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|Computer Science & Applications · January 2026

UGC NET Computer Science & Applications January 2026 Question PaperPaper 1 & Paper 2 with Answer Key

The complete UGC NET January 2026 question paper for Computer Science & Applications, covering Paper 1 (General Aptitude) and Paper 2 (Computer Science & Applications). Every question below is shown with its options and the correct answer, and a detailed explanation you can unlock by signing in. All 150 questions are free to practise.

UGC NET January 2026 at a glance

Exam	UGC NET January 2026
Papers	Paper 1 (General Aptitude) and Paper 2 (Computer Science & Applications)
Exam date(s)	January 2, 2026
Shift(s)	Shift 1
Paper 1 questions	50 questions · 60 minutes · 100 marks
Paper 2 questions	100 questions · 120 minutes · 200 marks
Total	150 questions
Marking scheme	+2 per correct answer · No negative marking

Paper 1

General Aptitude

UGC NET January 2026 · Computer Science & Applications sitting

50 questions60 min100 marks

Practice Now

Paper 2

Computer Science & Applications

UGC NET January 2026

100 questions120 min200 marks

Practice Now

Paper 1 — General Aptitude

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50 questions with the answer key and explanations.

Question 1
Arrange the educational objectives of the cognitive domain from lower-order to higher-order skills.
A. Evaluation
B. Synthesis
C. Analysis
D. Knowledge
E. Comprehension
1. AA, B, C, D, E
2. BD, E, C, B, A
3. CA, B, C, E, D
4. DB, C, D, E, A
Answer: (B) D, E, C, B, A
Explanation
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Question 2
Which of the following is NOT a criticism of the behaviourist model of teaching?
1. AIt does not take into account important internal processes that take place in the mind.
2. BIt is not appropriate for higher-order learning.
3. CIt is not appropriate for lower-order learning.
4. DLearning is sometimes beyond the association between stimulus and response.
Answer: (C) It is not appropriate for lower-order learning.
Explanation
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Question 3
Whether the results of a study can be generalized beyond the specific research context refers to:
1. AReliability
2. BReplicability
3. CInternal validity
4. DExternal validity
Answer: (D) External validity
Explanation
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Question 4
Assertion (A): Validity can never apply to any single proposition by itself.
Reason (R): Validity and invalidity are attributes of arguments.
1. ABoth A and R are correct and R is the correct explanation of A.
2. BBoth A and R are correct but R is NOT the correct explanation of A.
3. CA is correct but R is not correct.
4. DA is not correct but R is correct.
Answer: (A) Both A and R are correct and R is the correct explanation of A.
Explanation
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Question 5
Match List-I with List-II.
List-I: Source of Knowledge List-II: Accepted by
A. Yogaja Pratyaksa I. Vedantins
B. Kevalajnana II. Carvaka
C. Only Pratyaksa III. Jaina
D. Aparoksanubhuti IV. Nyaya
1. AA-I, B-II, C-III, D-IV
2. BA-II, B-III, C-IV, D-I
3. CA-III, B-I, C-IV, D-II
4. DA-IV, B-III, C-II, D-I
Answer: (D) A-IV, B-III, C-II, D-I
Explanation
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Question 6
Match List-I with List-II.
List-I List-II
A. Keyboard I. Memory
B. Monitor II. Processor
C. CPU III. Output device
D. ROM IV. Input device
1. AA-IV, B-III, C-II, D-I
2. BA-I, B-IV, C-II, D-III
3. CA-III, B-II, C-IV, D-I
4. DA-I, B-II, C-IV, D-III
Answer: (A) A-IV, B-III, C-II, D-I
Explanation
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Question 7
The SWAYAM was launched in __________.
1. A2015
2. B2017
3. C2019
4. D2020
Answer: (B) 2017
Explanation
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Question 8
National Education Policy 2020 has entrusted __________ with the responsibility to set up the standards of content, technology and pedagogy for online/digital teaching-learning.
1. ANational Educational Technology Forum
2. BCentral Institute of Educational Technology
3. CConsortium for Educational Communication
4. DIndira Gandhi National Open University
Answer: (A) National Educational Technology Forum
Explanation
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Question 9
Which education policy first recommended 6% of GDP as public expenditure on education in India?
1. ANational Policy on Education, 1968
2. BNational Policy on Education, 1986
3. CRevised NPE - Programme of Action, 1992
4. DNational Education Policy, 2020
Answer: (A) National Policy on Education, 1968
Explanation
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Question 10
The Black-foot disease is caused by the contamination of __________ in drinking water.
1. AArsenic
2. BCadmium
3. CMercury
4. DLead
Answer: (A) Arsenic
Explanation
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Question 11
In quantitative research strategy, the principal orientation to the role of theory in relation to research is:
1. AInductive
2. BObjectivism
3. CDeductive
4. DConstructionism
Answer: (C) Deductive
Explanation
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Question 12
Arrange the following in ascending order of their year of appearance, beginning with the oldest.
A. Radio
B. Television
C. Book
D. Prototypical Newspaper
E. Film
1. AE, C, D, A, B
2. BD, C, E, A, B
3. CC, D, E, A, B
4. DC, E, D, A, B
Answer: (C) C, D, E, A, B
Explanation
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Question 13
Arrange the following in ascending order of their moisture content.
A. Trash
B. Rubbish
C. Refuse
D. Garbage
1. AA, B, C, D
2. BB, A, C, D
3. CA, B, D, C
4. DD, C, B, A
Answer: (A) A, B, C, D
Explanation
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Question 14

Match List-I with List-II.

List-I: Convention/Protocol	List-II: Related with
A. Montreal Protocol	I. Global Warming
B. Stockholm Convention	II. Transboundary movements of hazardous waste
C. Kyoto Protocol	III. Ozone layer depletion
D. Basel Convention	IV. Persistent organic pollutants

AA-I, B-II, C-III, D-IV
BA-III, B-IV, C-I, D-II
CA-IV, B-III, C-I, D-II
DA-III, B-IV, C-II, D-I

Answer: (B) A-III, B-IV, C-I, D-II

Explanation

Question 15
Assertion (A): Eutrophication leads to depletion of dissolved oxygen in a lake.
Reason (R): Dissolved oxygen in water decreases with increase in temperature.
1. ABoth A and R are correct and R is the correct explanation of A.
2. BBoth A and R are correct but R is NOT the correct explanation of A.
3. CA is correct but R is not correct.
4. DA is not correct but R is correct.
Answer: (B) Both A and R are correct but R is NOT the correct explanation of A.
Explanation
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Question 16
Assertion (A): Higher Education Institutions in the country with appropriate credit transfer mechanisms are facilitating students to enroll themselves in pre-designed academic programmes.
Reason (R): Academic Bank of Credits (ABC) works on the principle of multiple entry-multiple exit as well as anytime, anywhere and any-level learning.
1. ABoth A and R are correct and R is the correct explanation of A.
2. BBoth A and R are correct but R is NOT the correct explanation of A.
3. CA is correct but R is not correct.
4. DA is not correct but R is correct.
Answer: (D) A is not correct but R is correct.
Explanation
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Question 17
Assertion (A): The influence of mass media is rarely direct.
Reason (R): According to Limited Effects theories, media effects are almost always mediated by group membership or relationships.
1. ABoth A and R are correct and R is the correct explanation of A.
2. BBoth A and R are correct but R is NOT the correct explanation of A.
3. CA is correct but R is not correct.
4. DA is not correct but R is correct.
Answer: (A) Both A and R are correct and R is the correct explanation of A.
Explanation
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Question 18
Which of the following is NOT a programming language?
1. APython
2. BC
3. CJava
4. DGoogle
Answer: (D) Google
Explanation
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Question 19
USB stands for:
1. AUniform Service Bus
2. BUniversal Serial Bus
3. CUniform Sector Buffer
4. DUniverse Service Bus
Answer: (B) Universal Serial Bus
Explanation
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Question 20
Arrange the following phrases in the vision of the National Education Policy, 2020 in the correct sequence of their appearance.
A. A conscious awareness of one's roles and responsibilities in a changing world
B. An education system rooted in Indian Ethos
C. Developing a deep sense of respect towards the fundamental duties and constitutional values
D. Transforming Bharat into an equitable and vibrant knowledge society
E. Instilling among learners a deep-rooted pride in being Indian
1. AB, C, A, D, E
2. BD, E, A, C, B
3. CE, B, C, A, D
4. DB, D, C, A, E
Answer: (D) B, D, C, A, E
Explanation
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Question 21
The incomes of A and B are in the ratio 3 : 2 and their expenses are in the ratio 5 : 3. If each of them saves Rs. 3,000, then A's income is:
1. ARs. 15,000
2. BRs. 12,000
3. CRs. 6,000
4. DRs. 18,000
Answer: (D) Rs. 18,000
Explanation
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Question 22
Complete the following series: 582, 581, 585, 576, 592, 567, ______
1. A601
2. B602
3. C603
4. D599
Answer: (C) 603
Explanation
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Question 23
A person who believes that all social, political, economic and cultural change is inevitably based on the development and diffusion of technology is a:
1. ATechnological Determinist
2. BCultural Determinist
3. CTechnological Sceptic
4. DEconomic Determinist
Answer: (A) Technological Determinist
Explanation
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Question 24
In everyday conversation, when people make reference to paying attention, planning, forgetting, guessing, daydreaming and so on, they are invoking:
1. APropaganda
2. BCognitive Concepts
3. CMisinformation
4. DDisinformation
Answer: (B) Cognitive Concepts
Explanation
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Question 25
Which of the following are TRUE?
A. For the Nyaya, knowledge of Abhava is acquired through Anupalabdhi Pramana.
B. The Nyaya makes a distinction between Laukika and Alaukika Pratyaksa.
C. The Nyaya reduces Arthapatti to Anumana.
D. The Nyaya accepts the validity of Upamana as a Pramana.
1. AA and B only
2. BB and C only
3. CB, C and D only
4. DA, B, C and D
Answer: (C) B, C and D only
Explanation
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Question 26
Which of the following memories/storage are volatile?
A. RAM
B. ROM
C. Cache
D. SSD
E. HDD
1. AA, C and D only
2. BB, D and E only
3. CA and C only
4. DA and B only
Answer: (C) A and C only
Explanation
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Question 27
Which of the following renewable energy resources contributes the highest proportion to the total installed power capacity in India?
1. ASolar
2. BNuclear
3. CWind
4. DHydro
Answer: (A) Solar
Explanation
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Question 28
Identify the correct mathematical statements.
A. A batsman makes a score of 87 runs in the 17th inning and thus increases his average by 3. Then his average after the 17th inning is 36.
B. The value of a number subtracted from the numerator and denominator of the fraction 6/7 to obtain a fraction equal to 3/4 is 3.
C. If 15% of 40 is greater than 25% of a number by 2, then the number is 16.
D. (1 + 1/2)(1 + 1/3)(1 + 1/4)...(1 + 1/120) = 60.5
1. AB, C and D only
2. BB and C only
3. CC and D only
4. DA and B only
Answer: (A) B, C and D only
Explanation
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Question 29
Match the words with their codes.
List-I: Word List-II: Code
A. CRAB I. 32
B. BARK II. 49
C. OVER III. 24
D. SING IV. 60
1. AA-II, B-IV, C-I, D-III
2. BA-III, B-II, C-I, D-IV
3. CA-III, B-I, C-IV, D-II
4. DA-IV, B-III, C-II, D-I
Answer: (C) A-III, B-I, C-IV, D-II
Explanation
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Question 30
"Sound is eternal, because it is caused" commits the fallacy (hetvabhasa) of:
1. AAsiddha
2. BSatpratipaksa
3. CSavyabhicara
4. DViruddha
Answer: (D) Viruddha
Explanation
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Question 31
During which of the following years was the manufacturing of product A approximately equal to the average production of that year?
The values represent the number (in thousands) of five types of products manufactured in different years.

Year A B C D E
2021 29 45 26 15 32
2022 32 52 58 26 35
2023 28 54 92 24 39
2024 26 58 86 27 34
2025 23 63 82 28 34
1. A2021
2. B2022
3. C2023
4. D2024
Answer: (A) 2021
Explanation
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Question 32
Manufacturing of product C in 2023 was approximately what percent of the total production of product B in 2022 and 2024 together?
The values represent the number (in thousands) of five types of products manufactured in different years.

Year A B C D E
2021 29 45 26 15 32
2022 32 52 58 26 35
2023 28 54 92 24 39
2024 26 58 86 27 34
2025 23 63 82 28 34
1. A84%
2. B80%
3. C90%
4. D79%
Answer: (A) 84%
Explanation
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Question 33
If we consider the combined production of products C and E in each year, what is the difference between their combined production in 2023 and 2025?
The values represent the number (in thousands) of five types of products manufactured in different years.

Year A B C D E
2021 29 45 26 15 32
2022 32 52 58 26 35
2023 28 54 92 24 39
2024 26 58 86 27 34
2025 23 63 82 28 34
1. A10,000
2. B15,000
3. C5,000
4. D20,000
Answer: (B) 15,000
Explanation
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Question 34
During which of the following years did product B have the lowest percentage of manufacturing with reference to the combined manufacturing of that year?
The values represent the number (in thousands) of five types of products manufactured in different years.

Year A B C D E
2021 29 45 26 15 32
2022 32 52 58 26 35
2023 28 54 92 24 39
2024 26 58 86 27 34
2025 23 63 82 28 34
1. A2021
2. B2022
3. C2023
4. D2024
Answer: (C) 2023
Explanation
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Question 35
Manufacturing of product D in 2021 is approximately less by what percent of its production in 2024?
The values represent the number (in thousands) of five types of products manufactured in different years.

Year A B C D E
2021 29 45 26 15 32
2022 32 52 58 26 35
2023 28 54 92 24 39
2024 26 58 86 27 34
2025 23 63 82 28 34
1. A50%
2. B60%
3. C66%
4. D44%
Answer: (D) 44%
Explanation
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Question 36
Choose the correct statements.
A. Diagnosis of the learners is done in the Inter-Active stage of teaching.
B. Feedback and testing of the learners are done in the Post-Active stage of teaching.
C. Fixing up goals and content are done in the Post-Active stage of teaching.
D. Selection of appropriate testing devices is done in the Post-Active stage of teaching.
E. Decision about strategies of teaching is done in the Post-Active stage of teaching.
1. AA, D and E only
2. BA, C and D only
3. CA, B and D only
4. DA, B and E only
Answer: (C) A, B and D only
Explanation
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Question 37

Match the lesson-planning concepts with their descriptions.

List-I	List-II
A. Concept Mapping	I. Graphical tools for organizing and representing knowledge
B. Five E-based Lesson Planning	II. Constructive Approach
C. Stimulus-response-based lesson planning	III. Systematic arrangement of subject matter
D. Unit Plan	IV. Behaviourist Approach

AA-III, B-II, C-I, D-IV
BA-I, B-II, C-IV, D-III
CA-III, B-I, C-II, D-IV
DA-IV, B-III, C-II, D-I

Answer: (B) A-I, B-II, C-IV, D-III

Explanation

Question 38
Identify the advantages of open-ended questions in survey research.
A. It is easy to process answers.
B. Respondents can answer in their own terms.
C. They enhance the comparability of answers.
D. They are useful for exploring new areas or areas in which the researcher has limited knowledge.
1. AA and B only
2. BA, B and C only
3. CB, C and D only
4. DB and D only
Answer: (D) B and D only
Explanation
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Question 39

Match the communication groups with their characteristics.

