UGC NET Computer Science PYQ, December 2014 (Paper II & III, Selected Questions)
All 51 questions of UGC NET Computer Science PYQ, December 2014 (Paper II & III, Selected Questions) with the correct answer from the official answer key and a solution. Read them free on UTpaper, or attempt the paper as a timed online test.
Q1UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
The Excess-3 decimal code is self-complementing code because
(1) The binary sum of a code and its 9's complement is equal to 9
(2) It is a weighted code
(3) Complement can be generated by inverting each bit pattern
(4) The binary sum of a code and its 10's complement is equal to 9
A(1)
B(2) and (3)
C(1) and (3)
DAll are correct.
Show answerHide answer
Correct answer: (C) (1) and (3)
Explanation
The Excess-3 decimal code is a self-complementing code because the binary sum of a code and its 9's complement is equal to 9 and the complement can be generated by inverting each bit pattern.
Q2UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
The BCD adder to add two decimal digits needs minimum of
A6 full adders and 2 half adders
B5 full adders and 3 half adders
C4 full adders and 3 half adders
D5 full adders and 2 half adders
Show answerHide answer
Correct answer: (C) 4 full adders and 3 half adders
Explanation
To add two BCD digits we need two 4-bit adders: the first adds the two BCD numbers and the second corrects the result to a valid BCD number. The first 4-bit adder needs 3 full adders and 1 half adder (giving A₃ A₂ A₁ A₀). In the correction step 0110 is added to A₃ A₂ A₁ A₀. A₀ is always added to 0, so no adder is needed; A₁ is added to 1 with no incoming carry, so a half adder is used; A₂ is added to 1 plus a carry, so a full adder is used; A₃ is added only to the carry, so a half adder is used. Total = 3 + 1 = 4 full adders and 1 + 2 = 3 half adders.
Q3UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
The size of the ROM required to build an 8-bit adder/subtracter with mode control, carry input, carry output and two's complement overflow output is given as :
A2¹⁶ × 8
B2¹⁸ × 10
C2¹⁶ × 10
D2¹⁸ × 8
Show answerHide answer
Correct answer: (B) 2¹⁸ × 10
Explanation
Total inputs to the ROM = 8 + 8 (two 8-bit numbers) + 1 (mode) + 1 (carry in) = 18, so the ROM has 2¹⁸ words. Outputs = 8 result bits + 1 carry out + 1 overflow = 10 bits. Hence the ROM size is 2¹⁸ × 10.
Q4UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
How many PUSH and POP operations will be needed to evaluate the following expression by reverse polish notation in a stack machine (A × B) + (C × D/E)?
A4 PUSH and 3 POP instructions
B5 PUSH and 4 POP instructions
C6 PUSH and 2 POP instructions
D5 PUSH and 3 POP instructions
Show answerHide answer
Correct answer: (B) 5 PUSH and 4 POP instructions
Explanation
The postfix form of the expression is AB*CD*E/+. Counting a PUSH for each of the 5 operands and a POP for each of the 4 operators (ignoring the pushes of intermediate results), the evaluation needs 5 PUSH and 4 POP instructions.
Q5UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Consider an array A[20, 10], assume 4 words per memory cell and the base address of array A is 100. What is the address of A [11, 5]? Assume row major storage.
A560
B565
C570
D575
Show answerHide answer
Correct answer: (A) 560
Explanation
Each cell needs 4 words. In row-major order, rows 0 to 10 contain 11 × 10 = 110 elements, and A[11][0] to A[11][4] add 5 more, so 115 elements precede A[11][5]. Address = 100 + 4 × 115 = 560.
Q6UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Convert the following infix expression into its equivalent post fix expression
(A + B^D)/(E – F) + G
AABD^ + EF –/G+
BABD+ ^EF –/G+
CABD+ ^EF/ –G+
DABD^ + EF /–G+
Show answerHide answer
Correct answer: (A) ABD^ + EF –/G+
Explanation
^ has the highest precedence, so B^D becomes BD^, and A + B^D becomes ABD^+. E – F becomes EF–. Dividing gives ABD^+EF–/, and adding G gives ABD^+EF–/G+.
