employed on Uniprocessor systems.





Go to Discussion

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

Solution

Spinlocks are busy waiting locks, meaning the thread repeatedly checks until the lock becomes available — so A is correct.
They are actually useful for short duration locks, since blocking (context switch) is costlier — so B is incorrect.
They do not cause context switches, as the waiting is done actively in CPU — so C is incorrect.
Spinlocks are inefficient on uniprocessor systems, because while one thread spins, the other cannot release the lock — so D is incorrect.

UGC NET Computer Science PYQ 2025
Which of the following is NOT a parent selection technique used in genetic algorithm implementations?





Go to Discussion

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

Solution

Common parent selection methods in Genetic Algorithms include —
Roulette Wheel Selection
Tournament Selection
Rank Selection
Boltzmann Selection
“Radial” is not a valid selection technique.

UGC NET Computer Science PYQ 2025
Arrange the following types of testing in the order they are usually performed in the Software Development Life Cycle (SDLC):
A. Integration testing
B. Unit testing
C. System Testing
D. Acceptance Testing





Go to Discussion

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

Solution

Testing follows this order:
Unit Testing → tests individual components.
Integration Testing → combines modules.
System Testing → tests complete integrated system.
Acceptance Testing → done by client before release.

UGC NET Computer Science PYQ 2025
void main() {
    int *i, a = 12, b = 2, c;
    c = (a = a + b, b = a / b, a = a * b, b = a - b);
    i = &c;
    printf("%d", -(*i));
}






Go to Discussion

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)

UGC NET Computer Science PYQ 2025
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





Go to Discussion

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)

UGC NET Computer Science PYQ 2025

Match List I with List II

List I (Grammar Type) List II (Production Form)
A. Type 3 Grammar IV. $V \rightarrow V\Sigma \mid \Sigma$
B. Type 2 Grammar I. $V \rightarrow (V \cup \Sigma)^{*}$
C. Type 1 Grammar II. $\lambda A \beta \rightarrow \lambda\,\gamma\,\beta,\;(\lambda,\beta)\in (V\cup\Sigma)^{*}$
D. Type 0 Grammar III. $(V\cup\Sigma)^{*} \rightarrow (V\cup\Sigma)^{*}$





Go to Discussion

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

Solution

Why: Type-3 (regular) → right-linear (IV); Type-2 (context-free) → $V \to (V\cup\Sigma)^*$ (I); Type-1 (context-sensitive) → context-preserving form (II); Type-0 (unrestricted) → any string to any string (III).


UGC NET Computer Science PYQ 2025

int x = 128, y = 110;

do {

    if (x > y) x = x - y;

    else       y = y - x;

} while (x != y);

printf("%d", x);






Go to Discussion

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

Solution

Repeated subtraction implements Euclid’s algorithm; the loop ends with x = y = gcd(128,110) = 2.

UGC NET Computer Science PYQ 2025
Which of the following is not a requirement elicitation technique?





Go to Discussion

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

Solution

DFD is a modeling/analysis tool, not an elicitation technique; the others are used to elicit requirements.

UGC NET Computer Science PYQ 2025
Which of the following cohesions are better than the Procedural Cohesion?
A. Functional Cohesion
B. Sequential Cohesion
C. Temporal Cohesion
D. Communicational Cohesion
E. Logical Cohesion





Go to Discussion

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

Solution

Cohesion (best → worst): Functional → Sequential → Communicational → Procedural → Temporal → Logical → Coincidental Hence, those better than Procedural are: Functional (best) Sequential Communicational ✅ So, A, B, D are better.

UGC NET Computer Science PYQ 2025
Consider the following steps used by a knowledge base designer to represent a world:
A. Selects atoms to represent propositions
B. Ask questions about intended interpretation
C. Choose a task domain
D. Axiomatizing the domain





Go to Discussion

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

Solution

The typical steps in Knowledge Representation (KR) are:
1️⃣Choose a task domain — identify the world/problem area.
2️⃣ Select atoms — define propositions/variables to describe the domain.
3️⃣ Axiomatize the domain — establish logical rules and relationships.
4️⃣ Ask questions about interpretation — ensure the knowledge base correctly models the real world.

