Introduction
UGC NET Exam is a very popular exam in India for people interested in the research domain. NET(National Eligibility Test) is conducted by NTA (NATIONAL TESTING AGENCY) to identify qualified candidates for Assistant Professor positions in colleges and universities, as well as Junior Research Fellowships (JRF). This article will discuss the UGC NET Exam of Aug 2016 Paper-III (Re-Test) - Part 1 solutions from questions 1 to 25. You can find the solutions to Q26 to 50 in Aug 2016 Paper-III (Re-Test) - Part 2 and Q51 to 75 in Aug 2016 Paper-III (Re-Test) - Part 3.
Questions
1.A ripple counter is a (n):
(A) Synchronous Counter (B) Asynchronous counter
(C) Parallel counter (D) None of the above
Answer: B
Solution:A ripple counter is an Asynchronous counter. For Detailed explanation, see Counters in Digital Logic.
2.8085 microprocessor has ............... bit ALU.
(A) 32 (B) 16
(C) 8 (D) 4
Answer: C
Solution: In 8085, ALU is of 8 bits. Other registers A, B, C, D, E, H, L are also 8 bit registers. However stack pointer and program counter are of 16 bits
3. The register that stores the bits required to mask the interrupts is ................
(A) Status register (B) Interrupt service register
(C) Interrupt mask register (D) Interrupt request register
Answer: C
Solution: Interrupt Request Register (IRR), In-Service Register (ISR), and Interrupt Mask Register (IMR) are the three status registers available (IMR).
IRR: Shows which interrupt request lines are currently active.
ISR: Interrupt Service Request Level.
IMR: An interrupt mask that shows which interrupts are active and which are not.
4. Which of the following in 8085 microprocessor performs
HL = HL + HL ?
(A) DAD D (B) DAD H
(C) DAD B (D) DAD SP
Answer: B
Solution:
DAD B performs HL=HL+BC
DAD H performs HL=HL+HL
DAD SP performs HL=HL+SP
DAD D performs HL=HL+DE
5. In ............... addressing mode, the operands are stored in the memory. The address of the corresponding memory location is given in a register which is specified in the instruction.
(A) Register direct (B) Register indirect
(C) Base indexed (D) Displacement
Answer: B
Solution:The operands are placed in memory in register indirect addressing mode. In the instruction, the address of the matching memory region is specified in a register.
6. The output of the following combinational circuit
is :
(A) X . Y (B) X + Y
(C) X Å Y (D) X Å Y
Answer: D
Solution: Output of first NAND gate (x'y')'=x+y
Output of second NAND gate (xy)'=x'+y'
p/p of final NAND gate ((x+y)(x'+y')) ' =(xy'+yx')'=(x⨁y)'=x XNOR y
7. Which of the following statements is/are True regarding some advantages that an object-oriented DBMS (OODBMS) offers over a relational database?
I. An OODBMS avoids the “impedance mismatch” problem.
II. An OODBMS avoids the “phantom” problem.
III. An OODBMS provides higher performance concurrency control than most relational databases.
IV. An OODBMS provides faster access to individual data objects once they have been read from disk.
(A) II and III only (B) I and IV only
(C) I, II, and III only (D) I, III and IV only
Answer: B
Solution: The "impedance mismatch" problem is avoided with an OODBMS. Correct.
The "phantom" problem is not avoided by using an OODBMS.
Most relational databases have superior performance concurrency control than an OODBMS. Because it is widely diffused.
Individual data items can be accessed more quickly using an OODBMS after they have been read from disc.
Correct
8. The Global conceptual Schema in a distributed database contains information about global relations. The condition that all the data of the global relation must be mapped into the fragments, that is, it must not happen that a data item which belongs to a global relation does not belong to any fragment, is called :
(A) Disjointness condition (B) Completeness condition
(C) Reconstruction condition (D) Aggregation condition
Answer: B
Solution: Completeness condition: It says that if a relation is fragmented, there should be no loss of information after fragmentation.
9. Suppose database table T1(P, R) currently has tuples {(10, 5), (15, 8), (25, 6)} and table T2 (A, C) currently has {(10, 6), (25, 3), (10, 5)}. Consider the following three relational algebra queries RA1, RA2 and RA3:
The number of tuples in the resulting table of RA1, RA2 and RA3 are given by:
(A) 2, 4, 2 respectively (B) 2, 3, 2 respectively
(C) 3, 3, 1 respectively (D) 3, 4, 1 respectively
Answer: D
Solution:
10.Consider the table R with attributes A, B and C. The functional dependencies that hold on R are : A → B, C → AB. Which of the following statements is/are True?
