Qus : 1
UGC NET Computer Science PYQ 2025
2
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:
1
A-II, B-III, C-I, D-IV 2
A-II, B-III, C-IV, D-I 3
A-II, B-IV, C-III, D-I 4
A-IV, B-III, C-I, D-II 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
Qus : 2
UGC NET Computer Science PYQ 2025
2
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:
1
A-II, B-III, C-I, D-IV 2
A-II, B-III, C-IV, D-I 3
A-III, B-II, C-I, D-IV 4
A-IV, B-II, C-I, D-III 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
Qus : 5
UGC NET Computer Science PYQ 2025
4
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: 1
B → A → C → D 2
C → B → A → D 3
C → A → B → D 4
A → D → B → C 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.
Qus : 6
UGC NET Computer Science PYQ 2025
2
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?
1
CD 2
EC 3
AE 4
AC 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}.
Qus : 7
UGC NET Computer Science PYQ 2025
4
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:
A and D only
A, B and D only
C and D only
B and C only
1
1 2
2 3
3 4
4 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.
Qus : 8
UGC NET Computer Science PYQ 2025
3
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?
(a) P→¬Q; (b) P ∧ ¬Q; (c) Q→¬P; (d) P→¬Q
(a) P→Q; (b) ¬P ∧ ¬Q; (c) ¬Q→P; (d) ¬P→¬Q
(a) P→¬Q; (b) ¬P ∧ ¬Q; (c) Q→¬P; (d) ¬P→¬Q
(a) P→¬Q; (b) P ∧ ¬Q; (c) Q↔P; (d) ¬P→¬Q
1
1 2
2 3
3 4
4 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
Qus : 10
UGC NET Computer Science PYQ 2025
4
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:
B, C, A, D
B, A, C, D
C, B, A, D
C, B, D, A
1
1 2
2 3
3 4
4 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:
(C) Characterize the structure of an optimal solution.
(B) Recursively define the value of an optimal solution.
(D) Compute the value of an optimal solution (usually bottom-up).
(A) Construct the optimal solution from the computed information.
Hence, the correct order is C → B → D → A .
Qus : 14
UGC NET Computer Science PYQ 2025
1
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:
1
A-IV, B-III, C-I, D-II 2
A-I, B-II, C-IV, D-III 3
A-III, B-IV, C-I, D-II 4
A-II, B-IV, C-I, D-III 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.
Qus : 15
UGC NET Computer Science PYQ 2025
1
Consider the following table defining the sample inputs and corresponding target values for a perceptron model.
Sample No
x1
x2
target
w1
w2
S1 0 0 0 0 0
S2 0 1 1
S3 1 0 1
S4 1 1 1
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 }
w1 = 0.1, w2 = 0.1
w1 = 0.0, w2 = 0.2
w1 = 0.0, w2 = 0.1
w1 = 0.2, w2 = 0.2
1
1 2
2 3
3 4
4 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)
Qus : 16
UGC NET Computer Science PYQ 2025
3
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:
1
E, A, C, B, D 2
D, B, A, E, C 3
B, C, D, A, E 4
C, A, B, D, E 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).
Qus : 17
UGC NET Computer Science PYQ 2025
3
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:
1
B, C, A, D 2
D, B, A, C 3
B, C, D, A 4
C, A, B, D 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.
Qus : 18
UGC NET Computer Science PYQ 2025
1
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)) )
1
1 2
2 3
3 4
4 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]) ).
Qus : 19
UGC NET Computer Science PYQ 2025
2
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:
1
$A \rightarrow III,\; B \rightarrow I,\; C \rightarrow II,\; D \rightarrow IV$ 2
$A \rightarrow III,\; B \rightarrow I,\; C \rightarrow IV,\; D \rightarrow II$ 3
$A \rightarrow III,\; B \rightarrow III,\; C \rightarrow I,\; D \rightarrow IV$ 4
$A \rightarrow IV,\; B \rightarrow III,\; C \rightarrow II,\; D \rightarrow I$ 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 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 .
Qus : 20
UGC NET Computer Science PYQ 2025
3
The longest common subsequence of $\{1,2,3,2,4,1,2\}$ and $\{2,4,3,1,2,1\}$ is
1
$2,1,2,3$ 2
$1,3,2,1$ 3
$2,3,2,1$ 4
$2,3,1,2,1$ 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.
Qus : 22
UGC NET Computer Science PYQ 2025
2
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
1
B, C, A, D 2
A, B, C, D 3
A, C, B, D 4
B, C, 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 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.
Qus : 23
UGC NET Computer Science PYQ 2025
2
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.
1
B, C, E only 2
B, D, E only 3
A, C, E only 4
A, C, D 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 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.