List-I: Types of groups in communication	List-II: Characteristic
A. Primary Group	I. Members interact with one another to benefit an audience
B. Self-Help Group	II. Goal is to improve members' skills
C. Learning Group	III. Membership is based on sentiment
D. Public Group	IV. All members have a common problem

AA-III, B-IV, C-II, D-I
BA-II, B-IV, C-III, D-I
CA-I, B-IV, C-II, D-III
DA-III, B-II, C-IV, D-I

Answer: (A) A-III, B-IV, C-II, D-I

Explanation

Question 40
Which of the following are TRUE?
A. For the Carvaka, Pratyaksa is the only pramana.
B. For the Carvaka, Sabda is the only pramana.
C. For the Carvaka, Anumana is not a pramana.
D. For the Jaina, Mati includes both perceptual and inferential knowledge.
1. AA and C only
2. BB and C only
3. CA and B only
4. DA, C and D only
Answer: (D) A, C and D only
Explanation
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Question 41

Match the sampling methods with their descriptions.

List-I	List-II
A. Snowball	I. Handpicked cases on the basis of specific suitability
B. Purposive	II. Equivalent of stratified sampling
C. Quota	III. Choosing the nearest individuals to serve as respondents
D. Convenience	IV. Identifying other respondents with the help of initial respondents

AA-I, B-III, C-II, D-IV
BA-IV, B-I, C-II, D-III
CA-IV, B-III, C-II, D-I
DA-I, B-II, C-IV, D-III

Answer: (B) A-IV, B-I, C-II, D-III

Explanation

Question 42
Arrange the following intervals of the standard normal variate in increasing order of probabilities.
A. 0 <= Z <= +1
B. +1 <= Z <= +2
C. +2 <= Z <= +3
D. +3 <= Z < +infinity
1. AD, C, B, A
2. BC, B, A, D
3. CB, A, C, D
4. DA, B, C, D
Answer: (A) D, C, B, A
Explanation
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Question 43
Arrange the following operating systems in the correct chronological order of their development.
A. Linux
B. MS-DOS
C. MAC OS
D. Windows
1. AD, B, A, C
2. BB, C, D, A
3. CA, B, C, D
4. DB, A, D, C
Answer: (B) B, C, D, A
Explanation
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Question 44
Arrange the following fractions in descending order.
I. 9/11
II. 8/9
III. 4/7
IV. 3/5
V. 8/15
1. AII > IV > I > V > III
2. BII > I > V > IV > III
3. CI > II > III > IV > V
4. DII > I > IV > III > V
Answer: (D) II > I > IV > III > V
Explanation
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Question 45

Match the HECI bodies with their functions.

Body under HECI	Function
A. NHERC	I. Accreditation of higher education institutions
B. NAAC	II. Single-point regulator for higher education sector
C. HEGC	III. Formulating a National Higher Education Qualification Framework
D. GEC	IV. Funding and financing of higher education

AA-II, B-I, C-IV, D-III
BA-III, B-II, C-IV, D-I
CA-I, B-IV, C-II, D-III
DA-IV, B-I, C-III, D-II

Answer: (A) A-II, B-I, C-IV, D-III

Explanation

Question 46
In simple terms, the first line of the passage means:
The use of the term dharma for law, nevertheless, was neither universal nor inevitable. This is borne out by the first century C.E. author Kautilya's compendium Treatise on Politics (Arthasastra). The significance of Kautilya's work for the history of law in India rests primarily on the fact that it provides a different and in many respects unique lens into that history. His treatise belongs to a distinct scholarly tradition with social and political priorities different from those represented by the science of dharma (dharmasastra), the primary discipline devoted to jurisprudential reflection within the Brahmanical scholastic tradition. Kautilya makes no attempt to reduce the variety of laws within society into the single category of dharma. Indeed, we do not find a single comprehensive term within the Treatise on Politics to refer to law as such, or even to the broad areas of religious and secular norms covered by the term dharma within the discourse of the science of dharma. What is clear, however, is that Kautilya, both formally and in obiter dicta, argues for the plurality of law; law is not one but multiple. Although his text is later than the earliest documents of the science of dharma, it nevertheless taps into an alternate intellectual history that probably ran parallel to the one represented by the science of dharma.
1. AThere was nothing unavoidable or locally acceptable about the use of the word 'law' for 'dharma', anyway.
2. BThere was nothing inevitable or meaningful about the use of the word 'dharma' for law, anyway.
3. CThere was nothing unavoidable or globally acceptable about the use of the word 'dharma' for 'law', anyway.
4. DThere was nothing permanent or unavoidable about the use of the word 'dharma' for law, anyway.
Answer: (C) There was nothing unavoidable or globally acceptable about the use of the word 'dharma' for 'law', anyway.
Explanation
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Question 47
The work of Kautilya is of significance because:
The use of the term dharma for law, nevertheless, was neither universal nor inevitable. This is borne out by the first century C.E. author Kautilya's compendium Treatise on Politics (Arthasastra). The significance of Kautilya's work for the history of law in India rests primarily on the fact that it provides a different and in many respects unique lens into that history. His treatise belongs to a distinct scholarly tradition with social and political priorities different from those represented by the science of dharma (dharmasastra), the primary discipline devoted to jurisprudential reflection within the Brahmanical scholastic tradition. Kautilya makes no attempt to reduce the variety of laws within society into the single category of dharma. Indeed, we do not find a single comprehensive term within the Treatise on Politics to refer to law as such, or even to the broad areas of religious and secular norms covered by the term dharma within the discourse of the science of dharma. What is clear, however, is that Kautilya, both formally and in obiter dicta, argues for the plurality of law; law is not one but multiple. Although his text is later than the earliest documents of the science of dharma, it nevertheless taps into an alternate intellectual history that probably ran parallel to the one represented by the science of dharma.
1. AIt works as a complete view of the history of law.
2. BIt works as an alternate view of the history of law in India.
3. CIt works as a scholar's view that is identical to the science of dharma.
4. DIt works as a perspective that fits into contemporary ideas on dharma of his time.
Answer: (B) It works as an alternate view of the history of law in India.
Explanation
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Question 48
The word "comprehensive" in the passage can be replaced by the word __________.
The use of the term dharma for law, nevertheless, was neither universal nor inevitable. This is borne out by the first century C.E. author Kautilya's compendium Treatise on Politics (Arthasastra). The significance of Kautilya's work for the history of law in India rests primarily on the fact that it provides a different and in many respects unique lens into that history. His treatise belongs to a distinct scholarly tradition with social and political priorities different from those represented by the science of dharma (dharmasastra), the primary discipline devoted to jurisprudential reflection within the Brahmanical scholastic tradition. Kautilya makes no attempt to reduce the variety of laws within society into the single category of dharma. Indeed, we do not find a single comprehensive term within the Treatise on Politics to refer to law as such, or even to the broad areas of religious and secular norms covered by the term dharma within the discourse of the science of dharma. What is clear, however, is that Kautilya, both formally and in obiter dicta, argues for the plurality of law; law is not one but multiple. Although his text is later than the earliest documents of the science of dharma, it nevertheless taps into an alternate intellectual history that probably ran parallel to the one represented by the science of dharma.
1. Aalternative
2. Bintriguing
3. Camazing
4. Dall-embracing
Answer: (D) all-embracing
Explanation
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Question 49
The passage tells the reader that Kautilya "makes no attempt to reduce the variety of laws within society to the single category of dharma". This demonstrates his argument for the __________ of law.
The use of the term dharma for law, nevertheless, was neither universal nor inevitable. This is borne out by the first century C.E. author Kautilya's compendium Treatise on Politics (Arthasastra). The significance of Kautilya's work for the history of law in India rests primarily on the fact that it provides a different and in many respects unique lens into that history. His treatise belongs to a distinct scholarly tradition with social and political priorities different from those represented by the science of dharma (dharmasastra), the primary discipline devoted to jurisprudential reflection within the Brahmanical scholastic tradition. Kautilya makes no attempt to reduce the variety of laws within society into the single category of dharma. Indeed, we do not find a single comprehensive term within the Treatise on Politics to refer to law as such, or even to the broad areas of religious and secular norms covered by the term dharma within the discourse of the science of dharma. What is clear, however, is that Kautilya, both formally and in obiter dicta, argues for the plurality of law; law is not one but multiple. Although his text is later than the earliest documents of the science of dharma, it nevertheless taps into an alternate intellectual history that probably ran parallel to the one represented by the science of dharma.
1. APlurality
2. BJurisprudence
3. CScientific View
4. DScholastic base
Answer: (A) Plurality
Explanation
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Question 50
Which of the following is concerned primarily with jurisprudential reflection according to Brahmanical scholastic tradition?
The use of the term dharma for law, nevertheless, was neither universal nor inevitable. This is borne out by the first century C.E. author Kautilya's compendium Treatise on Politics (Arthasastra). The significance of Kautilya's work for the history of law in India rests primarily on the fact that it provides a different and in many respects unique lens into that history. His treatise belongs to a distinct scholarly tradition with social and political priorities different from those represented by the science of dharma (dharmasastra), the primary discipline devoted to jurisprudential reflection within the Brahmanical scholastic tradition. Kautilya makes no attempt to reduce the variety of laws within society into the single category of dharma. Indeed, we do not find a single comprehensive term within the Treatise on Politics to refer to law as such, or even to the broad areas of religious and secular norms covered by the term dharma within the discourse of the science of dharma. What is clear, however, is that Kautilya, both formally and in obiter dicta, argues for the plurality of law; law is not one but multiple. Although his text is later than the earliest documents of the science of dharma, it nevertheless taps into an alternate intellectual history that probably ran parallel to the one represented by the science of dharma.
1. AArthasastra
2. BNitisastra
3. CDharmasastra
4. DTreatise on Politics
Answer: (C) Dharmasastra
Explanation
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Paper 2 — Computer Science & Applications

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100 questions with the answer key and explanations.

Question 1
Which of the following statements are correct?

A. The HAVING CLAUSE is an optional clause which tells Oracle to group rows based on distinct values that exist for specified columns.
B. Unions can be used in subqueries.
C. The intersect clause outputs only rows produced by both the queries intersected.
D. The statement containing a subquery is called a parent statement.
E. If a select statement is defined as a subquery, the innermost select statement gets executed first.
1. AA, B, C Only
2. BC, D, E Only
3. CB, C, E Only
4. DA, B, D Only
Answer: (B) C, D, E Only
Explanation
The HAVING clause filters groups that GROUP BY has already formed, it does not itself group rows by distinct column values, so statement A wrongly assigns GROUP BY's job to HAVING.

A. Wrongly describes HAVING — grouping rows by distinct values of specified columns is done by GROUP BY, HAVING only restricts which already-formed groups appear in the result.

B. Unions in subqueries — a subquery is restricted to a single SELECT expression under standard subquery rules, so it cannot itself contain a UNION of two SELECT statements.

C. INTERSECT clause — correct. INTERSECT is a set operator that returns only the rows that appear in both result sets being compared.

D. Parent statement — correct. The statement that encloses and uses the result of a subquery is standard terminology called the parent statement, also known as the outer query.

E. Innermost SELECT executes first — correct. Nested, non-correlated subqueries are resolved from the inside out, so the innermost SELECT executes first and its result feeds the enclosing query.

C, D and E hold while A and B do not, so C, D, E Only is correct.
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Question 2
Arrange the following steps of the MapReduce framework.

A. Input Splitting
B. Reducing
C. Mapping
D. Shuffling & Sorting
1. AD, A, C, B
2. BB, C, A, D
3. CA, C, D, B
4. DC, B, D, A
Answer: (C) A, C, D, B
Explanation
MapReduce processes data in four sequential stages. Input splitting divides the input into fixed-size chunks, mapping applies the map function to each split to produce key-value pairs, shuffling and sorting groups and orders those pairs by key across the cluster, and reducing aggregates the values for each key into the final output.

A-C-D-B — Input Splitting, Mapping, Shuffling & Sorting, Reducing — is the correct execution order.
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Question 3
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion (A): If L is regular, then its compliment L' is necessarily regular.
Reason (R): Complement of a language can be obtained by swapping final and non-final states in a DFA.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (A) Both A and R are correct and R is the correct explanation of A
Explanation
Regular languages are closed under complement. Given a DFA for L, swapping every accepting state to non-accepting and vice versa produces a DFA that accepts exactly the strings L rejects, which is the complement of L. Since a DFA always exists for any regular language, this complement is itself regular, so Assertion A holds.

Reason R states this construction correctly and restricts it to a DFA, where the swap works precisely because the DFA is deterministic and total, so every string reaches exactly one state. This construction is the actual reason A is true, so R correctly explains A.
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Question 4
A phase of a compiler which produces the symbol table is
1. ASyntax analysis
2. BSemantic analysis
3. CTarget code generation
4. DCode optimization
Answer: (A) Syntax analysis
Explanation
Symbol table entries for identifiers and declarations are created as the parser recognizes declaration constructs while building the parse tree, so syntax analysis is the phase that produces the symbol table.

A. Syntax analysis — correct. The parser creates symbol table entries while recognizing declarations in the grammar.

B. Semantic analysis — this phase uses and updates existing symbol table entries with type and scope information, it does not originate the table.

C. Target code generation — this phase consumes finalized symbol table information to generate machine code, long after the table already exists.

D. Code optimization — this phase transforms intermediate code for efficiency and does not create symbol table entries.
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Question 5
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion (A): HDFS follows a master-slave architecture with Name Node and Data Nodes.
Reason (R): Name Node stores metadata, while Data Nodes store actual data blocks.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (A) Both A and R are correct and R is the correct explanation of A
Explanation
HDFS is built as a master-slave system where a single NameNode acts as the master and multiple DataNodes act as slaves, which makes Assertion A correct.

Reason R correctly explains why this is a master-slave design. The NameNode holds only the metadata, such as the file-to-block mapping and directory structure, while the DataNodes hold and serve the actual data blocks, so the NameNode coordinates while the DataNodes carry out the storage work.
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Question 6
Which of the following statements are true if the universe of discourse for all variables consists of all integers?

A. ∀n∃m(n² < m)
B. ∃n∃m(n + m = 4 ∧ n − m = 2)
C. ∃n∃m(n² + m² = 5)
D. ∃m∀n(m + n = 0)
E. ∀m∃n(m + n = 0)
1. AA, B & C Only
2. BA, B, C & D Only
3. CA, B, C & E Only
4. DB, C, E & D Only
Answer: (C) A, B, C & E Only
Explanation
Each statement must be checked on its own quantifier structure before combining them.

A. True. For any integer n, choosing m = n² + 1 always makes n² < m, since integers are unbounded above.

B. True. Solving n + m = 4 and n − m = 2 together gives n = 3 and m = 1, both integers, so such a pair exists.

C. True. Taking n = 1 and m = 2 gives 1² + 2² = 5, so such integers exist.

D. False. This asks for one fixed m that satisfies m + n = 0 for every integer n, but no single m can be the negative of every n at once.

E. True. This asks whether, for every m, some n makes m + n = 0, and n = −m always works because n is chosen after m is fixed.

D and E look similar but the quantifier order reverses their meaning. E lets n depend on m, so it is always satisfiable, while D forces one m to work for all n, which is impossible. A, B, C and E hold, so A, B, C & E Only is correct.
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Question 7
In transactions, a write-item(X) command includes the following steps. Arrange the following steps in the correct sequence.

A. Copy item X from the program variable named X into its correct location in the buffer.
B. Store the updated block from the buffer back to disk.
C. Find the address of the disk block that contains item X.
D. Copy that disk block into a buffer in main memory.
1. AC, D, A, B
2. BA, C, D, B
3. CC, A, B, D
4. DD, B, C, A
Answer: (A) C, D, A, B
Explanation
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Question 8
The inorder and preorder traversal of a binary tree are d, b, e, a, f, c, g and a, b, d, e, c, f, g respectively. The postorder traversal of the binary tree is
1. Ad, e, b, f, g, c, a
2. Be, d, b, g, f, c, a
3. Ce, d, b, f, g, c, a
4. Dd, e, f, g, b, c, a
Answer: (A) d, e, b, f, g, c, a
Explanation
The first element of a preorder sequence is always the tree's root, so a is the root, and its position in the inorder sequence splits the tree into a left subtree {d, b, e} and a right subtree {f, c, g}.