I. The decomposition of R into R1(C, A) and R2(A, B) is lossless.
II. The decomposition of R into R1(A, B) and R2(B, C) is lossy.
(A) Only I (B) Only II
(C) Both I and II (D) Neither I nor II
Answer: C
Solution: R is lossless when decomposed into R1(C, A) and R2(A, B). As a result of C → A, A → B.. As a result, C AB can be deduced with no loss. R1(A, B) and R2(B, C) are lossy decompositions of R. We can derive A → B and C → B but we can't derive C → AB, therefore it's lossy. As a result.
11.Consider the following ORACLE relations:
One (x, y) = {<2, 5>, <1, 6>, <1, 6>, <1, 6>, <4, 8>, <4, 8>}
Two (x, y) = {<2, 55>, <1, 1>, <4, 4>, <1, 6>, <4, 8>, <4, 8>, <9, 9>, <1, 6>}
Consider the following two SQL queries SQ1 and SQ2:
SQ1 : SELECT * FROM One)
EXCEPT
(SELECT * FROM Two);
SQ2 : SELECT * FROM One)
EXCEPT ALL
(SELECT * FROM Two);
For each of the SQL queries, what is the cardinality (number of rows) of the result obtained when applied to the instances above?
(A) 2 and 1 respectively (B) 1 and 2 respectively
(C) 2 and 2 respectively (D) 1 and 1 respectively
Answer: B
Solution: SQ1: EXCEPT (SELECT * FROM Two); This will result in a single tuple (2, 5) because all duplicate tuples will be eliminated. SQ2: SELECT * FROM ONE) EXCEPT ALL (SELECT * FROM TWO); This will result in two tuples (2,5) and (1.6)> because duplicates will not be removed.
12. Which one of the following pairs is correctly matched in the context of database design?
List – I List – II
(Database term) (Definition)
I. Specialization A. Result of taking the union of two or more
disjoint (lower-level) entity sets to produce
a higher-level entity set.
II. Generalization B. Express the number of entities to which another
entity can be associated via a relationship set.
III. Aggregation C. Result of taking a subset of a higher-level
entity set to form a lower-level entity set.
IV. Mapping cardinalities D. An abstraction in which relationship sets (along
with their associated entity sets) are treated as
higher-level entity sets, and can participate in
relationships.
Codes :
I II III IV
(A) D A B C
(B) D C B A
(C) C D A B
(D) C A D B
Answer: D
Solution: When a subset of a higher-level entity set is used to create a lower-level entity set, the result is Specialization
When two or more disjoint(lower-level) entity sets are combined to generate a higher-level entity set, the result is Generalization
Relationship sets (together with their related entity sets) are viewed as higher-level entity sets that can participate in relationships in this concept.
Aggregation
The amount of entities with which one entity can be linked through a relationship set. Cartographic representation of cardinalities
13.Consider a raster grid having XY-axes in positive X-direction and positive upward Y-direction with Xmax = 10, Xmin = –5, Ymax = 11, and Ymin = 6. What is the address of memory pixel with location (5, 4) in raster grid assuming base address 1 (one)?
(A) 150 (B) 151
(C) 160 (D) 161
Answer: D
Solution: Raster graphics, often known as bitmap graphics, is a sort of digital image that represents an image by using tiny rectangular pixels, or picture elements, arranged in a grid configuration.
14. Consider a N-bit plane frame buffer with W-bit wide lookup table with W > N. How many intensity levels are available at a time?
(A) 2^N (B) 2^W
(C) 2^N+W (D) 2^N–1
Answer: A
Solution: A frame buffer is a chunk of computer memory that is large and contiguous. Each pixel in the rater has at least one memory bit; this quantity of memory is referred to as a bit plane. The image is built up one bit at a time in the frame buffer.
N-bit color Frame buffer:
Additional bit planes are used to incorporate color or grey scales into a frame buffer rater graphics device. Each pixel on the CRT has a pixel location in each of the N bit planes that control its intensity. Each of the N bit planes' binary values is stored in matching places in a register. The binary integer that results is regarded as a degree of intensity between 0 (dark) and 2n -1. (full intensity).
