UGC NET Computer Science Paper III June 2015 Part 1

Introduction

UGC NET or NTA-UGC-NET is the examination for determining the eligibility for the assistant professor and/or Junior Research Fellowship award in Indian universities and colleges. National Testing Agency examines on behalf of the University Grants Commission. In this article, we will cover questions 1 to 25. We have covered questions 26 to 50 in the second part of UGC NET Nov 2015 PAPER-III Part-2, and 51 to 75 in the third part of UGC NET Nov 2015 PAPER-III Part-3.

Questions

1. For the 8-bit word 00111001, the check bits stored with it would be 0111. Suppose when the word is read from memory, the check bits are calculated to be 1101. What is the data word that was read from memory?

(A) 10011001 (B) 00011001

Answer: B

Explanation: 00111001 for the 8-bit word, 0111 for the check bits

There are four bits, and their positions are

2^0 = 1; p1 = 1

2^1 = 2; p2 = 1

2^2 = 4; p4 = 1

2^3 = 8; p8 = 0.

The encoded text will be: d12 = 0; d11 = 0; d10 = 1; d9 = 1; d8 = 1; d7 = 1; d6 = 0; d5 = 0; d4 = 1; d3 = 1; d2 = 1; d1 = 1.

1101 is the number of bits to check at the opposite end.

1010 is the result of XOR(0111, 1101). Change the tenth bit.

d12 = 0; d11 = 0; d10 = 0; d9 = 1; d8 = 1; d7 = 1; d6 = 0 d5 = 0; d4 = 1; d3 = 1; d2 = 1; d1 = 1;

00011001 will be the data word read from memory.

2. Consider a 32 - bit microprocessor, with a 16 - bit external data bus, driven by an 8 MHz input clock. Assume that the microprocessor has a bus cycle whose minimum duration equals four input clock cycles. What is the maximum data transfer rate for this microprocessor?

(A) 8x106 bytes/sec (B) 4x106 bytes/sec

Answer: B

Explanation: Data sent per bus cycle = 2B in a 32-bit CPU with a 16-bit external data bus.

The bus clock is set to 8 MHz.

4 clock cycles is the minimum bus cycle duration.

8 MHz / 4 = 2 M/sec maximum bus cycle rate

2 M * 2 = 4 * 106 B/sec data transmission rate = bus cycle rate * data per bus cycle

3. The RST 7 instruction in 8085 microprocessor is equivalent to:

(A) CALL 0010 H (B) CALL 0034 H

Answer: C

Explanation: The interrupt and vector addresses for the 8085 microprocessor are as follows:

RST 7 CALL 0038 H

RST 6 CALL 0030 H

RST 4 CALL 0028 H

RST 3 CALL 0020 H

RST 2 CALL 0018 H

RST 1 CALL 0010 H

RST 0 CALL 0008 H

4. The equivalent hexadecimal notation for octal number 2550276 is:

(A) FADED (B) AEOBE

Answer: C

Explanation:

2550276 = 010 101 101 000 010 111 110

0 1010 = A 1101= D 0000= 0 1011= B 1110= E

5. The CPU of a system having 1 MIPS execution rate needs 4 machine cycles on average for executing an instruction. Fifty percent of the cycles use the memory bus. A memory read/write employs one machine cycle. For the execution of the programs, the system utilizes 90 percent of the CPU time. For block data transfer, an IO device is attached to the system while the CPU executes the background programs continuously. What is the maximum IO data transfer rate if the programmed IO data transfer technique is used?

(A) 500 Kbytes/sec (B) 2.2 Mbytes/sec

Answer: D

Explanation: Given, the CPU speed of 10 instructions/sec

1 CPU instruction 4 machine cycles

CPU utilization = 90%

Programmed 10: each byte transfer requires 4 cycles (instructions), in status, check status branch, R/W data memory maximum data transfer rate,

CPU Speed/4 = 10^6/4 = 250 kbytes/sec

6. The number of flip-flops required to design a modulo-272 counter is:

(A) 8 (B) 9

Answer: B

Explanation: If we have "n" flip-flops then the total number of combinations possible is 2".

So the total number of states possible is 2 " i.e. MOD 2".