Within the left subtree, preorder b, d, e makes b the root, and inorder d, b, e places d as its left child and e as its right child.

Within the right subtree, preorder c, f, g makes c the root, and inorder f, c, g places f as its left child and g as its right child.

Postorder visits left, then right, then the root at every level, giving d, e, b for the left subtree, f, g, c for the right subtree, and a last, so the full sequence is d, e, b, f, g, c, a.
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Question 9
Which of the following languages cannot be accepted by a pushdown automaton?

A. L = { aⁿbⁿcⁿ : n ≥ 0 }
B. L = { aᵖ : p is Prime }
C. L = { wcwᴿ : w ∈ (a, b)* }
D. L = { w ∈ (a, b)* : nₐ(w) = n_b(w) }
E. L = { aⁿbᵐc⁽ⁿ⁺ᵐ⁾ : n ≥ 0, m ≥ 0 }
1. AA, C & D Only
2. BB, D & E Only
3. CA & B Only
4. DC & D Only
Answer: (C) A & B Only
Explanation
A pushdown automaton has only one stack, so it can enforce at most one counting relationship between symbols at a time, not two independent relationships simultaneously.

A. Cannot be accepted. Matching aⁿbⁿcⁿ needs two independent equalities, n = (number of b's) and n = (number of c's), and a single stack cannot verify both at once, so this language is not context-free.

B. Cannot be accepted. This is a unary language, and every context-free language over a one-letter alphabet is actually regular, so it is context-free only if its exponents form an eventually periodic set. Prime numbers are not eventually periodic, so this language is neither regular nor context-free.

C. Can be accepted. A PDA pushes w onto the stack while reading it, then pops and matches symbols against wᴿ after seeing c, the standard construction for such center-marked languages.

D. Can be accepted. The stack tracks the running difference between the counts of a's and b's, pushing on one symbol and popping on the other, a classic PDA construction for equal-count languages.

E. Can be accepted. A PDA pushes once for each a and once for each b onto the same stack, then pops once per c, so it accepts exactly when the number of c's equals n + m.

Only A and B describe languages a PDA cannot accept, so A & B Only is correct.
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Question 10

Match List-I (Grammar types) with List-II (Production forms).

List-I (Grammar)	List-II (All productions are of the form)
A. Regular Grammar	I. A → aX, where a ∈ T and X ∈ V*
B. Unrestricted Grammar	II. A → xB, A → x or A → Bx, A → x where A, B ∈ V and x ∈ T*
C. Chomsky Normal Form	III. x → y, where x ∈ (V ∪ T)+ and y ∈ (V ∪ T)*
D. Greibach Normal Form	IV. A → BC or A → a, where A, B, C are in V and a is in T

AA-II, B-III, C-I, D-IV
BA-I, B-III, C-II, D-IV
CA-II, B-III, C-IV, D-I
DA-II, B-IV, C-I, D-III

Answer: (C) A-II, B-III, C-IV, D-I

Explanation

Each grammar type has a distinct restriction on the shape of its productions.

A-II — Regular grammar productions are right-linear (A → xB or A → x) or left-linear (A → Bx or A → x), with x a string of terminals, matching item II.

B-III — Unrestricted grammar places no shape restriction beyond x containing at least one variable, matching the most general form x → y in item III.

C-IV — Chomsky Normal Form restricts every production to exactly two variables (A → BC) or a single terminal (A → a), matching item IV.

D-I — Greibach Normal Form requires every production to start with exactly one terminal followed only by variables (A → aX), matching item I.

A-II, B-III, C-IV, D-I is therefore the correct match.

Question 11
Match List-I (Process/lifecycle models) with List-II.
List-I List-II
A. V-Model I. Trigger transition
B. Incremental process model II. Classic life cycle
C. Spiral Model III. Risk Driven
D. Concurrent Model IV. Core product
1. AA-IV, B-I, C-III, D-II
2. BA-II, B-III, C-IV, D-I
3. CA-II, B-I, C-III, D-IV
4. DA-II, B-IV, C-III, D-I
Answer: (D) A-II, B-IV, C-III, D-I
Explanation
Each process model is best known for one specific descriptive tag from the software engineering literature.

A-II — The V-Model is a variant of the classic waterfall life cycle that pairs each development stage with a corresponding testing stage.

B-IV — The incremental model delivers a core product in the first increment, then adds features in successive increments.

C-III — The spiral model is explicitly risk-driven, since each loop of the spiral begins with a risk analysis before proceeding.

D-I — The concurrent development model represents each software element as being in one of several states, moving between them through triggered event transitions.

A-II, B-IV, C-III, D-I is therefore the correct match.
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Question 12

Match List-I (CPU schedulers) with List-II.

List-I	List-II
A. Short term scheduler	I. Determine which processes are added to the system for execution and place them in the ready queue
B. Medium term scheduler	II. Select which process should be executed next and allocate CPU time
C. Long term scheduler	III. Manage the degree of multiprogramming by swapping processes between main memory and disk storage
D. I/O Scheduler	IV. Order the input/output requests from processes waiting for I/O operations

AA-I, B-IV, C-II, D-III
BA-III, B-II, C-I, D-IV
CA-II, B-III, C-I, D-IV
DA-I, B-II, C-III, D-IV

Answer: (C) A-II, B-III, C-I, D-IV

Explanation

Each scheduler operates at a different stage of a process's lifecycle.

A-II — The short-term (CPU) scheduler runs most frequently and selects which ready process gets the CPU next.

B-III — The medium-term scheduler swaps processes between main memory and disk to control the degree of multiprogramming.

C-I — The long-term (job) scheduler decides which new processes are admitted into the system and placed on the ready queue.

D-IV — The I/O scheduler orders pending input/output requests from waiting processes.

A-II, B-III, C-I, D-IV is therefore the correct match.

Question 13
If a transaction tₐ holds a lock on some data item d in shared and intention-exclusive (SIX) mode, then which of the following is true for another transaction tᵦ?
1. Atᵦ can lock d in shared (S) mode
2. Btᵦ can lock d in exclusive (X) mode
3. Ctᵦ can lock d in intention-exclusive (IX) mode
4. Dtᵦ can lock d in intention-shared (IS) mode
Answer: (D) tᵦ can lock d in intention-shared (IS) mode
Explanation
Multi-granularity locking uses a compatibility matrix for the modes IS, IX, S, SIX and X, and a held SIX lock is compatible with only one other mode besides itself, which is IS.

A. Shared (S) — incompatible. SIX already implies an exclusive lock is held somewhere below in the hierarchy, so a plain shared request conflicts with it.

B. Exclusive (X) — incompatible. X conflicts with every other mode, including SIX.

C. Intention-exclusive (IX) — incompatible. IX signals an intent to write at a finer granularity, which conflicts with the exclusive component of SIX.

D. Intention-shared (IS) — correct. IS only signals an intent to read at a finer granularity, which does not conflict with the shared or exclusive components of SIX.

Only intention-shared mode is compatible with a held SIX lock.
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Question 14
In a multi-agent system (MAS), what are the key characteristics of individual agents?

A. They must be identical in their capabilities and goals.
B. They are autonomous and act independently.
C. They interact with each other and their environment.
D. They are controlled by a single, centralized authority at all times.
1. AA, B & C Only
2. BB, C & D Only
3. CB & C Only
4. DA & D Only
Answer: (C) B & C Only
Explanation
A multi-agent system is defined by agents that operate on their own and communicate, not by uniformity or central control.

A. False. Agents in a MAS are typically heterogeneous, with different capabilities and goals suited to their roles.

B. True. Autonomy, the ability to act without direct external control, is a defining property of an agent.

C. True. Agents interact with each other and with their environment to coordinate or compete toward their goals.

D. False. MAS are generally decentralized, which is precisely what distinguishes them from a single centrally controlled system.

Only B and C hold, so B & C Only is correct.
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Question 15
How many comparisons are required in the worst case to find the second smallest element of a list of n elements?
1. An + ⌈log n⌉ − 1
2. Bn + ⌈log n⌉
3. C⌈log n⌉
4. Dn + ⌈log n⌉ − 2
Answer: (D) n + ⌈log n⌉ − 2
Explanation
Finding the second smallest efficiently uses a tournament (knockout) method. Comparing elements in a single-elimination tournament to find the minimum takes n − 1 comparisons, and during that tournament the eventual minimum is directly compared against exactly ⌈log₂ n⌉ other elements, one at each round it wins.

The second-smallest element must be among those ⌈log₂ n⌉ elements that lost directly to the minimum, so finding the smallest among that shortlist takes ⌈log₂ n⌉ − 1 further comparisons.

(n − 1) + (⌈log₂ n⌉ − 1) = n + ⌈log n⌉ − 2, which is the worst-case comparison count.
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Question 16
Consider the following statements about reinforcement learning.

A. The adaptive dynamic programming agent leaves the transition model between states utilizes to solve the corresponding Markov decision process using dynamic programming.
B. Temporal difference needs a transition model to perform its updates.
C. The prioritized sweeping heuristic focuses on adjusting states with successors that have undergone significant changes in utility estimates.
D. The approach of modified policy iteration involves adopting a simplified value utility estimates following each change to the learned model.
1. AA, B & C Only
2. BA, B & D Only
3. CA, C & D Only
4. DB, C & D Only
Answer: (C) A, C & D Only
Explanation
Adaptive dynamic programming (ADP), temporal difference (TD) learning, prioritized sweeping and modified policy iteration are distinguished mainly by whether they use a transition model and how they update value estimates.

A. True. An ADP agent learns the transition model of the environment and then solves the resulting Markov decision process using standard dynamic programming.

B. False. Temporal difference learning is model-free by design, updating value estimates directly from observed transitions and rewards without ever building a transition model.

C. True. Prioritized sweeping directs computation toward states whose successors have recently had large changes in their utility estimates, since updating those states first propagates useful information fastest.

D. True. Modified policy iteration replaces full, exact policy evaluation with a small number of simplified value-update passes after each change to the learned model, trading some accuracy for speed.

A, C and D hold while B misdescribes temporal difference learning as needing a model, so A, C & D Only is correct.
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Question 17
Arrange the following steps for Pulse Code Modulation (PCM).

A. Sampling
B. Low-Pass filter
C. Encoding
D. Quantization
1. AA, B, C, D
2. BB, C, D, A
3. CB, A, D, C
4. DD, C, B, A
Answer: (C) B, A, D, C
Explanation
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Question 18
Arrange the following tasks of a lexical analyzer in the correct sequence.

A. Token Generation.
B. Removal of comments/White sphere.
C. Buffering and reading source characters.
D. Return token stream to parser.
1. AA, B, C, D
2. BB, A, C, D
3. CC, B, A, D
4. DB, A, D, C
Answer: (C) C, B, A, D
Explanation
A lexical analyzer first buffers and reads the raw source characters, strips out comments and white space that carry no lexical meaning, groups the remaining characters into tokens, and then passes the resulting token stream on to the parser.

C-B-A-D — Buffering and reading source characters, Removal of comments/white space, Token Generation, Return token stream to parser — is the correct sequence.
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Question 19
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion (A): Let R ⊆ A × A be a binary relation as defined below: A = N × N; ((a,b),(c,d)) ∈ R if only if a ≤ c or b ≤ d. This binary relation R is a partial order.
Reason (R): A binary relation will be a partial order if this satisfy reflexive, anti-symmetric and transitive property.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (D) A is not correct but R is correct
Explanation
A partial order must be reflexive, antisymmetric and transitive, and Reason R states this standard definition correctly.

Checking Assertion A against this definition on A = N × N with ((a,b),(c,d)) ∈ R iff a ≤ c or b ≤ d.

Reflexivity holds trivially, since a ≤ a is always true.

Antisymmetry fails. Take (1,5) and (5,1). Since 1 ≤ 5, (1,5) R (5,1) holds. For the reverse pair, 5 ≤ 1 is false but 1 ≤ 5 is true, so (5,1) R (1,5) also holds. Both distinct pairs relate to each other in both directions, which violates antisymmetry.

Since the relation is not antisymmetric, it is not a partial order, so Assertion A is false while Reason R correctly states the definition of a partial order.
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Question 20
Which of the following statements are true for link state routing?

A. Link state Routing is suffer from count-to-infinity problem.
B. Link state Routing uses flooding in the network to distribute link state packets.
C. Each router creates its own link state packet.
D. Every router has exactly the same link state database.
E. Each router finds its own shortest paths to the other routers by using distance vector routing.
1. AA & E Only
2. BB, C & D Only
3. CA, B, C & D Only
4. DA, B & E Only
Answer: (B) B, C & D Only
Explanation
Link-state routing and distance-vector routing solve the same problem with very different mechanisms, and several of these statements swap one algorithm's property for the other's.

A. False. Count-to-infinity is a distance-vector problem, caused by routers slowly updating hop counts through each other, and link-state routing avoids it because every router computes paths from a complete, identical topology map.

B. True. Each router floods its link state packet to every other router in the network so that everyone converges on the same topology view.

C. True. Every router is responsible for creating the link state packet that describes its own directly connected links.

D. True. Because flooding delivers every router's link state packet to every other router, all routers end up building an identical link state database once the network converges.

E. False. Once a router has the complete topology database, it computes shortest paths using Dijkstra's algorithm, not distance vector routing, which is a separate algorithm that never builds a full topology map.

B, C and D hold, so B, C & D Only is correct.
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Question 21
Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R

Assertion (A): Higher cyclomatic complexity indicates a more complex program.
Reason (R): It is calculated based on the number of independent execution paths in the program.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (A) Both A and R are correct and R is the correct explanation of A
Explanation
Cyclomatic complexity is a direct measure of a program's structural complexity, so a higher value does indicate a more complex program, confirming Assertion A.

Reason R correctly explains why. Cyclomatic complexity is computed from the number of linearly independent paths through the program's control flow graph, and more independent paths mean more distinct execution routes to understand and test, which is exactly what makes a program more complex. Reason R therefore correctly explains Assertion A.
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Question 22
Arrange the following prefix expressions in ascending order on the basis of their values.

A. − * 2 / 8 4 3
B. ↑ − * 3 6 * 4 2 5
C. + − ↑ 3 2 ↑ 2 3 / 6 − 4 2
D. * + 3 + 3 ↑ 3 + 1 1 3
1. AA, D, C, B
2. BA, C, D, B
3. CA, B, C, D
4. DC, A, D, B
Answer: (B) A, C, D, B
Explanation
Each prefix expression must be evaluated inside out, applying every operator to its two operands as they resolve.

Expression A: / 8 4 = 2, then * 2 2 = 4, then − 4 3 = 1. So A = 1.

Expression B: * 3 6 = 18, * 4 2 = 8, then − 18 8 = 10, then ↑ 10 5 = 100000. So B = 100000.

Expression C: ↑ 3 2 = 9 and ↑ 2 3 = 8, so − 9 8 = 1. Also − 4 2 = 2, so / 6 2 = 3. Adding the two parts, + 1 3 = 4. So C = 4.

Expression D: + 1 1 = 2, then ↑ 3 2 = 9, then + 3 9 = 12, then + 3 12 = 15, then * 15 3 = 45. So D = 45.

Ranking the four values 1, 4, 45 and 100000 from smallest to largest gives A (1), C (4), D (45), B (100000), so A, C, D, B is the correct ascending order.
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Question 23

Match List-I (Network-layer address protocols) with List-II.