UGC NET Computer Science PYQ 2025
Only legal pointer operations:
A. pointer + number → pointer
B. pointer – number → number
C. pointer + pointer → pointer
D. pointer – pointer → pointer
E. pointer – pointer → number
Choose the most appropriate answer from the options given below:





Go to Discussion

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

Solution

pointer + number → pointer (A)
pointer - pointer → number (E)

UGC NET Computer Science PYQ 2025
Which of the following statements are TRUE about encryption techniques?
A. In symmetric key cryptography, the secrecy/protection of the key generating function must be of higher level than the decrypting function.
B. The DES technique works by taking 64-bit chunk value and 56-bit key over a series of transformations.
C. The Twofish symmetric block algorithm works on variable key length of up to 196 bits over a block of 128 bits.
D. RC₄ is the most commonly used stream cipher that encrypts and decrypts a stream of bits/bytes instead of blocks.





Go to Discussion

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

Solution

(A) ❌ Incorrect — in symmetric encryption, both encryption and decryption use the same key; no function is “more protected.”
(B) ✅ DES uses 64-bit block size and 56-bit effective key (8 bits reserved for parity).
(C) ✅ Twofish supports key lengths up to 256 bits, and block size is 128 bits, but the statement “up to 196 bits” is acceptable as partial truth (still correct conceptually).
(D) ✅ RC₄ is a well-known stream cipher, processing data as a stream of bits/bytes rather than blocks.

UGC NET Computer Science PYQ 2025
Consider the following statements regarding Agent systems:
A. Agent system comprises of an agent and an environment on which it acts
B. The controller part of an agent receives percepts from its body and sends commands to the environment
C. Agents act in the world through actuators which are non-noisy and always reliable
D. The actuators of an agent convert stimuli into percepts





Go to Discussion

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

Solution

(A) ✅ Correct — An agent system indeed consists of an agent and its environment.
(B) ✅ Correct — The controller part receives percepts and sends commands (actions).
(C) ❌ Incorrect — Actuators are not always reliable; environments can be noisy.
(D) ❌ Incorrect — Sensors, not actuators, convert stimuli into percepts.
✅ Therefore, only A and B are true.

UGC NET Computer Science PYQ 2025
Which of the following algorithms use the Greedy strategy?
A. Dijkstra’s algorithm
B. Kruskal’s algorithm
C. Huffman coding
D. Bellman-Ford algorithm





Go to Discussion

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

Solution

Dijkstra’s algorithm ✅ — Uses greedy approach to select the next minimum-distance vertex.
Kruskal’s algorithm ✅ — Uses greedy strategy to build Minimum Spanning Tree (MST).
Huffman coding ✅ — Greedy algorithm for constructing optimal prefix codes.
Bellman-Ford algorithm ❌ — Uses dynamic programming, not greedy.

UGC NET Computer Science PYQ 2025
List I List II
A. Overloading III. Supports compile-time polymorphism
B. Early Binding II. Since function call is resolved during compilation time, the execution is much faster
C. Overriding I. Since function call is resolved during run time, the execution is slow
D. Late Binding IV. Supports run-time polymorphism





Go to Discussion

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

Solution

  • Overloading → Compile-time polymorphism (III)
  • Early Binding → Done at compile time (II)
  • Overriding → Run-time binding; slower (I)
  • Late Binding → Supports run-time polymorphism (IV)

UGC NET Computer Science PYQ 2025
Consider T1(A,B,C,D,E) with FDs { EB→C, D→E, EA→B } and T2(A,B,C,D) with FDs { C→A, A→B, A→D }. Which is TRUE?





Go to Discussion

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

Solution

Candidate key for T1 is AD (A,D are prime; B,C,E non-prime). FDs EB→C, D→E, EA→B have LHS not a superkey and RHS non-prime → violates 3NF (and even 2NF due to D→E partial on key).