Here 2" = 272

So, n = log2(272)

n=9

7. Let E1 and E2 be two entities in the E-R diagram with simple single-valued attributes. R1 and R2 are two relationships between E1 and E2 where R1 is one-many and R2 is many-many. R1 and R2 do not have any attributes of their own. How many a minimum number of tables are required to represent this situation in the Relational Model?

(A) 4 (B) 3

Answer: B

Explanation: Three tables are required

For R1, you only need the tables for E1 and E2 because it's a one-to-many situation. E1 will be referenced via a foreign key in E2. As a result, numerous E2 tuples will refer to the same E1 tuple.

For R2, you'll need the tables E1 and E2, as well as a third table for the mappings, which can't be seen in either E1 or E2. There will be references to both E1 and E2 in the third table, say E3.

8. The STUDENT information in a university is stored in the relation

STUDENT (Name, Sex, Marks, DEPT_Name)

Consider the following SQL Query SELECT DEPT_Name from STUDENT where SEX='M' group by DEPT_Name having avg (Marks)>(SELECT avg (Marks) from STUDENT). It returns the Name of the Department for which:

(A) The Average marks of Male students are more than the average marks of students in the same Department

(B) The average marks of male students are more than the average marks of the students in the University

(D) The average marks of students are more than the average marks of male students in the University

Answer: B

Explanation: The SQL query:

SELECT DEPT Name from STUDENT

where SEX = 'M' group by DEPT Name

having avg (Marks)>SELECT avg (Marks) from STUDENT

returns the results. Male students receive higher average grades than female students at the university.

9. Select the ‘False’ statement from the following statements about Normal Forms:

(A) Lossless preserving decomposition into 3NF is always possible

(B) Lossless preserving decomposition into BCNF is always possible

(D) BCNF is stronger than 3NF

Answer: B

Explanation: Because 3NF preserves lossless decomposition and functional dependency, option(A) is correct.

BCNF ensures lossless decomposition but does not retain functional dependency, hence Option(B) is false.

Because any relationship with two attributes is in BCNF, Option(C) is true.

Option(D) is correct since BCNE ensures that a relation has fewer duplicate values than 3NF.

10. The Relation

Vendor Order (V_no, V_ord_no, V_name, Qty_sup, unit_price)

is in 2NF because :

(A) Non-key attribute V_name is dependent on V_no which is part of the composite key

(B) Non-key attribute V_name is dependent on Qty_sup

(D) key attribute V_ord_no is dependent on primary_key unit price

Answer: A

Explanation: If there is no partial dependency in a relation, it is NF: If the LHS of a functional dependency is either the primary key or a non-key attribute, or if all of a relation's attributes are prime attributes, the relation is in 2NF. Option(A) is incorrect since the Non-key attribute V name is dependent on V no, which is a component of the composite key, but there is a partial dependency due to V no.

Option(B) is also incorrect because there is the possibility of a partial dependency because Qty sup is a primus attribute. Option(C) and Option(D) are correct because they are both correct. The primary key determines the key properties.

11. The relation schemas R1 and R2 form a Lossless join decomposition of R if and only if:

(a) R1⋂R2→(R1-R2)

(b) R1→R2

(d) R2→R1⋂R2

(A) (a) and (b) happens (B) (a) and (d) happens

Answer: C

Explanation: If and only if there is a common attribute that is the main key in any of the two decomposed relations, the decomposition is said to be lossless. This criterion is satisfied in this case by statements (a) and (c). As a result, option (C) is right.

12. In the indexed scheme of blocks to a file, the maximum possible size of the file depends on:

(A) The number of blocks used for the index, and the size of the index

(B) Size of Blocks and size of Address

(D) Size of Block

Answer: A

Explanation: The maximum potential file size in an indexed scheme of blocks to a file is determined by the number of blocks used for the index and the index's size.

13. Give the number of principal vanishing point(s) along with their direction for the standard perspective transformation:

(A) Only one in the direction K

(B) Two in the directions I and J

(D) Only two in the directions J and K

Answer: A

Explanation: Because N.I = 0 and N.J=0, there are no vanishing points in the directions I and J. However, N.K = K.K = 1 because the view plane normal N is vector K. As a result, there is just one main vanishing point, which is in the K direction.