The DAC converts this to an analogue voltage between 0 and the electron gun's maximum voltage. There are a total of 2N intensity levels available. The diagram below shows a system with three bit planes and eight (23) intensity levels. For a particular raster resolution, each bit plane requires the full complement of memory; for example, a 3-bit plane frame buffer for a 1024 X 1024 raster requires 3,145,728 (3 X 1024 X1024) memory bits.
15. Consider the Breshenham’s line generation algorithm for a line with gradient greater than one, current point (xi, yi) and decision parameter, di. The next point to be plotted (xi+1, yi+1) and updated decision parameter, di+1, for di < 0 are given as .................
(A) xi+1 = xi+1
yi+1 = yi
di+1 = di + 2 dy
(B) xi+1 = xi
yi+1 = yi + 1
di+1 = di + 2 dx
(C) xi+1 = xi
yi+1 = yi + 1
di+1 = di + 2 (dx – dy)
(D) xi+1 = xi + 1
yi+1 = yi + 1
di+1 = di + 2 (dy – dx)
Answer: B
Solution: When di is zero, we choose east pixel and increment X with no change in Y for the next points, then add east pixel value to old di for new di.
When di is greater than zero, we choose the north east pixel and increment both X and Y, as well as adding the value of the north pixel to di.
When di = 0, we can choose any of them to plot the next set of points.
As a result, option A satisfies di 0.
16.A point P(2, 5) is rotated about a pivot point (1, 2) by 60°. What is the new transformed point P'?
(A) (1, 4) (B) (–1, 4)
(C) (1, – 4) (D) (– 4, 1)
Answer: B
Solution:
17.In perspective projection (from 3D to 2D), objects behind the centre of projection are projected upside down and backward onto the view-plane. This is known as ..............
(A) Topological distortion (B) Vanishing point
(C) View confusion (D) Perspective foreshortening
Answer: C
Solution: In perspective projection (from 3D to 2D), objects behind the centre of projection are projected upside down and backward onto the view-plane. This is known as view confusion.
18. The Liang-Barsky line clipping algorithm uses the parametric equation of a line from (x1,y1) to (x2,y2) along with its infinite extension which is given as :
x = x1 + Dx.u
y = y1 + Dy.u
Where Dx = x2 – x1, Dy = y2 – y1, and u is the parameter with 0 ≤ u ≤ 1. A line AB with end points A(–1, 7) and B(11, 1) is to be clipped against a rectangular window with xmin=1, xmax=9, ymin=2, and ymax=8. The lower and upper bound values of the parameter u for the clipped line using Liang-Barsky algorithm is given as:
(A) (0, 2/3) (B) (1/6, 5/6)
(C) (0, 1/3) (D) (0, 1)
Answer: B
Solution:
first we will calculate ∆x and ∆y:
i.e. ∆x = x2– x1
= 11 – ( – 1)
= 11 + 1
= 12.
∆y = y2– y1
= 1 – 7
= – 6
Now P1 = -∆x = – 12
P2 = ∆x = 12
P3 = -∆y = 6
P4 = ∆y = – 6
Q1 = x1 – xmin = – 1 – 1 = -2
Q2 = xmax – x1 = 9 – ( – 1) = 9 + 1 = 10.
Q3 = y1 – ymin = 7 – 2 = 5.
Q4 = ymax – y1 = 8 -7 = 1.
P1, P4 < 0 and P2, P3 > 0.
Initially: t1 = 0, t2 = 1
t1 = max(0, Q1 / P1, Q4 / P4)
= max(0, 2 / 12, 1 / -6)
= 1 / 6.
t2 = min(1, Q1 / P1, Q4 / P4)
= min(1, 10 / 12, 5 /6).
= 5 / 6.
i.e. u ranges between (1 / 6, 5 / 6).
19. Match the following with reference to Functional programming history:
a. Lambda calculus i. Church, 1932
b. Lambda calculus as
programming language ii. Wordsworth, 1970
c. Lazy evaluation iii. Haskel, 1990
d. Type classes iv. Mecarthy, 1960
Codes :
a b c d
(A) iv i iii ii
(B) i iv ii iii
(C) iii ii iv i
(D) ii i iv iii
Answer: B
Solution: Alonzo Church introduced lambda calculus in 1932.