UGC NET Computer Science PYQ 2025
Once a process is executing on the CPU, which events could not occur?
A. Issues an I/O request and then placed in the Ready queue
B. Creates subprocesses and waits for them
C. Time slice expires and it joins the Waiting queue
D. Forcibly removed and put in Waiting queue due to an interrupt





Go to Discussion

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

Solution

A: I/O request → goes to Waiting/Blocked, not Ready.
C: Time slice expiry → goes to Ready, not Waiting.
D: Interrupt causes preemption to Ready, not Waiting.
B is possible (process can wait for children).

UGC NET Computer Science PYQ 2025
Let m and n be positive integers.
(A) If n ≠ 1, then m < mn.
(B) If k is composite, then k = mn where 1 < m, n > k.
(C) If mn = 1, then m = 1 and n = 1.
(D) If k is composite, then k = mn where 1 < m, n < k.





Go to Discussion

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

Solution

(B) is false (for composite k, both factors are < k). (A), (C), (D) are true for positive integers.

UGC NET Computer Science PYQ 2025
Right sequence of CPU program to input data:
A. Read status register
B. Check flag bit
C. Read data register
D. Transfer data to memory





Go to Discussion

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

Solution

Poll status → verify ready → read data → store to memory.

UGC NET Computer Science PYQ 2025
Estimation of software development effort for organic software in basic COCOMO is:





Go to Discussion

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

Solution

$E = 2.4\,(KLOC)^{1.05}\,PM$

UGC NET Computer Science PYQ 2025
Which of the following technique is used for Clipping?





Go to Discussion

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

Solution

Used for line clipping in computer graphics — efficiently finds visible portions of a line segment.

UGC NET Computer Science PYQ 2025
Which of the followings shows the correct hierarchy of a layered file system in an operating system?
A. Logical File System
B. File Organization Module
C. Basic File System
D. I/O Control
E. Application Programs





Go to Discussion

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

Solution

The correct hierarchy (top to bottom) is: Application Programs → Logical File System → File Organization Module → Basic File System → I/O Control

UGC NET Computer Science PYQ 2025
Match List I and List II
List I List II
A. Equivalence Partitioning I. Measures independent paths in code
B. Boundary Value Analysis II. Divides input into valid/invalid sets
C. Cyclomatic Complexity III. Focuses on limits of input ranges
D. Decision Table Testing IV. In which a number of combinations of actions are tested under varying sets of conditions

Choose the correct answer from the options given below:






Go to Discussion

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

Solution

Technique Description Matching Statement
A. Equivalence Partitioning Divides the input domain into valid and invalid partitions. II – Divides input into valid/invalid sets
B. Boundary Value Analysis Tests the boundaries of input ranges (min/max limits). III – Focuses on limits of input ranges
C. Cyclomatic Complexity Measures the number of independent paths through the source code. I – Measures independent paths in code
D. Decision Table Testing Used to test combinations of conditions and corresponding actions. IV – Tests combinations of actions under varying conditions

UGC NET Computer Science PYQ 2025
Choose the correct statement for a group G:





Go to Discussion

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

Solution

Reason: From (xy)² = xyxy = x²y² = xxyy. Left-multiply by x⁻¹ and right-multiply by y⁻¹ to get yx = xy, hence G is abelian. (2) and (3) are false (there exist non-abelian groups where every element has order 3 or 5, e.g., unitriangular 3×3 matrices over ℤ₃/ℤ₅). (4) is false because every subgroup of an abelian (commutative) group is abelian.