14. Consider a triangle A(0, 0), B(1, 1), C(5, 2). The triangle has to be rotated by an angle of 45deg about the point P(-1, -1). What shall be the coordinates of the new triangle?

(A) A'=(1, √2-1), B'=(-1, 2√2-1), C'=(3√2-1, (9/2)√2-1)

(B) A'=(1, √2-1), B'=(2√2-1, -1), C'=(3√2-1, (9/2)√2-1)

(D) A'=(√2-1, -1), B'=(-1, 2√2-1), C'=(3√2-1, (9/2)√2-1)

Answer: C

Explanation: The composite rotation matrix would be = Positive translation * Rotation (45degree) * Negative translation

15. The process of dividing an analog signal into a string of discrete outputs, each of constant amplitude, is called:

(A) Strobing (B) Amplification

Answer: D

Explanation: Quantization is the process of splitting an analog signal into a series of discrete outputs, each with a constant amplitude.

Amplification is the process of amplifying an analog signal.

Strobing is the term for irregular behavior, jerk, or loss of continuity in an analog signal.

Conditioning is the process of modifying an analog signal in order to get it ready for the next stage of processing.

16. Which of the following is not a basic primitive of the Graphics Kernel System(GKS)?

(A) POLYLINE (B) POLY DRAW

Answer: B

Explanation: There is nothing like POLYDRAW in the Graphics Kernel System's fundamental primitive (GKS).

17. Which of the following statements is/are incorrect?

(a) Mapping the coordinates of the points and lines that form the picture into the appropriate coordinates on the device or workstation is known as viewing transformation.

(b) The right-handed cartesian co-ordinates system in whose coordinates we describe the picture is known as the world coordinate system.

(c) The coordinate system that corresponds to the device or workstation where the image is to be displayed is known as the physical device coordinate system.

(d) Left-handed coordinate system in which the display area of the virtual display device corresponds to the unit(|x|) square whose lower left-handed corner is at the origin of the coordinate system, is known as a normalized device coordinate system.

(A) (a) only (B) (a) and (b)

Answer: D

Explanation: Viewing transformation is the process of converting the coordinates of the points and lines that make up a picture into the proper coordinates on the device or workstation. Correct

The world coordinate system is a right-handed cartesian co-ordinates system in which we describe the picture's coordinates.

Correct

The physical device coordinate system correct is the coordinate system that matches to the device or workstation where the image is to be shown.

A normalized device coordinate system is a left-handed coordinate system in which the virtual display device's display area corresponds to the unit (|x|) square whose lower left-hand corner is at the origin of the coordinate system. Incorrect

18. Match the following

List-I List-II

(a) Flood Gun (i) An electron gun designed to flood the entire screen with electrons.

(b) Collector (ii) Partly energized by flooding gun, stores the charge generated by the writing gun

(d) Phosphorus grains (iv) Used in memory-tube display and similar to those used in standard CRT

(e) Writing Gun System (v) Used in memory-tube display and basically the same as the electron gun used in a conventional CRT.

Codes:

(a) (b) (c) (d) (e)

(A) (i) (ii) (iii) (iv) (v)

(B) (ii) (iii) (i) (iv) (v)

(D) (iv) (v) (i) (ii) (iii)

Answer: A

Explanation:

Flood Gun is an electron gun that shoots electrons across the entire screen.

The charge generated by the writing gun is stored in the collector, which is partially electrified by the flooding gun.

The collector is discharged into the earth.

Phosphorus granules comparable to those used in normal CRTs are employed in memory – tube displays.

The writing gun system is similar to the electron gun in a standard CRT and is utilised in memory–tube displays.

19. Minimal deterministic finite automaton for the language L={ 0n | n≥0, n≠4 } will have:

(A) 1 final state among 5 states

(B) 4 final states among 5 states

(D) 5 final states among 6 states

Answer: D

Explanation:

20. The regular expression corresponding to the language L where

L={x∈{0,1}* | x ends with 1 and does not contain substring 00 } is:

(A) (1 + 01)* (10 + 01) (B) (1 + 01)* 01

Answer: C

Explanation:

A)it does not contain a substring, but it also terminates with 0, therefore it violates the question's criterion, so A is false.