Qus : 24
UGC NET Computer Science PYQ 2025
3
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:
1
A→III, B→I, C→II, D→IV 2
A→III, B→II, C→IV, D→I 3
A→IV, B→III, C→II, D→I 4
A→II, B→I, C→III, D→IV 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 → $\mu_A(x) + \mu_B(x) - \mu_A(x)\mu_B(x)$
Qus : 26
UGC NET Computer Science PYQ 2025
3
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:
1
A–II, B–III, C–I, D–IV 2
A–II, B–III, C–IV, D–I 3
A–III, B–II, C–I, D–IV 4
A–IV, B–II, C–I, D–III 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
Qus : 27
UGC NET Computer Science PYQ 2025
2
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:
1
A→IV, B→III, C→I, D→II 2
A→III, B→IV, C→I, D→II 3
A→I, B→III, C→IV, D→II 4
A→II, B→IV, C→I, D→III 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
Qus : 29
UGC NET Computer Science PYQ 2025
3
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
1
A, B, C, D, E
2
B, A, D, A, E 3
B, C, A, D, E 4
C, B, A, D, E 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).
Qus : 30
UGC NET Computer Science PYQ 2025
4
Consider the following DFA that generates set of strings over Σ = {a, b, c}.
Identify the best description of the language for the DFA.
1
L = (a* + b* + c*)* 2
L = (a + b + c)(abc)(a + b + c)* 3
L = {set of strings, all starting with ‘a, b, c’ but ending with ‘c’} 4
L = {set of strings having even count (including 0) of substring ‘abc’} 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.
Qus : 32
UGC NET Computer Science PYQ 2025
1
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):
1
C, D, E, B, A 2
C, A, D, E, B 3
C, D, A, E, B 4
D, C, A, B, E 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).
Qus : 33
UGC NET Computer Science PYQ 2025
2
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.
1
A–III, B–II, C–IV, D–I 2
A–IV, B–II, C–I, D–III 3
A–II, B–I, C–IV, D–II 4
A–IV, B–I, C–III, D–II 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
Qus : 34
UGC NET Computer Science PYQ 2025
2
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
1
A, B only 2
B, C only 3
C, D only 4
A, B, 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 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.
Qus : 35
UGC NET Computer Science PYQ 2025
4
Consider the following table about processes, their burst time and arrival time
Process Burst Time Arrival Time P1 9 0 P2 30 0 P3 4 0 P4 8 2 P5 11 6
Now, which process finishes second last in the Gantt chart for:
non-preemptive SJF, and
Round Robin (time quantum = 10)?
1
SJF : $P_4$ ; RR : $P_4$ 2
SJF : $P_5$ ; RR : $P_4$ 3
SJF : $P_4$ ; RR : $P_3$ 4
SJF : $P_5$ ; RR : $P_5$ 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 Scheduling Execution 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
Qus : 36
UGC NET Computer Science PYQ 2025
1
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
1
D, A, B, C 2
D, B, A, C 3
B, C, A, D 4
B, C, 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 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 .
Qus : 39
UGC NET Computer Science PYQ 2025
4
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
1
B, A, C, E, D 2
C, B, A, E, D 3
C, B, D, A, E 4
C, A, D, B, E 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).
Qus : 42
UGC NET Computer Science PYQ 2025
1
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
1
A–II, B–I, C–IV, D–III 2
A–III, B–I, C–II, D–IV 3
A–II, B–III, C–IV, D–I 4
A–III, B–II, C–IV, 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 A → IV → Conditional statement
B → I → Inverse (negating both sides)
C → II → Converse (reversing condition and result)
D → III → Contrapositive (negating and reversing)
Qus : 47
UGC NET Computer Science PYQ 2025
3
The tight asymptotic bound for the recurrence:
$T(n) = 2T(n/4) + \sqrt{n}$
1
$\Theta(\sqrt{n})$ 2
$\Theta(n \log n)$ 3
$\Theta(\sqrt{n} \log n)$ 4
$\Theta(n \log \sqrt{n})$ 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
$\boxed{T(n) = \Theta(\sqrt{n} \log n)}$
Qus : 50
UGC NET Computer Science PYQ 2025
2
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:
1
A → C → B → D 2
C → A → D → B 3
C → A → B → D 4
A → D → B → C 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.
Qus : 51
UGC NET Computer Science PYQ 2025
2
In a relational database, which one of the following is CORRECT:
1
A relation with only two attributes is always in BCNF.
2
If all attributes of a relation are prime attributes, then the relation is in BCNF. 3
Every relation has at least one non-prime attribute. 4
BCNF decomposition preserves functional dependencies. 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.
Qus : 52
UGC NET Computer Science PYQ 2025
3
Match List I with List II
List I (Hashing Collision Handling Method) List II (Strategy) A. Chaining I. Check next slot B. Linear Probing II. Use second hash function C. Quadratic Probing III. Linked list at index D. Double Hashing IV. Skip slots using quadratic step
1
A–II, B–III, C–I, D–IV
2
A–II, B–III, C–IV, D–I 3
A–III, B–I, C–IV, D–II 4
A–IV, B–II, C–I, D–III 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)
Qus : 56
UGC NET Computer Science PYQ 2025
1
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.
1
A Only 2
D Only 3
B, D Only
4
A, C and D 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 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.