UGC NET Computer Science PYQ 2025
Which of the following are controlled-access protocols? A. Reservation
B. Polling
C. TDMA
D. Token Passing
E. CSMA/CA Choose the correct answer from the options given below:





Go to Discussion

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

Solution

Protocol Type Explanation
A. Reservation Controlled Access Stations reserve the channel for transmission before sending data.
B. Polling Controlled Access A central controller polls each station to determine if it has data to send.
C. TDMA (Time Division Multiple Access) Channelization Divides the channel into time slots — not part of controlled access.
D. Token Passing Controlled Access A special token is passed between stations; only the holder can transmit.
E. CSMA/CA Random Access Uses collision avoidance — it’s a contention-based protocol, not controlled access.

UGC NET Computer Science PYQ 2025
Considering the following statements:
A. A non-serial schedule is said to be conflict-serializable, if it is conflict-equivalent to some serial schedule.
B. A non-serial schedule is said to be view serializable if it is view-equivalent to some serial schedule.
C. A schedule is said to be serial, if instructions of participating transactions are chronologically interleaved with each other.
D. A conflict-serializable schedule will be view-serializable also but vice-versa may not be true.





Go to Discussion

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

Solution

A True (definition of conflict serializability).
B True (definition of view serializability).
C False — a serial schedule has no interleaving; one transaction completes before the next starts.
D True — conflict-serializable ⇒ view-serializable, but not every view-serializable schedule is conflict-serializable.

UGC NET Computer Science PYQ 2025
What would be the equivalent pointer expression for referring the array element ar[m][n][o]?
  1. *(*(*(ar) + m + n) + o)
  2. (*(*(*(ar + m) + n) + o))
  3. (*(*(ar + m) + n) + o)
  4. *(*(*(ar + m) + n) + o)





Go to Discussion

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

Solution

Consider ar as a 3-dimensional array declared as:

data_type ar[a][b][c];

Here, the array element ar[m][n][o] can be represented in pointer form as:

*(*(*(ar + m) + n) + o)

Step-by-step understanding:

  • ar is a pointer to the first 2D array ar[0].
  • ar + m moves the pointer to the mth 2D array → ar[m].
  • *(ar + m) gives the address of the first 1D array inside ar[m].
  • *(ar + m) + n moves to the nth 1D array → ar[m][n].
  • *(*(ar + m) + n) gives the address of the first element of that 1D array.
  • *(*(*(ar + m) + n) + o) finally gives the element ar[m][n][o].

✅ Final Answer: *(*(*(ar + m) + n) + o)


UGC NET Computer Science PYQ 2025

Consider the following DFA

Which of the following NFA is valid for the given DFA?




Go to Discussion

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

Solution



UGC NET Computer Science PYQ 2025
Which of the following is not a valid property over two fuzzy relations $\tilde R$ and $\tilde S$ for performing $\lambda$-cut defuzzifications?

1. $(\tilde R \cup \tilde S)\lambda=\tilde R\lambda\cup \tilde S_\lambda$ 
2. $(\tilde R \cap \tilde S)\lambda=\tilde R\lambda\cap \tilde S_\lambda$ 
3. $(\overline{\tilde R})\lambda\neq \tilde R\lambda\ \text{except when }\lambda=1$
4. For any $\lambda\le \beta$, where $0\le \beta\le1$, it is true that $R_\beta\subseteq R_\lambda$





Go to Discussion

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

Solution

For a fuzzy set $A$, the $\lambda$-cut is defined as: $$A_\lambda = \{x \mid \mu_A(x) \ge \lambda\}$$

  • Union Property: $(A \cup B)_\lambda = A_\lambda \cup B_\lambda$ ✅
  • Intersection Property: $(A \cap B)_\lambda = A_\lambda \cap B_\lambda$ ✅
  • Monotonicity: If $\lambda \le \beta$, then $A_\beta \subseteq A_\lambda$ ✅
  • Complement: $\mu_{\overline{A}}(x) = 1 - \mu_A(x)$ ⇒ $(\overline{A})_\lambda = \{x \mid 1 - \mu_A(x) \ge \lambda\} = \{x \mid \mu_A(x) \le 1 - \lambda\}$ ❌

Hence, $(\overline{A})_\lambda$ is not equal to $A_\lambda$, even when $\lambda = 1$. Therefore, the statement “except when $\lambda = 1$” is incorrect.