List-I	List-II
A. Address resolution protocol	I. Map physical address to IP address
B. Reverse address resolution protocol	II. Used by host and router to send notification of datagram problem back to the sender
C. Internet control message protocol	III. Help a multicast router identify hosts in a LAN
D. Internet Group message protocol	IV. Map IP address to physical address

AA-I, B-II, C-III, D-IV
BA-II, B-I, C-IV, D-III
CA-IV, B-II, C-III, D-I
DA-IV, B-I, C-II, D-III

Answer: (D) A-IV, B-I, C-II, D-III

Explanation

Each protocol resolves one direction of the address mapping or one specific network-layer notification role.

A-IV — Address Resolution Protocol maps a known IP address to the corresponding physical (MAC) address.

B-I — Reverse Address Resolution Protocol does the opposite, mapping a known physical address to an IP address.

C-II — Internet Control Message Protocol carries error and diagnostic notifications, such as an undeliverable datagram, back to the sending host or router.

D-III — Internet Group Management Protocol lets a multicast router discover which hosts on a LAN belong to a multicast group.

A-IV, B-I, C-II, D-III is therefore the correct match.

Question 24

Match List-I (Algorithm design techniques) with List-II.

List-I	List-II
A. Dynamic programming	I. Floyd Warshall shortest path
B. Greedy	II. Huffman coding
C. Back tracking	III. Hamiltonian cycle problem
D. Branch and bound	IV. Travelling salesman (TSP) using lower bound

AA-II, B-III, C-IV, D-I
BA-IV, B-III, C-II, D-I
CA-I, B-II, C-III, D-IV
DA-III, B-IV, C-I, D-II

Answer: (C) A-I, B-II, C-III, D-IV

Explanation

Each algorithmic technique is best illustrated by one of its classic applications.

A-I — The Floyd-Warshall all-pairs shortest path algorithm is a textbook dynamic programming algorithm, building solutions from smaller subproblems.

B-II — Huffman coding greedily merges the two lowest-frequency nodes at each step, making it a classic greedy algorithm.

C-III — The Hamiltonian cycle problem is classically solved by backtracking, building a path and abandoning a branch as soon as it cannot be extended.

D-IV — The travelling salesman problem, when solved by pruning partial tours using a computed lower bound, is a standard branch-and-bound application.

A-I, B-II, C-III, D-IV is therefore the correct match.

Question 25
Which of the following is FALSE?
1. AEvery field is an integral domain.
2. BEvery integral domain is field.
3. CFinite integral domain is field.
4. DThe ring Zₚ of integers module p is field iff p is prime.
Answer: (B) Every integral domain is field.
Explanation
A field requires every nonzero element to have a multiplicative inverse, which automatically rules out zero divisors, so every field is also an integral domain. This makes the first statement true, not the false one being asked for.

The reverse direction fails. The integers Z form an integral domain, since they have no zero divisors, but Z is not a field because an element like 2 has no multiplicative inverse within Z. So every integral domain is a field is the false statement.

The remaining two statements are standard true results. A finite integral domain is always a field, because multiplication by any nonzero element is an injective, and therefore surjective, map on a finite set, which produces the needed inverse. Zₚ is a field exactly when p is prime, since only then does every nonzero residue have a multiplicative inverse modulo p.
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Question 26
Consider the following set of processes, all of which arrive at time 0 in the order P1, P2, P3, P4 and P5, with the CPU burst time given in milliseconds. Based on the table, what is the average waiting time (in milliseconds) using priority scheduling? Assume a lower priority number indicates higher priority.
Process Burst Time (ms) Priority
P1 10 3
P2 1 1
P3 2 4
P4 1 5
P5 5 2
1. A8
2. B8.2
3. C8.3
4. D8.4
Answer: (B) 8.2
Explanation
In priority scheduling with a lower number meaning higher priority, the processes run in the order of their priority values rather than their arrival order.

Ordering P1-P5 by priority gives the execution sequence P2 (priority 1), P5 (priority 2), P1 (priority 3), P3 (priority 4), P4 (priority 5).

Since all processes arrive at time 0, each process's waiting time equals the sum of the burst times of every process scheduled before it.

P2 waits 0 ms, P5 waits 1 ms (after P2), P1 waits 6 ms (after P2 + P5 = 1 + 5), P3 waits 16 ms (after P2 + P5 + P1 = 6 + 10), and P4 waits 18 ms (after P2 + P5 + P1 + P3 = 16 + 2).

Average waiting time = (0 + 1 + 6 + 16 + 18) / 5 = 41 / 5 = 8.2 ms.
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Question 27
Assertion (A): The order of the group (Z₅, ×) divides the order of the group (Z₁₃, ×).
Reason (R): The order of a subgroup of a group divides the order of the group.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (B) Both A and R are correct but R is NOT the correct explanation of A
Explanation
The multiplicative group (Z₅, ×) consists of the nonzero residues {1, 2, 3, 4} under multiplication mod 5, giving order 4. The multiplicative group (Z₁₃, ×) consists of {1, 2, ..., 12} under multiplication mod 13, giving order 12.

Since 4 divides 12, Assertion A is numerically true.

Reason R states Lagrange's theorem, that the order of a subgroup always divides the order of its parent group. This is a true general statement.

However, R does not explain A here. Lagrange's theorem applies only when one group is an actual subgroup of the other, sharing the same underlying set and operation. (Z₅, ×) is not a subset of (Z₁₃, ×) at all, since its elements are residues mod 5 combined under multiplication mod 5, a different operation from multiplication mod 13. The divisibility in A is a coincidental numerical fact, not an instance of the subgroup relationship that R describes.

So both A and R are correct as standalone statements, but R is not the correct explanation of A.
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Question 28
A signal travels through a transmission medium and its power is reduced to one-half of its original value, so that P₂ = 1/2 P₁. What is the power loss in attenuation?
1. A-6 dB
2. B-3 dB
3. C-1.5 dB
4. D-1 dB
Answer: (B) -3 dB
Explanation
Attenuation in decibels for a power ratio uses the formula
dB = 10 log₁₀(P₂ / P₁).

Substituting P₂ = 0.5 P₁ gives
dB = 10 log₁₀(0.5) ≈ 10 × (-0.301) ≈ -3.01 dB.

The negative sign shows a loss, so halving the power gives an attenuation of about -3 dB, matching option B.

Option A (-6 dB) comes from mistakenly using the amplitude/voltage formula dB = 20 log₁₀(P₂/P₁) instead of the power formula, which doubles the result to about -6.02 dB. Options C and D do not correspond to any standard relationship for a 50% power loss and are too small in magnitude.
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Question 29
Which logic is designed to address the problem of referential opacity in propositional attitudes?
1. ARegular Logic
2. BModal Logic
3. CFirst-order logic
4. DPredicate logic
Answer: (B) Modal Logic
Explanation
Referential opacity arises in propositional-attitude contexts such as belief, knowledge, and desire, where substituting equal terms inside the attitude can change the truth value of the whole statement, unlike in standard extensional logic.

Modal logic extends classical logic with operators for necessity and possibility, and its extensions such as epistemic and doxastic logic are specifically built to represent propositional attitudes like belief and knowledge without collapsing under substitution of co-referring terms. This makes modal logic the framework designed to handle referential opacity.

First-order logic and predicate logic are extensional systems, so they cannot correctly represent attitude contexts where substitution of equals fails, and there is no established logic called Regular Logic that addresses this problem.
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Question 30
Which of the following statements about DFS are correct?
A. It can detect cycles in a graph.
B. It can be used to find connected components.
C. It works for both directed and undirected graphs.
D. Guarantees shortest path in unweighted graphs.
1. AA, B, C Only
2. BA, B, C, D
3. CC, D Only
4. DB, C Only
Answer: (A) A, B, C Only
Explanation
Depth-first search explores as far as possible along each branch before backtracking, which gives it several well-known properties.

A. It can detect cycles in a graph. — correct. A back edge to an ancestor still on the recursion stack signals a cycle in both directed and undirected graphs.

B. It can be used to find connected components. — correct. Restarting DFS from every unvisited vertex and grouping vertices reached in each run identifies the connected components.

C. It works for both directed and undirected graphs. — correct. The traversal only needs an adjacency structure, so it applies to either graph type with minor bookkeeping differences.

D. Guarantees shortest path in unweighted graphs. — incorrect. DFS can return a much longer path than necessary because it dives deep before exploring nearby vertices, so breadth-first search, not DFS, guarantees the shortest path in an unweighted graph.
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Question 31
Which of the following logical implications are tautologies?
A. [~p ∧ (p ∨ q)] → q
B. ~p → (p → q)
C. [p ∧ (p → q)] → q
D. ~(p → q) → q
1. AA & B Only
2. BB & D Only
3. CA, B & C Only
4. DB, C & D Only
Answer: (C) A, B & C Only
Explanation
An implication is a tautology when it evaluates to true for every combination of truth values of its variables.

A. [~p ∧ (p ∨ q)] → q — this is true in every row: whenever ~p ∧ (p ∨ q) is true, p must be false and q must be true to make p ∨ q true, so q is guaranteed. It is a tautology.

B. ~p → (p → q) — whenever p is false, the consequent p → q is automatically true, and whenever p is true, the antecedent ~p is false, making the whole implication true either way. It is a tautology.

C. [p ∧ (p → q)] → q — this is modus ponens, one of the most basic valid inference forms, and it holds in every row of its truth table. It is a tautology.

D. ~(p → q) → q — taking p = true and q = false makes p → q false, so ~(p → q) is true, but q itself is false, giving a true antecedent and a false consequent. The implication is false in this case, so D is not a tautology.

Only A, B and C hold in every case, so the correct set is A, B & C Only.
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Question 32
Match List-I (Compiler input) with List-II (Compiler phase).
LIST-I (Input) LIST-II (Compiler Phase)
A. Syntax Tree I. Code Generator
B. Character Stream II. Syntax Analyzer
C. Intermediate Representation III. Semantic Analyzer
D. Token Stream IV. Lexical Analyzer
1. AA-II, B-III, C-IV, D-I
2. BA-III, B-IV, C-I, D-II
3. CA-III, B-I, C-IV, D-II
4. DA-I, B-IV, C-II, D-III
Answer: (B) A-III, B-IV, C-I, D-II
Explanation
Each pairing follows from what each compiler phase consumes as its input.

A-III, B-IV, C-I, D-II.

A. Syntax Tree — III. Semantic Analyzer. The parser produces the syntax tree, and the semantic analyzer consumes it to check types and scope.

B. Character Stream — IV. Lexical Analyzer. The lexical analyzer is the first phase, reading the raw source code character by character.

C. Intermediate Representation — I. Code Generator. The code generator takes the intermediate representation and translates it into target code.

D. Token Stream — II. Syntax Analyzer. The lexical analyzer hands its stream of tokens to the syntax analyzer, which parses them against the grammar.
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Question 33
Assertion (A): The control points have only local control over the shape of the B-Splines.
Reason (R): Bezier curves do have local control.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (C) A is correct but R is not correct
Explanation
A B-spline curve is built from piecewise polynomial segments, so moving one control point reshapes only the nearby segments of the curve while the rest of the curve stays unchanged. This local-control property makes Assertion A correct.

Bezier curves behave the opposite way. Every control point contributes to a single blending function that spans the entire curve, so moving any one control point reshapes the whole curve. Bezier curves therefore have global control, not local control, which makes Reason R incorrect.

So A is correct but R is not correct.
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Question 34
Which of the following statements are true for digital-to-digital encoding schemes?
A. Unipolar encoding uses one level of value.
B. Unipolar encoded signal contains a direct current component (a component with zero frequency).
C. In NRZ-L, the level of the signal is dependent upon the state of the bit.
D. In NRZ-I, the signal is not inverted of a 1 is encountered.
E. RZ encoding scheme removes both DC components and problem of synchronization.
1. AA, B, C & E Only
2. BA, C, D & E Only
3. CB, C, D & E Only
4. DA, B, D & E Only
Answer: (A) A, B, C & E Only
Explanation
Evaluate each statement against the standard digital-to-digital encoding rules.

A. Unipolar encoding uses one level of value. — correct. Unipolar signaling represents one bit value with a single non-zero voltage level and the other bit value with zero voltage, so only one active signal level is used.

B. Unipolar encoded signal contains a direct current component. — correct. Because the signal never swings to a negative voltage, its average value is nonzero, which shows up as a DC component in the frequency spectrum.

C. In NRZ-L, the level of the signal is dependent upon the state of the bit. — correct. NRZ-L stands for non-return-to-zero level, where one voltage level directly represents bit 1 and the other level represents bit 0.

D. In NRZ-I, the signal is not inverted of a 1 is encountered. — incorrect. The defining rule of NRZ-I (non-return-to-zero invert) is the opposite: the signal level inverts whenever a 1 occurs and stays the same for a 0, so this statement misstates the standard rule.

E. RZ encoding scheme removes both DC components and problem of synchronization. — correct. Because the signal returns to zero at the middle of every bit interval, it produces a transition in every bit period, which supplies a clocking reference and avoids the long DC runs that unipolar and NRZ schemes suffer from.

So the true statements are A, B, C and E.
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Question 35
Assertion (A): 1-persistent CSMA provides low delay under low load conditions.
Reason (R): When the channel is free, stations transmit immediately without any random delay, ensuring quick access and minimal delay.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (A) Both A and R are correct and R is the correct explanation of A
Explanation
In 1-persistent CSMA, a station that wants to transmit senses the channel and, as soon as the channel becomes idle, transmits with probability 1, that is, immediately and without waiting any random backoff period.

Under low load, few stations are competing for the channel at any given time, so the chance that two stations sense the channel idle and transmit at the same instant is small. This lets stations gain access and finish transmitting quickly, giving low delay, so Assertion A is correct.

Reason R correctly describes this exact mechanism: transmitting immediately once the channel is sensed free, with no random delay, is precisely why access is quick under low contention. So R is both correct and the correct explanation of A.
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Question 36
Match List-I (UML diagram types) with List-II.
LIST-I LIST-II
A. Class diagram I. subset
B. Package diagram II. relationship
C. Object diagram III. dependencies
D. Component diagram IV. static structure
1. AA-II, B-III, C-IV, D-I
2. BA-IV, B-I, C-III, D-II
3. CA-IV, B-I, C-II, D-III
4. DA-III, B-IV, C-II, D-I
Answer: (C) A-IV, B-I, C-II, D-III
Explanation
Each pairing follows from what each UML diagram type is used to show.

A-IV, B-I, C-II, D-III.

A. Class diagram — IV. Static structure. It models the static structure of a system through classes, attributes, and their relationships.

B. Package diagram — I. Subset. It groups related model elements together, carving out a subset of the overall model.

C. Object diagram — II. Relationship. It is a snapshot of specific object instances and the relationships (links) between them at one point in time.

D. Component diagram — III. Dependencies. It shows the dependencies among the software components that make up a system.
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Question 37
Which of the following statements are true?
A. Painters' algorithm is also called depth sort algorithm.
B. Gourad shading is an interpolating shading method.
C. Gourad shading is an averaging shading method.
D. The Z-buffer algorithm finds the largest depth value Z.
1. AA, B Only
2. BB, C Only
3. CC, D Only
4. DA, D Only
Answer: (A) A, B Only
Explanation
A. Painters' algorithm is also called depth sort algorithm. — correct. The algorithm sorts polygons by depth and paints them from farthest to nearest, which is exactly why it is also known as the depth-sort algorithm.

B. Gourad shading is an interpolating shading method. — correct. Gouraud shading computes intensity at each vertex and then linearly interpolates those intensity values across the surface of the polygon.

C. Gourad shading is an averaging shading method. — incorrect. Averaging is only used as an intermediate step to compute a vertex normal from the surrounding polygon normals, but the shading method itself is classified by intensity interpolation, not by averaging.

