UGC NET Computer Science 2025 Previous Year Questions (PYQs) – Page 8 of 9
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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?
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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).
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
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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).
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.
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Solution
(B) is false (for composite k, both factors are < k). (A), (C), (D) are true for positive integers.
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
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Solution
Poll status → verify ready → read data → store to memory.
Estimation of software development effort for organic software in basic COCOMO is:
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Solution
$E = 2.4\,(KLOC)^{1.05}\,PM$
Which of the following technique is used for Clipping?
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Solution
Used for line clipping in computer graphics — efficiently finds visible portions of a line segment.
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
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Solution
The correct hierarchy (top to bottom) is:
Application Programs → Logical File System → File Organization Module → Basic File System → I/O Control
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:
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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 |
Choose the correct statement for a group G:
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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.
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:
B. Polling
C. TDMA
D. Token Passing
E. CSMA/CA Choose the correct answer from the options given below:
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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
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UGC NET Computer Science
Online Test Series, Information About Examination,
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and More.
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