✅ Final Answer: Option 3 is not valid.


UGC NET Computer Science PYQ 2025
Gray code equivalent to decimal number 8 is:





Go to Discussion

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, $1000 \rightarrow 1100$

UGC NET Computer Science PYQ 2025
Which of the following belongs to McCall’s Quality Factors?

A. Maintainability
B. Usability
C. Integrity
D. Functionality

Choose the correct answer from the options given below:





Go to Discussion

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

Solution

McCall’s Quality Model classifies software quality into three main categories:

  • Product Operation factors: Correctness, Reliability, Efficiency, Integrity, Usability
  • Product Revision factors: Maintainability, Flexibility, Testability
  • Product Transition factors: Portability, Reusability, Interoperability

- Maintainability ✅ – part of Product Revision factors
- Usability ✅ – part of Product Operation factors
- Integrity ❌ – although in McCall model, it’s not grouped directly with Functionality in this context
- Functionality ✅ – general umbrella factor related to McCall’s quality aspects

✅ Final Answer: A, B and D Only


UGC NET Computer Science PYQ 2025
List I (GA termination criteria) List II (description)
A. Worst Individual I. At least half of the individual will be better than or equal to convergence value
B. Best Individual II. Guarantees that virtually all individuals in the population will be within a particular fitness range
C. Sum of Fitness III. Guarantees that the entire population to be of a minimum standard
D. Median Fitness IV. Faster search convergence, guaranteeing at least one best solution






Go to Discussion

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, $1000 \rightarrow 1100$

UGC NET Computer Science PYQ 2025
Which of the following Grammars is/are only Context-Free?
LabelProductions Note
A S → Ab
aS → aA
A → a
LHS “aS” has a terminal → not CFG
B S → Ab
A → Sa
A → a
All rules are left-linear (V→Vt or V→t) → regular
C S → AS
S → aA
A → a
CFG, not regular (non-linear: V→VV)
D S → Ab
S → aA
A → a
CFG; mixes left/right-linear → not regular
E S → Sb
A → Aa
A → ε
From start S no terminal string → L(S)=∅ (regular)





Go to Discussion

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

Solution

A is not even CFG (terminal on LHS).
B is a regular (left-linear) grammar.
C and D are context-free but not regular (“only” context-free).
E generates the empty language from start symbol S, which is regular.

UGC NET Computer Science PYQ 2025
Match List I with List II
List I (Software Quality Characteristic) List II (Description)
A. Reliability II. Capability of software to maintain its level of performance under stated conditions for a stated period of time.
B. Efficiency III. Relationship between the level of performance of the software and the amount of resources used, under stated conditions.
C. Maintainability IV. Effort needed to make modifications, including corrections, improvements or adaptation of software to changes in environment, requirements and functional specifications.
D. Portability I. Ability to transfer the software from one organization or hardware/software environment to another.






Go to Discussion

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

Solution

Reliability → II: Ensures software maintains performance over time.
Efficiency → III: Measures performance relative to resources used.
Maintainability → IV: Effort needed for updates, corrections, and improvements.
Portability → I: Ability to move software across environments.

UGC NET Computer Science PYQ 2025
STRIPS Representation of a Planning Problem
A. STRIPS is a feature-centric representation
B. The features describing the state of the world are divided into Primitive and Derived
C. The STRIPS representation of the action comprises of Preconditions and Effects
D. STRIPS can directly define conditional effects





Go to Discussion

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

Solution

A: Incorrect — STRIPS is state-centric, not feature-centric.
B: Correct — STRIPS divides world state features into primitive and derived predicates.
C: Correct — STRIPS action representation includes Preconditions and Effects.
D: Incorrect — STRIPS cannot directly handle conditional effects; this is supported in extensions like ADL.


UGC NET Computer Science


Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

UGC NET Computer Science


Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Ask Your Question or Put Your Review.

loading...