D. The Z-buffer algorithm finds the largest depth value Z. — incorrect. The Z-buffer algorithm keeps the smallest depth value at each pixel, since the point nearest to the viewer is the one that should be visible, not the farthest one.

So the correct statements are A and B only.
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Question 38
Dialog control is a function of which layer?
1. ANetwork Layer
2. BTransport Layer
3. CSession Layer
4. DPresentation Layer
Answer: (C) Session Layer
Explanation
Dialog control refers to managing whose turn it is to transmit in a conversation between two communicating systems, deciding whether communication is half-duplex or full-duplex.

The session layer of the OSI model is responsible for establishing, managing, and terminating sessions between applications, and dialog control is one of its core functions.

A. Network Layer — handles routing and logical addressing, not turn-taking between applications.

B. Transport Layer — handles end-to-end delivery and reliability, not who transmits next.

C. Session Layer — correct. It manages dialog control between communicating applications.

D. Presentation Layer — handles data translation, encryption, and compression, not dialog turn-taking.
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Question 39
Arrange the following sequence related to the FASTUS system.
A. Basic-group handling
B. Structure merging
C. Tokenization
D. Complex phrase handling
E. Complex word handling
1. AC, A, D, E, B
2. BE, A, C, D, B
3. CC, E, A, D, B
4. DC, E, D, A, B
Answer: (C) C, E, A, D, B
Explanation
FASTUS is a cascaded finite-state system for information extraction, and each stage builds on the output of the previous one.

C, E, A, D, B.

C. Tokenization — breaks the raw text into individual tokens, the first step.

E. Complex word handling — recognizes multi-word expressions, proper names, and other fixed word groups from the tokens.

A. Basic-group handling — identifies simple noun and verb groups from the processed words.

D. Complex phrase handling — combines the basic groups into larger noun and verb phrases.

B. Structure merging — the final stage, merging the recognized phrases into complete domain-level event structures.
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Question 40
If the set S = { [a 0; 0 b] : a, b ∈ Z } is a subring of the ring M₂ of 2×2 matrices over the integers, then S is a
1. Aleft ideal but not right ideal
2. Bright ideal but not left ideal
3. Cboth left and right ideal
4. Dneither left ideal nor right ideal
Answer: (D) neither left ideal nor right ideal
Explanation
Let D = [[a, 0], [0, b]] be an arbitrary element of S and let M = [[p, q], [r, s]] be an arbitrary element of M₂(Z).

Computing D·M gives [[ap, aq], [br, bs]], which has off-diagonal entries aq and br that are nonzero whenever q or r is nonzero, so D·M does not stay inside S for every choice of M.

Computing M·D gives [[pa, qb], [ra, sb]], which has off-diagonal entries qb and ra that are again nonzero in general, so M·D also does not stay inside S for every choice of M.

Since multiplying a diagonal matrix by an arbitrary matrix of M₂(Z), on either side, can introduce nonzero off-diagonal entries, S fails the absorption condition needed for both a left ideal and a right ideal. So S is neither a left ideal nor a right ideal, even though it is a valid subring on its own.
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Question 41
If r₁ and r₂ are regular expressions, then which of the following are correct?
A. L(r₁ + r₂) = L(r₁) ∪ L(r₂)
B. L(r₁ · r₂) = L(r₁)L(r₂)
C. L((r₁)) = L(r₁)
D. L(r₁*) = (L(r₁))*
1. AA & B Only
2. BB, C & D Only
3. CA, C & D Only
4. DA, B, C & D
Answer: (D) A, B, C & D
Explanation
Each statement is a direct restatement of how regular expression operators map to language operations.

A. The union operator + between two regular expressions produces the union of the two languages, so L(r₁ + r₂) = L(r₁) ∪ L(r₂) holds by definition.

B. Concatenating two regular expressions produces the concatenation of their languages, so L(r₁ · r₂) = L(r₁)L(r₂) holds by definition.

C. Parentheses only group a regular expression for precedence and do not change what it denotes, so L((r₁)) = L(r₁) holds trivially.

D. The Kleene star operator on a regular expression produces the Kleene closure of its language, so L(r₁*) = (L(r₁))* holds by definition.

All four statements correctly describe how regular expression syntax corresponds to language operations, so all of A, B, C and D are correct.
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Question 42

Match List-I (C++ file mode parameters) with List-II.

LIST-I (File mode parameter)	LIST-II (Meaning)
A. ios::ate	I. Go to end of file on opening
B. ios::in	II. Open fails if the file already exists
C. ios::nocreate	III. Open file for reading only
D. ios::noreplace	IV. Open fails if the file does not exist

AA-III, B-IV, C-II, D-I
BA-I, B-III, C-II, D-IV
CA-II, B-IV, C-I, D-III
DA-I, B-III, C-IV, D-II

Answer: (D) A-I, B-III, C-IV, D-II

Explanation

Each pairing follows from what each C++ file mode flag does when opening a file.

A-I, B-III, C-IV, D-II.

A. ios::ate — I. Go to end of file on opening. It opens a file and immediately positions the get and put pointers at the end.
B. ios::in — III. Open file for reading only. It opens a file strictly for reading.
C. ios::nocreate — IV. Open fails if the file does not exist. It refuses to create a new file, so the open fails when the file is missing.
D. ios::noreplace — II. Open fails if the file already exists. It refuses to overwrite an existing file, so the open fails when the file is present.

Question 43
Which of the following statements are true?
A. The ++ operator can be overloaded to perform decrementation of a variable.
B. It is necessary that a constructor in a class should always be public.
C. Size of an object is equal to the sum of sizes of data members and member functions within the class.
D. A static data member is useful when all objects of a class are to share a common item of information.
E. The this pointer always contains the address of the object using which the member function is being called.
1. AA, B, C Only
2. BB, C, E Only
3. CA, D, E Only
4. DB, D, E Only
Answer: (C) A, D, E Only
Explanation
A. The ++ operator can be overloaded to perform decrementation of a variable. — correct. Operator overloading in C++ only binds a symbol to a function definition, and the language does not enforce that the overloaded body must match the operator's conventional meaning, so a programmer can legally define operator++ to decrement.

B. It is necessary that a constructor in a class should always be public. — incorrect. Constructors can be declared private or protected, which is exactly how patterns such as the singleton pattern prevent direct external instantiation.

C. Size of an object is equal to the sum of sizes of data members and member functions within the class. — incorrect. Member functions are stored once as shared code, not copied into every object, so only the data members contribute to an object's size, along with any compiler-added overhead such as vtable pointers.

D. A static data member is useful when all objects of a class are to share a common item of information. — correct. A static data member has a single shared copy across all objects of the class, so it is the natural way to hold information common to every instance.

E. The this pointer always contains the address of the object using which the member function is being called. — correct. Every non-static member function implicitly receives a this pointer bound to the specific object invoking it.

So the true statements are A, D and E only.
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Question 44
Match List-I (AI subfields) with List-II.
LIST-I LIST-II
A. Expert systems I. Medical Diagnosis
B. NLP II. Text Summarization
C. Computer Vision III. Image Classification
D. ML IV. Predictive Analysis
1. AA-I, B-II, C-III, D-IV
2. BA-II, B-I, C-III, D-IV
3. CA-II, B-III, C-I, D-IV
4. DA-III, B-II, C-I, D-IV
Answer: (A) A-I, B-II, C-III, D-IV
Explanation
Each pairing follows from the primary application area of each AI subfield.

A-I, B-II, C-III, D-IV.

A. Expert systems — I. Medical Diagnosis. Expert systems apply encoded domain rules to specific problems, and medical diagnosis (as in systems like MYCIN) is one of the earliest and most famous applications.

B. NLP — II. Text Summarization. Natural language processing works directly with text, making summarization a natural fit.

C. Computer Vision — III. Image Classification. Computer vision analyzes and interprets images, and image classification is one of its core tasks.

D. Machine Learning — IV. Predictive Analysis. Machine learning builds models from historical data to forecast outcomes, which is the basis of predictive analysis.
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Question 45
Assertion (A): Unsupervised learning algorithms are used for tasks like data summarization and exploratory data analysis.
Reason (R): Unsupervised learning requires a labelled dataset to find relationships and patterns in data.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (C) A is correct but R is not correct
Explanation
Unsupervised learning algorithms, such as clustering and dimensionality reduction, are commonly applied to summarize data and explore its structure without needing predefined output labels, so Assertion A is correct.

Reason R claims the opposite of how unsupervised learning is defined. Unsupervised learning specifically works on unlabelled data to discover hidden patterns and structure, while it is supervised learning that requires a labelled dataset. So R is incorrect.

A is correct but R is not correct.
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Question 46
Assertion (A): Kruskal's algorithm and Prim's algorithm always produce a minimum spanning tree (MST).
Reason (R): Every connected graph has a unique MST.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (C) A is correct but R is not correct
Explanation
Both Kruskal's algorithm and Prim's algorithm are greedy algorithms proven to always construct a minimum spanning tree of a connected, weighted graph, so Assertion A is correct.

Reason R is false. A connected graph has a unique MST only when all edge weights are distinct. When two or more edges share the same weight, a graph can have multiple different spanning trees that all achieve the same minimum total weight, so the MST is not guaranteed to be unique.

A is correct but R is not correct.
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Question 47
Specialization process does not allow
1. Adefining a set of subclasses of an entity type
2. Bestablishing additional specific attributes with each subclass
3. Cdefining a generalized entity type from the given entity types
4. Destablishing additional specific relationship types between each subclass and other entity types or other subclasses
Answer: (C) defining a generalized entity type from the given entity types
Explanation
Specialization is a top-down process that starts from one entity type and breaks it into more specific subclasses, each of which can be given its own extra attributes and relationships, which is exactly what options A, B and D describe.

Defining a single generalized entity type from a set of given entity types is the reverse, bottom-up process, known as generalization, not specialization.

So specialization does not allow defining a generalized entity type from the given entity types.
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Question 48
Which planning approach is used for an environment with no observations?
1. AHierarchical planning
2. BReactive planning
3. CConformant planning
4. DContingency Planning
Answer: (C) Conformant planning
Explanation
Conformant planning is designed for environments where the agent has no sensing capability at all, so it cannot observe the current state during execution.

Because no observations are available, a conformant plan must be built in advance as a fixed sequence of actions that is guaranteed to reach the goal regardless of the actual initial state or outcome, with no branching based on feedback.

A. Hierarchical planning — decomposes a task into abstract subtasks, unrelated to observability.

B. Reactive planning — selects actions directly from current percepts without deliberative lookahead, which still needs some observation.

C. Conformant planning — correct. It works with zero observability by committing to a fixed action sequence in advance.

D. Contingency planning — assumes partial observability and builds a plan with branches selected using sensed information.
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Question 49
Arrange the following events in the correct order when an interrupt occurs during instruction execution.
A. Save program counter (PC) and status register
B. Fetch interrupt vector from memory
C. Execute the interrupted instruction completely
D. Transfer control to interrupt service routine (ISR)
E. Restore PC and register after ISR
1. AA, B, C, D, E
2. BE, D, C, B, A
3. CC, A, B, D, E
4. DA, B, E, C, D
Answer: (C) C, A, B, D, E
Explanation
An interrupt is only acted on at an instruction boundary, and the processor then follows a fixed save-dispatch-restore sequence.

C, A, B, D, E.

C. Execute the interrupted instruction completely — the processor finishes the instruction already in progress before reacting to the interrupt.

A. Save program counter (PC) and status register — the context is saved so the interrupted program can be restored later.

B. Fetch interrupt vector from memory — the processor locates the address of the appropriate interrupt service routine.

D. Transfer control to interrupt service routine (ISR) — execution jumps to that routine.

E. Restore PC and register after ISR — once the ISR finishes, the saved PC and status register are restored so normal execution can resume.
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Question 50
Which of the following is the correct order of increasing time complexity of the given algorithms?
A. Tower of Hanoi with n disks
B. Binary search given n sorted numbers
C. Heap sort given n numbers, worst case
D. Addition of two n × n matrices
1. AB, C, D, A
2. BA, B, C, D
3. CD, C, B, A
4. DC, A, D, B
Answer: (A) B, C, D, A
Explanation
Each algorithm has a standard, well-known time complexity.

Binary search on n sorted numbers — O(log n).

Heap sort, worst case — O(n log n).

Addition of two n × n matrices — O(n²), since every one of the n² entries needs one addition.

Tower of Hanoi with n disks — O(2ⁿ), since the number of moves doubles with each additional disk.

Ordering these by growth rate from smallest to largest gives log n < n log n < n² < 2ⁿ, so the increasing order is Binary search, Heap sort, Matrix addition, Tower of Hanoi — that is, B, C, D, A.
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Question 51
Arrange the following interframe time interval in increasing order.

A. PIFS (PCF Interframe Spacing)
B. DIFS (DCF Interframe Spacing)
C. SIFS (Short Interframe Spacing)
D. EIFS (Extended Interframe Spacing)
1. AA, B, C, D
2. BB, C, A, D
3. CD, A, B, C
4. DC, A, B, D
Answer: (D) C, A, B, D
Explanation
IEEE 802.11 defines four interframe spacing intervals that control how long a station waits before transmitting, and their durations increase in a fixed order.

SIFS (Short Interframe Space) is used for the highest-priority frames such as ACKs, so it must be the smallest gap.
PIFS (PCF Interframe Space) is slightly longer and lets the point coordinator gain priority access over ordinary DCF stations.
DIFS (DCF Interframe Space) is longer still and is the standard wait time for contention-based access.
EIFS (Extended Interframe Space) is the longest, used after a station receives a corrupted frame it cannot interpret, giving other stations time to recover.

Mapping the labels to this order gives C (SIFS), A (PIFS), B (DIFS), D (EIFS), which is C, A, B, D.
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Question 52
Which of the following are characteristics of RISC architecture?

A. Large instruction set with many adversing modes
B. Fixed instruction length
C. Emphasis on software rather than hardware
D. Simple instruction that execute in one cycle
1. AA, B Only
2. BB, C, D Only
3. CA, C, D only
4. DA, B, C, D
Answer: (B) B, C, D Only
Explanation
RISC (Reduced Instruction Set Computer) design keeps the hardware simple and pushes complexity into the compiler, which is the opposite philosophy of CISC.

A. Large instruction set with many adversing modes — this describes CISC, not RISC. RISC deliberately keeps the instruction set small with few addressing modes so the hardware stays simple.

B. Fixed instruction length — correct. RISC instructions are uniform in length so they can be fetched and decoded quickly in a pipeline.

C. Emphasis on software rather than hardware — correct. RISC shifts complexity to the compiler, letting the hardware stay minimal.

D. Simple instruction that execute in one cycle — correct. Most RISC instructions are designed to complete in a single clock cycle to support efficient pipelining.

Only B, C and D describe RISC, so the answer is B, C, D only.
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Question 53
Which of the following is the correct solution of the given recurrence relation?

T(n) = 3T(n/4) + n log n
1. Aθ(n log n)
2. Bθ(n² log n)
3. Cθ(n(log n)²)
4. Dθ(n log log n)
Answer: (A) θ(n log n)
Explanation
The Master theorem applies directly to T(n) = 3T(n/4) + n log n, with a = 3, b = 4 and f(n) = n log n.

Compare f(n) against n^(log_b a): log_4 3 ≈ 0.7925, so n^(log_4 3) ≈ n^0.7925.

Since n log n grows polynomially faster than n^0.7925 for large n, this falls under Master theorem Case 3.

Case 3 also needs the regularity condition a·f(n/b) ≤ c·f(n) for some c < 1. Checking it: 3·(n/4)log(n/4) = 0.75·n·log(n/4), which stays below 0.75·n log n for large n, so the condition holds.