Qus : 58
UGC NET Computer Science PYQ 2025
2
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
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 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.
Qus : 61
UGC NET Computer Science PYQ 2025
3
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. $(V\cup\Sigma)^{*} \rightarrow (V\cup\Sigma)^{*}$
1
A→II, B→III, C→IV, D→I 2
A→III, B→I, C→IV, D→II 3
A→IV, B→I, C→II, D→III 4
A→III, B→II, C→I, D→IV 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).
Qus : 65
UGC NET Computer Science PYQ 2025
1
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
1
C → A → D → B
2
A → C → D → B 3
C → D → A → B 4
A → D → C → B 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.
Qus : 67
UGC NET Computer Science PYQ 2025
4
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.
1
B, C only
2
A, D only 3
A, C only
4
B, C and D 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 (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.
Qus : 68
UGC NET Computer Science PYQ 2025
1
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
1
A, B only 2
B, D only 3
C, D only
4
B, C, D 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 (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.
Qus : 70
UGC NET Computer Science PYQ 2025
1
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
1
A–III, B–II, C–I, D–IV 2
A–IV, B–II, C–III, D–I 3
A–III, B–I, C–IV, D–II 4
A–IV, B–I, C–III, D–II 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)
Qus : 71
UGC NET Computer Science PYQ 2025
3
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?
1
T1 is in 3NF 2
T2 is in 3NF 3
T1 is not in 3NF 4
T1 is in 2NF 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).
Qus : 73
UGC NET Computer Science PYQ 2025
1
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.
1
(A), (C), (D) 2
(B), (C), (D) 3
(A), (B) 4
(A), (B), (C) 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.
Qus : 74
UGC NET Computer Science PYQ 2025
1
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
1
A, B, C, D 2
B, A, C, D 3
C, B, A, D 4
A, C, B, D 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.
Qus : 75
UGC NET Computer Science PYQ 2025
1
Estimation of software development effort for organic software in basic COCOMO is:
1
$E = 2.4\,(KLOC)^{1.05}\,PM$ 2
$E = 3.4\,(KLOC)^{1.06}\,PM$ 3
$E = 2.0\,(KLOC)^{1.05}\,PM$ 4
$E = 2.4\,(KLOC)^{1.07}\,PM$ 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 $E = 2.4\,(KLOC)^{1.05}\,PM$
Qus : 78
UGC NET Computer Science PYQ 2025
1
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:
1
A–II, B–III, C–I, D–IV 2
A–II, B–III, C–IV, D–I 3
A–III, B–II, C–I, D–IV 4
A–IV, B–II, C–I, D–III 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
Qus : 79
UGC NET Computer Science PYQ 2025
1
Choose the correct statement for a group G:
1
If for all x, y ∈ G, (xy)² = x²y² then G is commutative. 2
If for all x ∈ G, x³ = 1, then G is commutative. 1 is the identity element of G. 3
If for all x ∈ G, x⁵ = 1, then G is commutative. 1 is the identity element of G.
4
If G is commutative, the subgroup of G need not be commutative. 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.
Qus : 80
UGC NET Computer Science PYQ 2025
2
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:
1
A and D Only 2
A, B and D Only 3
C, D and E Only 4
A, D and E 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
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.
Qus : 81
UGC NET Computer Science PYQ 2025
3
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.
1
A, B, C, D 2
A, B, C Only 3
A, B, D Only
4
B, C, D 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 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.
Qus : 82
UGC NET Computer Science PYQ 2025
4
What would be the equivalent pointer expression for referring the array element ar[m][n][o]?
*(*(*(ar) + m + n) + o)
(*(*(*(ar + m) + n) + o))
(*(*(ar + m) + n) + o)
*(*(*(ar + m) + n) + o)
1
1 2
2 3
3 4
4 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)
Qus : 84
UGC NET Computer Science PYQ 2025
3
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 $R_\beta\subseteq R_\lambda$
1
1 2
2 3
3 4
4 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 .
Qus : 86
UGC NET Computer Science PYQ 2025
2
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:
1
A and D Only 2
A, B and D Only 3
C and D Only 4
A, 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
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
Qus : 87
UGC NET Computer Science PYQ 2025
1
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
1
A→III, B→IV, C→II, D→I 2
A→IV, B→III, C→I, D→II 3
A→II, B→I, C→III, D→IV 4
A→III, B→II, C→IV, 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 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 : 88
UGC NET Computer Science PYQ 2025
2
Which of the following Grammars is/are only Context-Free?
Label Productions 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)
1
A, B Only
2
C, D Only 3
C, E Only 4
A, B, D, E 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 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.
Qus : 89
UGC NET Computer Science PYQ 2025
1
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.
1
A–II, B–III, C–I, D–IV 2
A–II, B–III, C–IV, D–I 3
A–III, B–II, C–I, D–IV 4
A–IV, B–II, C–I, D–III 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.
Qus : 90
UGC NET Computer Science PYQ 2025
2
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
1
A, C Only
2
B, C Only 3
A, D Only
4
B, C, D 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 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.
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