Q7UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
A full binary tree with n leaves contains
An nodes
Blog₂ nodes
C2n – 1 nodes
D2ⁿ nodes
Show answerHide answer
Correct answer: (C) 2n – 1 nodes
Explanation
A full binary tree with n leaves contains n – 1 internal nodes, so total nodes = internal nodes + leaves = (n – 1) + n = 2n – 1.
Q8UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
You have to sort a list L. consisting of a sorted list followed by a few 'random' elements. Which of the following sorting method would be most suitable for such a task?
ABubble sort
BSelection sort
CQuick sort
DInsertion sort
Show answerHide answer
Correct answer: (D) Insertion sort
Explanation
For a sorted list followed by a few random elements: bubble sort and selection sort take O(n²) time, quick sort degrades to its O(n²) worst case on nearly sorted input, while insertion sort takes close to O(n) time. Hence insertion sort is most suitable.
Q9UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
How many distinct stages are there in DES algorithm, which is parameterized by a 56-bit key?
A16
B17
C18
D19
Show answerHide answer
Correct answer: (D) 19
Explanation
The DES algorithm works with a 64-bit block size, employs 16 processing rounds, and is driven by a 56-bit key (with 8 additional parity bits), making up a total of 19 distinct stages.
Q10UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
A computer program select an integer in the set {k : 1≤ k ≤10,00,000} at random and prints out the result. This process is repeated 1 million times. What is the probability that the value k = 1 appears in the printout at least once?
A0.5
B0.704
C0.632121
D0.68
Show answerHide answer
Correct answer: (C) 0.632121
Explanation
Probability of k = 1 in one trial = 1/1000000 = 10⁻⁶, so the probability of k ≠ 1 is 1 – 10⁻⁶. Probability that 1 is never printed in 10⁶ printouts = (1 – 10⁻⁶)^(10⁶) ≈ 1/e. Probability that 1 is printed at least once = 1 – (1 – 10⁻⁶)^(10⁶) ≈ 1 – 1/e = 0.632121.
Q11UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
If we define the functions f, g and h that map R into R by :
f(x) = x⁴, g(x) = √(x² + 1), h(x) = x² + 72, then the value of the composite functions ho(gof) and (hog)of are given as
Ax⁸ – 71 and x⁸ – 71
Bx⁸ – 73 and x⁸ – 73
Cx⁸ + 71 and x⁸ + 71
Dx⁸ + 73 and x⁸ + 73
Show answerHide answer
Correct answer: (D) x⁸ + 73 and x⁸ + 73
Explanation
gof(x) = √(x⁸ + 1), so ho(gof)(x) = (x⁸ + 1) + 72 = x⁸ + 73. hog(x) = (x² + 1) + 72 = x² + 73, so (hog)of(x) = x⁸ + 73. Hence both are x⁸ + 73.
Q12UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Which of the following is not a member of class?
AStatic function
BFriend function
CConst function
DVirtual function
Show answerHide answer
Correct answer: (B) Friend function
Explanation
A friend function of a class is a non-member function that is given the same access as member methods to the class's private and protected data. It is declared inside the class that grants access, but it is not a member of the class.
Q13UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
What will be the output of the following 'C' code?
main ( )
{ int x = 128 ;
printf("\n%d", 1 + x++);
}
A128
B129
C130
D131
Show answerHide answer
Correct answer: (B) 129
Explanation
x++ is a post-increment, so the old value 128 is used in the expression and printf prints 1 + 128 = 129; x becomes 129 only afterwards.
Q14UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
For the implementation of a paging scheme, suppose the average process size be x bytes, the page size be y bytes, and each entry requires z bytes. The optimum page size that minimizes the total overhead due to the page table and the internal fragmentation loss is given by :
Ax/2
Bxz/2
C√(2xz)
D√(xz)/2
Show answerHide answer
Correct answer: (C) √(2xz)
Explanation
Average internal fragmentation per process is y/2. A process has x/y pages and each page-table entry needs z bytes, so the page table takes xz/y bytes. Total overhead = xz/y + y/2. Setting d/dy (xz/y + y/2) = –xz/y² + 1/2 = 0 gives y = √(2xz).