Under Case 3, T(n) = Θ(f(n)) = Θ(n log n).
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Question 54
Let P and Q be two regular expressions over Σ. If P does not contain ε, then the equation in R namely, R = Q + RP has a unique solution given by:
1. AR = Q* P*
2. BR = Q P*
3. CR = P Q*
4. DR = Q* + P*R
Answer: (B) R = Q P*
Explanation
This equation is the standard form covered by Arden's theorem for solving language equations built from regular expressions.

Arden's theorem states that when P does not contain ε, the equation R = Q + RP has the unique solution R = QP*.

Substituting QP* back into the right-hand side confirms it: Q + (QP*)P = Q + QP*P, which reduces to the same closed form, verifying the solution.

So the unique solution is R = QP*.
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Question 55
Assertion (A): We can build an object from a class containing a pure virtual function.
Reason (R): A class containing a pure virtual function is called an abstract class.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (D) A is not correct but R is correct
Explanation
A class that declares a pure virtual function becomes an abstract class, and C++ does not allow an abstract class to be instantiated directly.

Assertion A claims an object can be built directly from such a class, but this is false because the compiler blocks any attempt to instantiate a class that still has an unimplemented pure virtual function.

Reason R correctly states the definition that a class containing a pure virtual function is called an abstract class, so this part is true.

Since A is false and R is true, the correct choice is that A is not correct but R is correct.
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Question 56
For implicit type conversion, arrange the following from a lower size type to a higher size type.

A. double
B. unsigned long int
C. unsigned int
D. long double
E. long int
1. AE, C, B, A, D
2. BC, B, E, D, A
3. CE, C, A, B, D
4. DC, E, B, A, D
Answer: (D) C, E, B, A, D
Explanation
C++ usual arithmetic conversions rank integer and floating types from smaller to larger capacity, and conversions always move from a lower-ranked type to a higher-ranked one during mixed-type expressions.

The standard low-to-high ordering used for these five types is unsigned int, long int, unsigned long int, double, long double.

unsigned int (C) sits just above int in rank.
long int (E) has a wider representable range than unsigned int on this ranking.
unsigned long int (B) extends long int further with an unsigned range.
double (A) and then long double (D) are the two floating types, with long double holding the widest range and precision of all five.

This gives the order C, E, B, A, D.
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Question 57
Consider a disk queue with requests for I/O to blocks on cylinders 98, 183, 37, 122, 14, 124, 65, 67 in that order. The disk head is initially at cylinder 53.

Using SSTF scheduling, arrange the following cylinders in the order the head visits them.

A. 67
B. 37
C. 65
D. 98
E. 14
1. AA, B, C, D, E
2. BC, A, E, B, D
3. CC, A, B, E, D
4. DA, C, D, B, E
Answer: (C) C, A, B, E, D
Explanation
SSTF (Shortest Seek Time First) always moves the head to whichever pending request is nearest to its current position, not the original one.

Starting at 53, work through the queue {98, 183, 37, 122, 14, 124, 65, 67} step by step.

From 53, the nearest request is 65 (distance 12), so the head moves to 65.
From 65, the nearest remaining request is 67 (distance 2), so the head moves to 67.
From 67, the nearest remaining request is 37 (distance 30, versus 31 for 98), so the head moves to 37.
From 37, the nearest remaining request is 14 (distance 23), so the head moves to 14.
From 14, the nearest remaining request is 98 (distance 84), so the head moves to 98.
The head then continues on to 122, 124 and finally 183.

The full service order is 65, 67, 37, 14, 98, 122, 124, 183.

Picking out only the labelled cylinders in this order gives 65(C), 67(A), 37(B), 14(E), 98(D), which is C, A, B, E, D.
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Question 58
Arrange the following steps of the Natural Language Processing pipeline in the correct order.

A. Syntax Analysis
B. Morphological Analysis
C. Pragmatics
D. Semantic Analysis
1. AA, B, C, D
2. BB, A, D, C
3. CB, D, A, C
4. DA, C, B, D
Answer: (B) B, A, D, C
Explanation
A standard NLP pipeline processes text in stages that build on each other, starting from the smallest linguistic unit and moving toward context.

Morphological analysis (B) comes first and breaks words into roots and affixes.
Syntax analysis (A) follows and checks how words combine grammatically into sentence structure.
Semantic analysis (D) comes next and interprets the meaning of the syntactically valid structure.
Pragmatics (C) is the final stage and interprets meaning in context, including intent and real-world knowledge beyond the literal sentence.

This gives the order B, A, D, C.
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Question 59
Arrange the following algorithms from the most efficient to the least efficient based on their time complexity.

A. Kruskal's Algorithm
B. Breadth First Search Algorithm
C. Bellman-Ford Algorithm
D. Dijkstra's Algorithm
E. Edmonds-Karp Algorithm
1. AB, D, A, E, C
2. BE, C, A, D, B
3. CB, D, A, C, E
4. DB, C, A, E, D
Answer: (C) B, D, A, C, E
Explanation
Comparing the standard asymptotic complexities for a graph with V vertices and E edges ranks these five algorithms by growth rate.

Breadth First Search (B) runs in O(V + E), the fastest of the five since it visits each vertex and edge only once.
Dijkstra's Algorithm (D) with a binary heap runs in O(E log V), adding a logarithmic factor for priority-queue operations.
Kruskal's Algorithm (A) runs in O(E log E), comparable to O(E log V) but with a heavier constant from sorting all edges.
Bellman-Ford (C) runs in O(VE), slower because it relaxes every edge across up to V-1 passes.
Edmonds-Karp (E) runs in O(VE²), the slowest here because it repeats a full BFS-based augmenting path search many times.

Ordering from most to least efficient gives B, D, A, C, E.
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Question 60
In artificial intelligence, the term "Turing Test" is primarily associated with:
1. ATesting neural network convergence
2. BMeasuring computational efficiency of an algorithm
3. CEvaluating a machine's ability to exhibit human-like behavior
4. DVerifying correctness of an AI search algorithm
Answer: (C) Evaluating a machine's ability to exhibit human-like behavior
Explanation
The Turing Test, proposed by Alan Turing, asks whether a human evaluator can distinguish a machine's text responses from a human's responses in conversation.

It does not measure raw computational speed, algorithmic correctness, or neural network training behavior. It specifically evaluates whether a machine can exhibit behavior indistinguishable from human intelligence.

So the Turing Test is about evaluating a machine's ability to exhibit human-like behavior.
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Question 61
Match List-I (IP addresses) with List-II (Address classes).
List-I List-II
A. 190.10.75.250 I. Class E IP address
B. 230.41.10.20 II. Class B IP address
C. 242.9.20.80 III. Class A IP address
D. 127.10.125.230 IV. Class D IP address
1. AA-I, B-II, C-III, D-IV
2. BA-II, B-I, C-IV, D-III
3. CA-II, B-IV, C-I, D-III
4. DA-III, B-I, C-II, D-IV
Answer: (C) A-II, B-IV, C-I, D-III
Explanation
IP address classes are determined by the value of the first octet.

A. 190.10.75.250 falls in the 128–191 range, which is Class B, so A pairs with II.
B. 230.41.10.20 falls in the 224–239 range, which is Class D (multicast), so B pairs with IV.
C. 242.9.20.80 falls in the 240–255 range, which is Class E, so C pairs with I.
D. 127.10.125.230 falls in the 0–127 range, which is conventionally taught as Class A, so D pairs with III.

This gives A-II, B-IV, C-I, D-III.
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Question 62
What is the space complexity of the CYK algorithm for the P table, where n is the number of words in the sentence and m is the number of non-terminal symbols in the grammar?
1. AO(n³)
2. BO(nm²)
3. CO(n²m)
4. DO(n²m²)
Answer: (C) O(n²m)
Explanation
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Question 63
Assertion (A): Neural networks are capable of approximating any continuous function to an arbitrary degree of accuracy.
Reason (R): This is explained by the universal approximation theorem.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (A) Both A and R are correct and R is the correct explanation of A
Explanation
The universal approximation theorem states that a feedforward neural network with at least one hidden layer of sufficient width and a suitable non-linear activation function can approximate any continuous function on a compact domain to any desired degree of accuracy.

Assertion A is a direct statement of this capability, and Reason R correctly names the theorem that explains it.

Since R is precisely the theoretical justification for A, R is the correct explanation of A.
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Question 64
Assertion (A): L = {aⁿbⁿcⁿ : n ≥ 0} is accepted by a linear bounded automaton.
Reason (R): Linear bounded automata recognize exactly the class of context-sensitive languages.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (A) Both A and R are correct and R is the correct explanation of A
Explanation
The language L = {aⁿbⁿcⁿ : n ≥ 0} is a classic example of a context-sensitive language that is not context-free.

Linear bounded automata (LBAs) are exactly the automata that recognize the class of context-sensitive languages, so any context-sensitive language, including this one, is accepted by some LBA.

Since Assertion A is true and Reason R states the precise theorem that explains why A holds, R correctly explains A.
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Question 65

Match List-I (Quadric surfaces) with List-II (Equations).

List-I (Quadric Surface)	List-II (Equation)
A. One-sheeted hyperboloid	I. x²/a² + y²/b² = cz
B. Two-sheeted hyperboloid	II. x²/a² + y²/b² - z²/c² = 1
C. Elliptic paraboloid	III. x²/a² - y²/b² = cz
D. Hyperbolic paraboloid	IV. z²/c² - x²/a² - y²/b² = 1

AA-I, B-II, C-III, D-IV
BA-II, B-III, C-IV, D-I
CA-II, B-IV, C-I, D-III
DA-IV, B-II, C-I, D-III

Answer: (C) A-II, B-IV, C-I, D-III

Explanation

Each quadric surface has a distinct standard equation determined by its sign pattern and which variable stands alone.

A one-sheeted hyperboloid has two positive squared terms and one negative term equal to 1, giving x²/a² + y²/b² - z²/c² = 1, so A pairs with II.
A two-sheeted hyperboloid flips this pattern with one positive term and two negative terms equal to 1, giving z²/c² - x²/a² - y²/b² = 1, so B pairs with IV.
An elliptic paraboloid has two positive squared terms equal to a linear term, giving x²/a² + y²/b² = cz, so C pairs with I.
A hyperbolic paraboloid has one positive and one negative squared term equal to a linear term, giving x²/a² - y²/b² = cz, so D pairs with III.

This gives A-II, B-IV, C-I, D-III.

Question 66
Match List-I (Algorithms) with List-II (Real-world applications).
List-I List-II
A. Dijkstra's Algorithm I. GPS route finding
B. Huffman Coding II. Data compression
C. KMP String Matching III. Text editor search function
D. Ford-Fulkerson IV. Network bandwidth optimization
1. AA-II, B-IV, C-I, D-III
2. BA-I, B-II, C-III, D-IV
3. CA-IV, B-II, C-III, D-I
4. DA-IV, B-III, C-II, D-I
Answer: (B) A-I, B-II, C-III, D-IV
Explanation
Each algorithm listed is best known for one classic real-world application.

Dijkstra's Algorithm (A) finds shortest paths in a weighted graph, which is exactly how GPS navigation systems compute routes, so A pairs with I.
Huffman Coding (B) builds a variable-length prefix code that minimizes expected code length, which underlies lossless data compression, so B pairs with II.
KMP string matching (C) efficiently finds a pattern inside text without backtracking, which is the algorithm behind fast text editor search, so C pairs with III.
Ford-Fulkerson (D) computes maximum flow in a flow network, which models network bandwidth optimization problems, so D pairs with IV.

This gives A-I, B-II, C-III, D-IV.
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Question 67
Which of the following is not true about constructors?
1. AWhen an object is created and initialized at the same time, a copy constructor gets called
2. BThey do not have return types, not even void
3. CThey can be virtual
4. DThey cannot be inherited, though a derived class can call the base class constructor
Answer: (C) They can be virtual
Explanation
Constructors in C++ follow a specific set of rules, and one of the most consistently tested facts is that a constructor can never be declared virtual.

A. When an object is created and initialized at the same time, a copy constructor gets called — this describes copy-initialization from another object of the same type, which correctly describes when the copy constructor runs.

B. They do not have return types, not even void — correct, constructors never specify a return type.

C. They can be virtual — this is false. A constructor cannot be virtual because virtual dispatch relies on the object's vtable pointer, which is only fully set up once construction is already underway, so the object does not yet exist in a state that supports dynamic dispatch during its own construction.

D. They cannot be inherited, though a derived class can call the base class constructor — correct, constructors are not inherited, but a derived class constructor can invoke a base class constructor through its initializer list.

Since virtual constructors are impossible in C++, statement C is the one that is not true.
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Question 68
Arrange the following WebApp design layers from top to bottom.

A. Navigation Design
B. Aesthetic Design
C. Interface Design
D. Content Design
E. Performance Design
1. AC, B, D, A, E
2. BC, B, D, E, A
3. CC, D, B, A, E
4. DC, D, A, B, E
Answer: (A) C, B, D, A, E
Explanation
Pressman's WebApp design model organizes design activities as layers arranged from the outermost, user-facing concern to the underlying implementation concern.

Interface design (C) is the top layer, describing how the application presents itself and interacts with the user.
Aesthetic design (B) comes next, covering the look-and-feel presentation built on that interface.
Content design (D) follows, covering the actual information, media and data the WebApp delivers.
Navigation design (A) comes after content, defining how users move between content and interface elements.
Performance design (E) sits at the base, covering the underlying components and infrastructure that make everything above it work.

This gives the order C, B, D, A, E.
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Question 69
Assertion (A): Instruction pipelining improves CPU throughput.
Reason (R): Pipelining decreases the execution time of each individual instruction.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (C) A is correct but R is not correct
Explanation
Pipelining overlaps the execution stages of multiple instructions so that a new instruction can start before the previous one finishes, which increases the number of instructions completed per unit time. This is throughput.

Assertion A is correct because this overlap is exactly why pipelining improves throughput.

Reason R claims pipelining decreases the execution time of each individual instruction, but this is false. Each instruction still takes roughly the same number of pipeline stages to complete, sometimes even more due to hazards and stalls. Pipelining does not make any single instruction finish faster — it only allows more instructions to be in flight at the same time.

Since A is true and R is false, the correct choice is that A is correct but R is not correct.
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Question 70

Match List-I (Horn-clause terms) with List-II.

List-I	List-II
A. Definite clause	I. Forward chaining
B. Horn clause	II. At most one is positive literal
C. Goal clause	III. No positive literals
D. Inference with Horn clause	IV. Exactly one is positive literal

AA-II, B-IV, C-III, D-I
BA-IV, B-III, C-I, D-III
CA-II, B-III, C-I, D-IV
DA-IV, B-II, C-III, D-I

Answer: (D) A-IV, B-II, C-III, D-I

Explanation

Horn-clause logic classifies clauses by how many positive literals they contain.

A definite clause has exactly one positive literal, so A pairs with IV.
A Horn clause has at most one positive literal, a broader category that includes definite clauses, so B pairs with II.
A goal clause is a Horn clause with no positive literals at all, so C pairs with III.
The standard inference procedure used with Horn clauses is forward chaining, so D pairs with I.

This gives A-IV, B-II, C-III, D-I.

Question 71

Match List-I (Optimization methods) with List-II (Problem types).

List-I	List-II
A. Branch and bound method	I. Integer programming problem
B. North-west corner rule	II. Quadratic programming problem
C. Lagrange multiplier method	III. Transportation problem
D. Wolfe's modified method	IV. Non-linear programming problem

AA-III, B-II, C-IV, D-I
BA-I, B-III, C-IV, D-II
CA-II, B-I, C-IV, D-III
DA-IV, B-I, C-II, D-III

Answer: (B) A-I, B-III, C-IV, D-II

Explanation

Each optimization technique listed is the standard method for one particular class of problem.