Mecarthy used lambda calculus as a programming language in 1960.
Wordsworth coined the term "lazy evaluation" in 1970.
Haskel introduced type classes in 1990.
20.Aliasing in the context of programming languages refers to:
(A) Multiple variables having the same location
(B) Multiple variables having the same identifier
(C) Multiple variables having the same value
(D) Multiple use of same variable
Answer: A
Explanation: Aliasing is a term used in computer programming to describe a situation in which the same memory address can be accessed using distinct identifiers. If a function takes two pointers A and B with the same value, for example, the name A aliases the name B.
21.Assume that the program ‘P’ is implementing parameter passing with ‘call by reference’. What will be printed by following print statements in P?
Program P()
{
x = 10;
y = 3;
funb (y, x, x)
print x;
print y;
}
funb (x, y, z)
{
y = y + 4;
z = x + y + z;
}
(A) 10, 7 (B) 31, 3
(C) 10, 3 (D) 31, 7
Answer: B
Solution: y = y + 4; → at &x 14 will be assigned
z = x + y + z; → Now &x will be assigned with 3 + 14 + 14 = 31.
22.The regular grammar for the language L = {anbm | n + m is even} is given by
(A) S → S1 | S2
S1 → a S1 | A1
A1 → b A1 | λ
S2 → aaS2 | A2
A2 → b A2 | λ
(B) S → S1 | S2
S1 → a S1 | a A1
S2 → aa S2 | A2
A1 → bA1 | λ
A2 → bA2 | λ
(C) S → S1 | S2
S1 → aaa S1 | aA1
S2 → aaS2 | A2
A1 → bA1 | λ
A2 → bA2 | λ
(D) S → S1 | S2
S1 → aa S1 | A1
S2 → aaS2 | aA2
A1 → bbA1 | λ
A2 → bbA2 | b
Answer: D
Solution:
L = {a^n b^m | n + m is even}
S = S1 | S2
S1 = aa S1| A1
S2 = aaS2| A2
A1 = bb A1| λ
A2 = bb A2| λ
23. Let Σ = {a, b} and language L = {aa, bb}. Then, the complement of L is
(A) {λ, a, b, ab, ba} È {w ϵ {a, b}* | |w| > 3}
(B) {a, b, ab, ba} È {w ϵ {a, b}* | |w| ≥ 3}
(C) {w ϵ { a, b}* | |w| > 3} È {a, b, ab, ba}
(D) {λ, a, b, ab, ba} È {w ϵ {a, b}* | |w| ≥ 3}
Answer: D
Solution: The strings generated by the language L = {aa, bb} will be of the same length. As a result, the complement will be the universal set of stringgs {aa, bb}. Strings {λ, a, b, ab, ba} and strings {w ∈ {a, b}* | |w| ≥ 3} are strings of length greater than or equal to 3.
24. Consider the following identities for regular expressions :
(a) (r + s)* = (s + r)*
(b) (r*)* = r*
(c) (r* s*)* = (r + s)*
Which of the above identities are true ?
(A) (a) and (b) only (B) (b) and (c) only
(C) (c) and (a) only (D) (a), (b) and ©
Answer: D
Solution: Any strings containing r or s, or both, will be generated by (r + s)*. DFA can be drawn for (r + s)*, which is the same as (s + r)*. It's a regular expression, after all.
(r*)* generates any string containing r, and its DFA may be easily drawn and is identical to r*. It's a regular expression as well.
(r* s*)* generates any strings that contain the letters r, s, or both. DFA for (r* s*)* is the same as DFA for (r + s)*. It's a regular expression, after all. All of the options are correct.
25. Suppose transmission rate of a channel is 32 kbps. If there are ‘8’ routes from source to destination and each packet p contains 8000 bits. Total end to end delay in sending packet P is ................
(A) 2 sec (B) 3 sec
(C) 4 sec (D) 1 sec
Answer: A
Solution: It is stated that a channel's transmission rate is 32 kbps, that there are eight paths from source to destination, and that each packet p has 8000 bits. Routes * packets / transmission rate = total delay. 8 * 8000 bits per second / 32 kbps = 8 * 8000 bits per second / 32000 bits per second = 2 seconds
24+ registered 