B)It is a partially accurate answer because, when compared to the c option, it does not generate all of the required strings, hence it is false.

C)It will generate all strings that finish in 1 and will not generate the substring 00. SO CORRECT

D)it will generate 00 as a substring, which violates the needed condition, resulting in a false result.

21. The transition function for the language L = {w|na(w) and nb(w) are both odd} is given by:

δ(q0, a)=q1 ; δ(q0, b)=q2

δ(q1, a)=q0 ; δ(q1, b)=q3

δ(q2, a)=q3 ; δ(q2, b)=q0

δ(q3, a)=q2 ; δ(q3, b)=q1

the initial and final states of the automata are:

(A) q0 and q0 respectively (B) q0 and q1 respectively

Answer: D

Explanation:

22. The clausal form of the disjunctive normal form ¬A∨¬B∨¬C∨D is:

(A) A ∧ B ∧ C ⇒ D (B) A ∨ B ∨ C ∨ D ⇒ true

Answer: A

Explanation: We know that P ⇒ Q = ¬P ∨ Q

therefore, A ∧ B ∧ C ⇒ D

= ¬(A ∧ B ∧ C) v D

= ¬A ∨ ¬B ∨ ¬C ∨ D.

23. Which of the following is false for the programming language PROLOG?

(A) A PROLOG variable can only be assigned to a value once

(B) PROLOG is a Strongly-Typed Language.

(D) the scope of a variable in PROLOG is a Single Query

Answer: B

Explanation: PROLOG variables can only be assigned to a value once. PROLOG is a strongly typed programming language. False

A variable's scope in PROLOG is a single sentence or rule. True

A variable's scope in PROLOG is a single query. True

24. Which one of the following is true?

(A) The resolvent of two Horn clauses is not a Horn clause.

(B) The resolvent of two Horn clauses is a Horn Clause.

(C) If we resolve a negated goal G against a fact or rule A to get Clause C then C has a positive literal and non-null-goal.

(D) If we resolve a negated goal G against a fact or rule A to get clause C then C has a positive literal or null goal.

Answer: B

Explanation: A Horn clause is not the resolvent of two Horn clauses. False

A Horn clause is the resolvent of two Horn clauses. True

Clause C has a positive literal or non-null aim if we resolve a negated goal G against a fact or rule A. False

Clause C has a positive literal or null objective if we resolve a negated goal G against a fact or rule A. False

25. Which transmission technique guarantees that data packets will be received by the receiver in the same order in which they were sent by the sender?

(A) Broadcasting (B) Unicasting

Answer: D

Explanation: Because the path is fixed and all packets transit through it in the same sequence, the circuit switching transmission approach ensures that data packets are received by the receiver in the same order as they were sent by the sender.

FAQs

What is the purpose of the UGC NET Exam?

UGC NET is an abbreviation for University Grants Commission National Eligibility Test. It is a national-level test used to evaluate eligibility for lectureships and Junior Research Fellowships (JRF) in Indian institutions and colleges.

Can I apply for UGC NET after completing a PG diploma?

A PGDM is equivalent to a master's degree. So, if you hold a PG Diploma from a recognized university, you are eligible to apply for the exam.

What is the minimum age for applying to the UGC NET Exam?

According to the UGC NET Eligibility, there is no age limit for assistant professors, but candidates for JRF must be at least 31 years old.

Will distance learning be accepted for the UGC NET?

Yes, distance education is acceptable for the UGC NET Exam, provided you are enrolled in a UGC accredited university/institute.

What is the UGC NET syllabus?

The NTA administers the UGC NET Exam in two parts: General Paper-1 (common to all topics) and Paper-2 (subject-specific). The syllabus for each Paper 2 subject varies. You can view the entire UGC NET Syllabus.

Conclusion

In this article, we have extensively discussed some of the previous year's UGC NET questions. We have covered questions 1 to 25 of UGC NET June 2015 PAPER-III Part-1.

We hope that this blog has helped you enhance your knowledge regarding the UGC NET questions and if you would like to learn more, check out our articles on UGC NET Nov 2015 PAPER-III Part-2, UGC NET Nov 2015 PAPER-III Part-3, NET Application form, PSU Recruitment through GATE, and GATE Books For CSE.

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