The branch and bound method (A) systematically explores and prunes a search tree of candidate integer solutions, making it the classic method for integer programming problems, so A pairs with I.
The north-west corner rule (B) constructs an initial basic feasible solution for a transportation problem, so B pairs with III.
The Lagrange multiplier method (C) finds stationary points subject to equality constraints and is the standard technique for non-linear programming problems, so C pairs with IV.
Wolfe's modified simplex method (D) extends the simplex method to handle a quadratic objective function, making it the standard method for quadratic programming problems, so D pairs with II.

This gives A-I, B-III, C-IV, D-II.

Question 72
Arrange the following steps in the order of execution while proving that a language is non-regular using the Pumping Lemma.

A. Split the string into xyz satisfying the pumping lemma conditions
B. Assume the language is regular
C. Choose a string of length greater than or equal to the pumping length
D. Show a contradiction when pumping y
1. AB, C, D, A
2. BC, D, A, B
3. CB, C, A, D
4. DA, B, C, D
Answer: (C) B, C, A, D
Explanation
Proving a language is not regular with the pumping lemma follows a fixed logical sequence, since each step depends on the one before it.

First, assume the language is regular (B), since the proof works by contradiction against this assumption.
Next, choose a string in the language of length at least the pumping length guaranteed by the lemma (C).
Then split that string into parts x, y, z satisfying the pumping lemma's length and repetition conditions (A).
Finally, show that pumping the y portion produces a string that is not in the language, contradicting the assumption (D).

This gives the order B, C, A, D.
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Question 73
Match List-I (Boolean algebra laws) with List-II.
List-I List-II
A. Domination law I. x + 1 = 1
B. Associative law II. x + (yz) = (x + y)(x + z)
C. Distributive law III. x + xy = x
D. Absorption law IV. x + (y + z) = (x + y) + z
1. AA-I, B-II, C-IV, D-III
2. BA-I, B-IV, C-II, D-III
3. CA-III, B-IV, C-II, D-I
4. DA-III, B-II, C-IV, D-I
Answer: (B) A-I, B-IV, C-II, D-III
Explanation
Each Boolean algebra law has a distinct standard form.

The domination law states that x + 1 = 1, since ORing anything with 1 always yields 1, so A pairs with I.
The associative law states that x + (y + z) = (x + y) + z, since grouping does not affect the result of repeated OR operations, so B pairs with IV.
The distributive law states that x + (yz) = (x + y)(x + z), since OR distributes over AND, so C pairs with II.
The absorption law states that x + xy = x, since the xy term is absorbed once x is already true, so D pairs with III.

This gives A-I, B-IV, C-II, D-III.
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Question 74
Assertion (A): Thrashing occurs when a process spends more time on paging than executing.
Reason (R): It happens when the degree of multiprogramming is too low and frames are insufficient.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (C) A is correct but R is not correct
Explanation
Thrashing occurs when a system spends most of its time swapping pages in and out rather than executing processes, so Assertion A correctly describes the symptom.

Reason R claims this happens when the degree of multiprogramming is too low, but this is the opposite of the actual cause. Thrashing occurs when the degree of multiprogramming is increased too high, so that the combined working sets of all active processes exceed the number of available frames, forcing constant page faults.

Since A is true and R states an incorrect cause, the correct choice is that A is correct but R is not correct.
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Question 75
A normalized matrix maps a 3D window specified by coordinates (1, 1, 1) and (4, 5, 6) onto a viewport specified by coordinates (2, 4, 2) and (8, 10, 8). Which of the following matrices correctly represents this transformation?
1. A2 0 0 0
  0 3/2 0 5/2
  0 0 6/5 4/5
  0 0 0 1
2. B2 0 0 0
  0 3 0 5
  0 0 6 4
  0 0 0 1
3. C2 1 0 1
  1 3 4 5
  0 0 2 4
  0 1 0 0
4. D1 0 0 1
  2 3 0 5
  1 0 4 5
  0 0 0 1
Answer: (A) 2 0 0 0
0 3/2 0 5/2
0 0 6/5 4/5
0 0 0 1
Explanation
A window-to-viewport transform scales each axis to match the viewport's extent, then translates so the window's minimum corner (1, 1, 1) lands on the viewport's minimum corner (2, 4, 2).

Scale factors, viewport extent divided by window extent on each axis:
sx = (8-2)/(4-1) = 2
sy = (10-4)/(5-1) = 3/2
sz = (8-2)/(6-1) = 6/5

Translation offsets, solved so (1,1,1) maps to (2,4,2):
tx = 2 - sx(1) = 0
ty = 4 - sy(1) = 5/2
tz = 2 - sz(1) = 4/5

Substituting these into the scale-then-translate matrix gives:

2 0 0 0
0 3/2 0 5/2
0 0 6/5 4/5
0 0 0 1

This matches option A exactly.

As a check, the window's max corner (4, 5, 6) should land on the viewport's max corner (8, 10, 8): 2(4) = 8, 3/2(5) + 5/2 = 10, 6/5(6) + 4/5 = 8, confirming the matrix is correct.
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Question 76

Match List-I (OS and distributed-systems terms) with List-II.

List-I	List-II
A. System Call	I. A process requests a function that is located in a different computer system.
B. Preemption	II. A request that a program makes to the operating system.
C. Real time application	III. The forced deallocation of the CPU from a process.
D. Remote procedure call	IV. A program that responds to activities in an external system within a maximum time determined by that external system.

AA-I, B-II, C-III, D-IV
BA-II, B-III, C-IV, D-I
CA-IV, B-III, C-II, D-I
DA-II, B-III, C-I, D-IV

Answer: (B) A-II, B-III, C-IV, D-I

Explanation

A system call is a request that a program makes to the operating system, so A pairs with II.
Preemption is the forced deallocation of the CPU from a running process, so B pairs with III.
A real-time application must respond to activities in an external system within a maximum time set by that external system, so C pairs with IV.
A remote procedure call lets a process invoke a function located on a different computer system, so D pairs with I.

A-II, B-III, C-IV, D-I is therefore the correct match.

Question 77
If Kₙ is complete simple graph of n vertices and Kₘ,ₙ is complete bipartite graph then arrange the following non-planner graph in ascending order on the bases of the minimum number of crossings.

A. K₆
B. K₇
C. K₄,₄
D. K₅,₅
1. AA, B, C, D
2. BC, D, A, B
3. CA, C, B, D
4. DB, D, A, C
Answer: (C) A, C, B, D
Explanation
The crossing number cr(G) of a graph is the minimum number of edge crossings achievable in any drawing of that graph, and for small complete and complete-bipartite graphs it is known exactly (the Zarankiewicz result, proven for these cases).

cr(K₆) = ¼ × ⌊6/2⌋ × ⌊5/2⌋ × ⌊4/2⌋ × ⌊3/2⌋ = ¼ × 3 × 2 × 2 × 1 = 3
cr(K₇) = ¼ × ⌊7/2⌋ × ⌊6/2⌋ × ⌊5/2⌋ × ⌊4/2⌋ = ¼ × 3 × 3 × 2 × 2 = 9
cr(K₄,₄) = ⌊4/2⌋ × ⌊3/2⌋ × ⌊4/2⌋ × ⌊3/2⌋ = 2 × 1 × 2 × 1 = 4
cr(K₅,₅) = ⌊5/2⌋ × ⌊4/2⌋ × ⌊5/2⌋ × ⌊4/2⌋ = 2 × 2 × 2 × 2 = 16

Ordering these values from smallest to largest gives K₆ (3) < K₄,₄ (4) < K₇ (9) < K₅,₅ (16), which is the sequence A, C, B, D.
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Question 78
Assertion (A): OLAP systems are optimized for complex queries and analysis.
Reason (R): OLTP systems are designed for transaction processing with large read-only queries.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (C) A is correct but R is not correct
Explanation
OLAP (Online Analytical Processing) systems are indeed built for complex, multidimensional analytical queries, so Assertion A is correct.

Reason R is not correct as stated. OLTP (Online Transaction Processing) systems are designed to handle a high volume of short, simple read and write transactions such as order entry or fund transfers, not large read-only queries. Large, read-heavy analytical queries over historical data are the hallmark of OLAP and data-warehouse workloads, not OLTP.

Because A is true and R misdescribes OLTP, the correct choice is that A is correct but R is not correct.
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Question 79
Which of the following is correct about Real time operating system?

A. It permits creation of multiple processes within an application.
B. It does not permit priorities to be assigned to processes.
C. It permits a programmer to define interrupts and interrupt processing routines.
D. It does not use priority-driven or deadline-oriented scheduling.
E. It provides fault tolerance and graceful degradation capabilities.
1. AA, B & D Only
2. BB, C & D Only
3. CA, C & E Only
4. DB, D & E Only
Answer: (C) A, C & E Only
Explanation
A real-time operating system (RTOS) is built to guarantee timely, predictable responses to events, and checking each statement against that goal resolves the set.

A. It permits creation of multiple processes within an application. — correct, an RTOS supports multitasking so an application can split its work across several cooperating processes or tasks.

B. It does not permit priorities to be assigned to processes. — wrong, priority assignment is central to an RTOS since tasks with tighter deadlines must be able to preempt less urgent ones.

C. It permits a programmer to define interrupts and interrupt processing routines. — correct, RTOS kernels expose interrupt service routine hooks so time-critical events can be handled directly.

D. It does not use priority-driven or deadline-oriented scheduling. — wrong, priority-driven and deadline-oriented scheduling (for example rate-monotonic or earliest-deadline-first) are the defining scheduling techniques of an RTOS.

E. It provides fault tolerance and graceful degradation capabilities. — correct, RTOS designs for critical systems build in fault tolerance so the system degrades gracefully instead of failing outright.

Only A, C and E hold, matching A, C & E Only.
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Question 80
Which of the following are limitations of Greedy algorithms?

A. They always fail for NP hard problem.
B. They may not give the optimal solution for all problems.
C. They are faster than dynamic programming in most cases.
D. They make local choices without looking ahead.
1. AA, C Only
2. BB, C Only
3. CA, B, C, D
4. DB, D Only
Answer: (D) B, D Only
Explanation
A limitation is a genuine weakness of a technique, which rules out any statement that instead describes an advantage or makes an overstated claim.

A. They always fail for NP hard problem. — wrong, greedy algorithms do not always fail on NP-hard problems, they often give fast approximate solutions and in special cases even exact optimal ones.

B. They may not give the optimal solution for all problems. — correct, greedy algorithms optimize locally at each step, and that local view can miss the true global optimum, which is exactly why they are not always correct.

C. They are faster than dynamic programming in most cases. — wrong as a limitation, speed relative to dynamic programming is an advantage of greedy methods, not a weakness.

D. They make local choices without looking ahead. — correct, this short-sightedness is the structural reason greedy solutions can end up suboptimal.

The genuine limitations are B and D only.
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Question 81
In relation to Index structures for files which of the following are true?

A. An index is an integral part of the data file.
B. An index can be created in conjunction with virtually any primary record organization.
C. An index cannot be created dynamically.
D. An index can be used to enforce a key constraint on an attribute.
E. An index can be created dynamically.
1. AA, B, C Only
2. BB, C, D Only
3. CB, D, E Only
4. DA, C, E Only
Answer: (C) B, D, E Only
Explanation
An index is a separate auxiliary structure built on top of a data file, and checking each statement against that fact resolves the set.

A. An index is an integral part of the data file. — wrong, an index is a distinct structure stored apart from the data file that only points into it.

B. An index can be created in conjunction with virtually any primary record organization. — correct, indexes can be layered on top of sequential, heap, or hashed file organizations alike.

C. An index cannot be created dynamically. — wrong, and directly contradicted by statement E.

D. An index can be used to enforce a key constraint on an attribute. — correct, a unique index is a standard mechanism for enforcing key uniqueness on an attribute.

E. An index can be created dynamically. — correct, indexes can be built or dropped at runtime as access patterns change.

Since C and E directly contradict each other, only one can be true, and standard index theory confirms E is the correct one. The true statements are B, D and E only.
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Question 82
Arrange the following applet life cycle methods in sequence.

A. start()
B. paint()
C. init()
D. stop()
1. AA, B, C, D
2. BC, A, B, D
3. CA, C, B, D
4. DB, A, C, D
Answer: (B) C, A, B, D
Explanation
An applet's lifecycle begins when it is loaded and ends when the browser leaves or reloads the page, and the JVM calls its lifecycle methods in a fixed order.

init() runs first, setting up the applet's initial state when it is loaded.
start() runs next, beginning or resuming the applet's execution right after initialization.
paint() then renders the applet's visual output on screen.
stop() runs last, when the applet becomes inactive, such as when the browser navigates away from the page.

The resulting order is init() → start() → paint() → stop(), which is C, A, B, D.
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Question 83
Assertion (A): In supervised learning, the model is trained using labelled data.
Reason (R): Supervised learning algorithms find patterns in data to predict outcomes without any prior knowledge of the correct output.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (C) A is correct but R is not correct
Explanation
Supervised learning trains a model on a dataset where every input is paired with a known, correct output label, so Assertion A is correct.

Reason R is not correct. It describes finding patterns without any prior knowledge of the correct output, which is the definition of unsupervised learning, not supervised learning. Supervised learning is defined by the opposite property, since it explicitly uses the correct output labels during training to guide the model toward accurate predictions.

Since A is true and R actually describes a different learning paradigm, the correct choice is that A is correct but R is not correct.
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Question 84
Assertion (A): Asymmetric key cryptography is significantly slower than symmetric key cryptography.
Reason (R): Asymmetric key cryptography is based on complex mathematical structure to secure data.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (A) Both A and R are correct and R is the correct explanation of A
Explanation
Asymmetric (public-key) cryptography relies on computationally expensive operations such as large-integer modular exponentiation or elliptic-curve point multiplication, which makes it markedly slower than symmetric algorithms that use simple substitution and permutation on fixed-size blocks. Assertion A is therefore correct.

Reason R is also correct. Asymmetric schemes such as RSA and elliptic-curve cryptography derive their security from hard mathematical problems, such as integer factorization or the discrete logarithm problem, and these same complex structures are what make the underlying operations computationally heavy.

Because the heavy mathematical machinery described in R is precisely why asymmetric cryptography runs slower, R correctly explains A.
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Question 85
Match List-I (Network protocols) with List-II (OSI/TCP-IP layers).
List-I List-II
A. SMTP I. Application layer
B. BGP II. Transport layer
C. TCP III. Data Link Layer
D. PPP IV. Network Layer
1. AA-I, B-II, C-III, D-IV
2. BA-II, B-I, C-IV, D-III
3. CA-IV, B-II, C-III, D-I
4. DA-I, B-IV, C-II, D-III
Answer: (D) A-I, B-IV, C-II, D-III
Explanation
SMTP operates at the Application layer since it defines how mail clients and servers exchange messages, so A pairs with I.
TCP is a core Transport-layer protocol responsible for reliable end-to-end delivery, so C pairs with II.
PPP is a Data Link-layer protocol used to establish point-to-point serial connections, so D pairs with III.
BGP is the protocol routers use to exchange reachability information between autonomous systems, and it is classified here as the remaining Network-layer routing protocol, so B pairs with IV.

A-I, B-IV, C-II, D-III is the only option that correctly places SMTP at Application, TCP at Transport, and PPP at Data Link, making it the internally consistent match.
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Question 86
Which of the following statements are correct regarding task elaboration in interface design?