Q15UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
In a demand paging memory system, page table is held in registers. The time taken to service a page fault is 8 m. sec. if an empty frame is available or if the replaced page is not modified, and it takes 20 m. sec. if the replaced page is modified. What is the average access time to service a page fault assuming that the page to be replaced is modified 70% of the time?
A11.6 m. sec.
B16.4 m. sec.
C28 m. sec.
D14 m. sec.
Show answerHide answer
Correct answer: (B) 16.4 m. sec.
Explanation
Average time to service a page fault = 0.7 × 20 + 0.3 × 8 = 14 + 2.4 = 16.4 m. sec.
Q16UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Consider the following justifications for commonly using the two-level CPU scheduling:
I. It is used when memory is too small to hold all the ready processes
II. Because its performance is same as that of the FIFO.
III. Because it facilitates putting some set of processes into memory and a choice is made from that.
IV. Because it does not allow to adjust the set of in-core processes
Which of the following is true?
AI, III and IV
BI and II
CIII and IV
DI and III
Show answerHide answer
Correct answer: (D) I and III
Explanation
Two-level scheduling is used to schedule efficiently when some processes are swapped out. A lower-level scheduler (e.g. round robin) selects only among the processes in memory, while a higher-level scheduler swaps processes in and out of memory. Hence I and III are correct.
Q17UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
A specific editor has 200 K of program text, 15 K of initial stack, 50 K of initialized data, and 70 K of bootstrap code. If five editors are started simultaneously, how much physical memory is needed if shared text is used?
A1135 K
B335 K
C1065 K
D320 K
Show answerHide answer
Correct answer: (B) 335 K
Explanation
The program text is shared, so it is not multiplied by 5. As given in the book, total memory = 200 + 15 + 50 + 70 = 335 K.
Q18UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Which of the following conditions does not hold good for a solution to a critical section problem?
ANo assumptions may be made about speeds or the number of CPUs.
BNo two processes may be simultaneously inside their critical sections.
CProcesses running outside its critical section may block other processes
DProcesses do not wait forever to enter its critical section.
Show answerHide answer
Correct answer: (C) Processes running outside its critical section may block other processes
Explanation
A correct solution must not let a process running outside its critical section block other processes. So condition (c) does not hold good for a solution to the critical section problem.
Q19UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Which of the following is true?
I. Implementation of self-join is possible is SQL with table alias
II. Outer-join operation is basic operation in relational algebra
III. Natural join and outer join operations are equivalent.
AI and II are correct
BII and III are correct
COnly III is correct
DOnly I is correct
Show answerHide answer
Correct answer: (D) Only I is correct
Explanation
Only I is correct: a self join is implemented on a single table using alias names. Outer join is not a basic relational algebra operation, and natural join and outer join are not equivalent.
Q20UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
What kind of mechanism is to be taken into account for converting a weak entity set into strong entity set in entity relationship diagram?
AGeneralization
BAggregation
CSpecialization
DAdding suitable attributes
Show answerHide answer
Correct answer: (D) Adding suitable attributes
Explanation
A weak entity does not have sufficient attributes to form a primary key; it is drawn as a double rectangle in an ER diagram. A weak entity set can be converted into a strong entity set simply by adding suitable attributes.
Q21UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
The best normal form of relation scheme R(A, B, C, D) along with the set of functional dependencies F = { AB → C, AB → D, C → A, D →B} is :
ABoyce-Codd Normal form
BThird Normal form
CSecond Normal form
DFirst Normal form
Show answerHide answer
Correct answer: (B) Third Normal form
Explanation
The candidate keys are AB, CD, BC and AD, so A, B, C and D are all prime attributes and the relation is in 3NF. It is not in BCNF because in C → A and D → B the left side is not a superkey. Hence the best normal form is 3NF.
Q22UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Identify the minimal key for relational scheme R(A, B, C, D, E) with functional dependencies F = {A → B, B → C, AC → D}
AA
BAE
CBE
DCE
Show answerHide answer
Correct answer: (B) AE
Explanation
{A}⁺ = {A, B, C, D}, which does not include E, so A alone is not a key. {AE}⁺ = {A, B, C, D, E}, and nothing can be removed from AE, so AE is the minimal key.