A. Alpha testing is done at the developer's site in a controlled environment.
B. Beta testing is done at the end-user site with developer presence.
C. Acceptance testing may be very formal and last for weeks.
D. Alpha testing always replaces Beta testing.
1. AA, B, C Only
2. BA, B, D Only
3. CB, C, D only
4. DA, B, C, D
Answer: (A) A, B, C Only
Explanation
Alpha and Beta testing are distinguished by where they happen and whether the developer is present, and checking each statement against that standard software engineering picture is what resolves this item.

A. Alpha testing is done at the developer's site in a controlled environment. — correct, Alpha testing is conducted at the developer's site with the developer able to control and monitor the environment.

B. Beta testing is done at the end-user site with developer presence. — as worded, this is not correct. Beta testing is conducted at one or more end-user sites, but specifically without the developer present, in an environment the developer cannot control, which is what makes it a realistic field test.

C. Acceptance testing may be very formal and last for weeks. — correct, acceptance testing can range from a few informal test cases to a planned, formal series of tests run over an extended period.

D. Alpha testing always replaces Beta testing. — wrong, Alpha and Beta testing serve different purposes, and rigorous engineering practice normally conducts both rather than treating one as a substitute for the other.

Taken literally, only A and C are correct statements on their own, but among the given combinations, A, B, C Only is the closest match because it contains both genuinely correct statements and adds only one incorrect one, whereas the remaining options also misclassify D or drop a correct statement.
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Question 87

Match List-I (Coupling types) with List-II.

List-I	List-II
A. Content coupling	I. Modifies internal data of another component
B. Control Coupling	II. Pass control flag to influence logic
C. Stamp Coupling	III. Uses entire structure/class as argument
D. Data Coupling	IV. Passes long list of data arguments.

AA-I, B-II, C-IV, D-III
BA-II, B-IV, C-I, D-III
CA-I, B-III, C-IV, D-II
DA-I, B-II, C-III, D-IV

Answer: (D) A-I, B-II, C-III, D-IV

Explanation

Coupling measures how tightly one module depends on another, and each type here has a distinct, well-defined signature.

Content coupling is the tightest form, where one module directly modifies another module's internal data, so A pairs with I.
Control coupling occurs when one module passes a control flag that dictates the logic path taken inside another, so B pairs with II.
Stamp coupling occurs when an entire data structure or class is passed as an argument even though only part of it is needed, so C pairs with III.
Data coupling, the loosest and most desirable form, is left to pair with IV, passing data arguments between modules.

A-I, B-II, C-III, D-IV is therefore the correct match.

Question 88
What does a superblock object represent in the virtual file system of a Linux OS?
1. Aan open file
2. Ban entire file system
3. Cindividual directory entry
4. Dan individual file
Answer: (B) an entire file system
Explanation
The Linux Virtual File System (VFS) uses four principal object types, and the superblock is the one that represents a whole mounted file system, holding metadata such as its type, size, and status.

An open file is represented by a file object, an individual file's metadata is held in an inode object, and a directory entry is represented by a dentry object, each of which is distinct from the superblock.

Because the superblock stores information describing an entire mounted file system, an entire file system is the correct answer.
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Question 89
Consider a B⁺ tree in which the maximum number of keys in a node is 5. What is the minimum number of keys in any non-root node?
1. A1
2. B2
3. C3
4. D4
Answer: (B) 2
Explanation
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Question 90
Assertion (A): Depth first search can be used to perform a topological sort of a directed acyclic graph.
Reason (R): A topological sort of a directed acyclic graph with vertex set V and edge set E is a linear ordering of its vertices such that if the graph contains an edge from u to v, then u appears before v in the ordering.
1. ABoth A and R are correct and R is the correct explanation of A
2. BBoth A and R are correct but R is NOT the correct explanation of A
3. CA is correct but R is not correct
4. DA is not correct but R is correct
Answer: (A) Both A and R are correct and R is the correct explanation of A
Explanation
Running depth-first search on a directed acyclic graph and listing vertices in reverse order of their finishing times produces a valid topological sort, because a vertex finishes only after every vertex reachable from it has already finished, which places it before its successors once the order is reversed. Assertion A is correct.

Reason R correctly states the defining property of a topological sort, that for every edge (u, v), vertex u must appear before v in the ordering.

R is the correct explanation of A because it is precisely this ordering property, guaranteed by DFS finishing times on an acyclic graph, that the DFS-based method is exploited to produce.
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Question 91
Based on the passage, including the 10% contingency margin, what is the project cost?
A project is planned for an enterprise accounting system with an estimated size of 60 KLOC. The organization applies Basic COCOMO in Semi-Detached mode, using these parameters:
- Effort = 3.0 x (KLOC)^1.0 person-months
- Development time = 2.5 x (Effort)^0.5 months
- Average cost per person-month (PM) = $10,000
- A contingency margin of 10% on effort is mandatory
- The client demands a detailed schedule if staff is capped at 15 engineers
1. A$ 1.80 M
2. B$ 1.98 M
3. C$ 2.4 M
4. D$ 2.36 M
Answer: (B) $ 1.98 M
Explanation
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Question 92
Based on the passage, what is the estimated development time in months?
A project is planned for an enterprise accounting system with an estimated size of 60 KLOC. The organization applies Basic COCOMO in Semi-Detached mode, using these parameters:
- Effort = 3.0 x (KLOC)^1.0 person-months
- Development time = 2.5 x (Effort)^0.5 months
- Average cost per person-month (PM) = $10,000
- A contingency margin of 10% on effort is mandatory
- The client demands a detailed schedule if staff is capped at 15 engineers
1. A28 Months
2. B26 Months
3. C34 Months
4. D12 Months
Answer: (C) 34 Months
Explanation
COCOMO's schedule equation is applied to the nominal effort estimate, the 180 PM figure computed before the contingency margin is added, since the contingency margin covers extra staffing risk rather than changing the underlying size-based schedule model.

Development time = 2.5 × (Effort)^0.5
Development time = 2.5 × (180)^0.5 = 2.5 × 13.42 ≈ 33.5 months

Rounding 33.5 to the nearest whole month gives 34 months, the closest listed option. Using the contingency-adjusted effort of 198 PM instead gives 2.5 × (198)^0.5 ≈ 35.2 months, which still rounds closest to 34 among the given choices, so 34 months is the answer under either basis.
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Question 93
Based on the passage, what is the implied average staffing level?
A project is planned for an enterprise accounting system with an estimated size of 60 KLOC. The organization applies Basic COCOMO in Semi-Detached mode, using these parameters:
- Effort = 3.0 x (KLOC)^1.0 person-months
- Development time = 2.5 x (Effort)^0.5 months
- Average cost per person-month (PM) = $10,000
- A contingency margin of 10% on effort is mandatory
- The client demands a detailed schedule if staff is capped at 15 engineers
1. A8 Engineers
2. B5 Engineers
3. C10 Engineers
4. D12 Engineers
Answer: (B) 5 Engineers
Explanation
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Question 94
Based on the passage, if the staff is capped at 15 engineers, what is the revised calendar duration?
A project is planned for an enterprise accounting system with an estimated size of 60 KLOC. The organization applies Basic COCOMO in Semi-Detached mode, using these parameters:
- Effort = 3.0 x (KLOC)^1.0 person-months
- Development time = 2.5 x (Effort)^0.5 months
- Average cost per person-month (PM) = $10,000
- A contingency margin of 10% on effort is mandatory
- The client demands a detailed schedule if staff is capped at 15 engineers
1. A20.0 Months
2. B18.4 Months
3. C13.2 Months
4. D22.0 Months
Answer: (C) 13.2 Months
Explanation
Capping the team size means the full committed effort, including the mandatory 10% contingency margin the organization must actually staff for, has to be delivered by a fixed number of engineers, so this calculation uses the contingency-inclusive effort of 198 PM rather than the nominal 180 PM used for the COCOMO schedule equation above.

Revised duration = Effort / Number of engineers
Revised duration = 198 PM / 15 = 13.2 months

13.2 months matches one of the listed options exactly, confirming that the contingency-inclusive effort is the correct basis for this figure. Using the nominal 180 PM effort instead gives 180 / 15 = 12.0 months, which is not one of the listed options, reinforcing that the capped-staff duration is meant to be read off the full committed effort.
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Question 95
Based on the passage, what is the initial effort, in person-months (PM), before adding the contingency margin?
A project is planned for an enterprise accounting system with an estimated size of 60 KLOC. The organization applies Basic COCOMO in Semi-Detached mode, using these parameters:
- Effort = 3.0 x (KLOC)^1.0 person-months
- Development time = 2.5 x (Effort)^0.5 months
- Average cost per person-month (PM) = $10,000
- A contingency margin of 10% on effort is mandatory
- The client demands a detailed schedule if staff is capped at 15 engineers
1. A240 PM
2. B260 PM
3. C180 PM
4. D200 PM
Answer: (C) 180 PM
Explanation
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Question 96
Based on the passage, how much CPU time is consumed by programmed I/O?
A system transfers data blocks of 2000 bytes.

With programmed I/O, each byte transfer takes 6 CPU cycles.
With direct memory access (DMA), the block transfer takes 3000 cycles in total for the DMA controller, in addition to a CPU setup overhead of 200 cycles.
The CPU clock speed is 1 GHz, so one clock cycle equals 1 nanosecond.
1. A12 μs
2. B20 μs
3. C24 μs
4. D30 μs
Answer: (A) 12 μs
Explanation
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Question 97
Based on the passage, which approach is more CPU-efficient for large block transfers?
A system transfers data blocks of 2000 bytes.

With programmed I/O, each byte transfer takes 6 CPU cycles.
With direct memory access (DMA), the block transfer takes 3000 cycles in total for the DMA controller, in addition to a CPU setup overhead of 200 cycles.
The CPU clock speed is 1 GHz, so one clock cycle equals 1 nanosecond.
1. AProgrammed I/O
2. BDMA
3. CBoth programmed I/O and DMA
4. DDepends on block size
Answer: (B) DMA
Explanation
DMA is designed to free the CPU from handling each byte of a transfer individually. The DMA controller moves the entire block on its own, and the CPU only pays a brief 200-cycle setup cost, compared with 12,000 cycles of CPU involvement under programmed I/O for the same 2000-byte block.

Because the CPU is occupied for far fewer cycles with DMA, DMA is the more CPU-efficient approach for large block transfers.
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Question 98
Based on the passage, how many CPU cycles are spent with DMA?
A system transfers data blocks of 2000 bytes.

With programmed I/O, each byte transfer takes 6 CPU cycles.
With direct memory access (DMA), the block transfer takes 3000 cycles in total for the DMA controller, in addition to a CPU setup overhead of 200 cycles.
The CPU clock speed is 1 GHz, so one clock cycle equals 1 nanosecond.
1. A200 Cycles
2. B3000 Cycles
3. C3200 Cycles
4. D4000 Cycles
Answer: (A) 200 Cycles
Explanation
With DMA, the CPU only performs the initial setup work and then hands off the actual byte-by-byte transfer to the DMA controller entirely. The 3000 cycles described for the block transfer are the DMA controller's own bus cycles for moving the data, not CPU cycles, so they do not count against the CPU's workload.

CPU cycles spent with DMA = 200 cycles (setup overhead only)

The CPU is occupied for only the 200-cycle setup overhead while DMA handles the transfer.
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Question 99
Based on the passage, how many CPU cycles are spent using programmed I/O?
A system transfers data blocks of 2000 bytes.

With programmed I/O, each byte transfer takes 6 CPU cycles.
With direct memory access (DMA), the block transfer takes 3000 cycles in total for the DMA controller, in addition to a CPU setup overhead of 200 cycles.
The CPU clock speed is 1 GHz, so one clock cycle equals 1 nanosecond.
1. A2000
2. B6000
3. C12000
4. D20000
Answer: (C) 12000
Explanation
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Question 100
Based on the passage, how much CPU time is consumed by DMA overhead?
A system transfers data blocks of 2000 bytes.

With programmed I/O, each byte transfer takes 6 CPU cycles.
With direct memory access (DMA), the block transfer takes 3000 cycles in total for the DMA controller, in addition to a CPU setup overhead of 200 cycles.
The CPU clock speed is 1 GHz, so one clock cycle equals 1 nanosecond.
1. A0.2 μs
2. B2.0 μs
3. C3.0 μs
4. D12.0 μs
Answer: (A) 0.2 μs
Explanation
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Every Computer Science & Applications PYQ on one topic

Artificial Intelligence 133 Computer System Architecture 133 Data Communication and Computer Networks 170 Data Structures and Algorithms 143 Database Management Systems 137 Discrete Structures and Optimization 173 Programming Languages and Computer Graphics 141 Software Engineering 146 System Software and Operating System 156 Theory of Computation and Compilers 167

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All 2026 sessions All Computer Science & Applications PYQs Computer Science & Applications topic-wise practice Paper 1 topic-wise practice

Practice Paper 1 from other subjects

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Paper 1 · English Literature sittings Paper 1 · Economics sittings Paper 1 · Management sittings Paper 1 · Commerce sittings

List-I: Source of Knowledge	List-II: Accepted by
A. Yogaja Pratyaksa	I. Vedantins
B. Kevalajnana	II. Carvaka
C. Only Pratyaksa	III. Jaina
D. Aparoksanubhuti	IV. Nyaya

List-I	List-II
A. Keyboard	I. Memory
B. Monitor	II. Processor
C. CPU	III. Output device
D. ROM	IV. Input device

Year	A	B	C	D	E
2021	29	45	26	15	32
2022	32	52	58	26	35
2023	28	54	92	24	39
2024	26	58	86	27	34
2025	23	63	82	28	34

List-I	List-II
A. V-Model	I. Trigger transition
B. Incremental process model	II. Classic life cycle
C. Spiral Model	III. Risk Driven
D. Concurrent Model	IV. Core product

LIST-I (Input)	LIST-II (Compiler Phase)
A. Syntax Tree	I. Code Generator
B. Character Stream	II. Syntax Analyzer
C. Intermediate Representation	III. Semantic Analyzer
D. Token Stream	IV. Lexical Analyzer

LIST-I	LIST-II
A. Class diagram	I. subset
B. Package diagram	II. relationship
C. Object diagram	III. dependencies
D. Component diagram	IV. static structure

LIST-I	LIST-II
A. Expert systems	I. Medical Diagnosis
B. NLP	II. Text Summarization
C. Computer Vision	III. Image Classification
D. ML	IV. Predictive Analysis

List-I	List-II
A. 190.10.75.250	I. Class E IP address
B. 230.41.10.20	II. Class B IP address
C. 242.9.20.80	III. Class A IP address
D. 127.10.125.230	IV. Class D IP address

List-I	List-II
A. Dijkstra's Algorithm	I. GPS route finding
B. Huffman Coding	II. Data compression
C. KMP String Matching	III. Text editor search function
D. Ford-Fulkerson	IV. Network bandwidth optimization

List-I	List-II
A. Domination law	I. x + 1 = 1
B. Associative law	II. x + (yz) = (x + y)(x + z)
C. Distributive law	III. x + xy = x
D. Absorption law	IV. x + (y + z) = (x + y) + z

Process	Burst Time (ms)	Priority
P1	10	3
P2	1	1
P3	2	4
P4	1	5
P5	5	2

List-I	List-II
A. SMTP	I. Application layer
B. BGP	II. Transport layer
C. TCP	III. Data Link Layer
D. PPP	IV. Network Layer

UGC NET Computer Science & Applications January 2026 Question PaperPaper 1 & Paper 2 with Answer Key

UGC NET January 2026 at a glance

General Aptitude

Computer Science & Applications

Paper 1 — General Aptitude

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Question 29

Question 30

Question 31

Question 32

Question 33

Question 34

Question 35

Question 36

Question 37

Question 38

Question 39

Question 40

Question 41

Question 42

Question 43

Question 44

Question 45

Question 46

Question 47

Question 48

Question 49

Question 50

Paper 2 — Computer Science & Applications

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24