Q23UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Match the following :
List–I
A. Call control protocol
B. A-bis
C. BSMAP
D. CDMA
List–II
i. Interface between Base Transceiver Station (BTS) and Base Station Controller (BSC)
ii. Spread spectrum
iii. Connection management
iv. Works between Mobile Switching Centre (MSC) and Base Station Subsystem (BSS)
Codes :
A B C D
Aiii iv i ii
Biii i iv ii
Ci ii iii iv
Div iii ii i
Show answerHide answer
Correct answer: (B) iii i iv ii
Explanation
Correct match: A. Call control protocol – iii. Connection management; B. A-bis – i. Interface between BTS and BSC; C. BSMAP – iv. Works between MSC and BSS; D. CDMA – ii. Spread spectrum.
Q24UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
An analog signal has a bit rate of 6000 bps and a baud rate of 2000 baud. How many data elements are carried by each signal element?
A0.336 bits/baud
B3 bits/baud
C120,00,000 bits/baud
DNone of the above
Show answerHide answer
Correct answer: (B) 3 bits/baud
Explanation
The signal sends 2000 signal elements (baud) per second and 6000 bits per second, so each signal element carries 6000/2000 = 3 bits/baud.
Q25UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Which of the following algorithms is not a broadcast routing algorithm?
AFlooding
BMulti destination routing
CReverse path forwarding
DAll of the above
Show answerHide answer
Correct answer: (D) All of the above
Explanation
Note: Flooding, multi-destination routing and reverse path forwarding are all broadcast routing algorithms, so none of them is 'not' a broadcast routing algorithm; the book keys (d). Flooding generates too many packets and consumes too much bandwidth but can be used for broadcasting. In multi-destination routing each packet contains a list of destinations, and reverse path forwarding approximates spanning-tree broadcast even when routers know nothing about the spanning tree.
Q26UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Which e-business model allows consumers to name their own price for products and services?
AB2B
BB2G
CC2C
DC2B
Show answerHide answer
Correct answer: (D) C2B
Explanation
The C2B model, also called a reverse auction or demand collection model, enables buyers to name their own (often binding) price for a specific good or service. The website collects the bids and offers them to participating sellers. C2C is consumer to consumer, and B2B and B2G have nothing to do with consumer price setting.
Q27UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
_____ model is designed to bring prices down by increasing the number of customers who buy a particular product at once.
AEconomic Order Quantity
BInventory
CData Mining
DDemand-Sensitive Pricing
Show answerHide answer
Correct answer: (D) Demand-Sensitive Pricing
Explanation
Inventory and data mining have nothing to do with price. Economic Order Quantity (EOQ) is the order quantity that minimizes total holding and ordering costs. The demand-sensitive pricing model brings prices down as more customers buy the product at once.
Q28UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
_____are applied throughout the software process.
AFramework activities
BUmbrella activities
CPlanning activities
DConstruction activities
Show answerHide answer
Correct answer: (B) Umbrella activities
Explanation
The umbrella activities in a software development life cycle process include the following: software project management, formal technical reviews, software quality assurance and software configuration management. Umbrella activities occur throughout the software process and focus primarily on project management, tracking and control.
Q29UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Which one of the following set of attributes should not be encompassed by effective software metrics?
ASimple and computable
BConsistent and objective
CConsistent in the use of units and dimensions
DProgramming language dependent
Show answerHide answer
Correct answer: (D) Programming language dependent
Explanation
Effective software metrics should be simple, computable, consistent, objective and consistent in the use of units and dimensions. They should not be programming language dependent, otherwise they cannot measure software consistently.
Q30UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Which one of the following is used to compute cyclomatic complexity?
AThe number of regions–1
BE – N + 1 , where E is the number of flow graph edges and N is the number of flow graph nodes
CP – 1, where P is the number of predicate nodes in the flow graph G.
DP + 1, where P is the number of predicate nodes in the flow graph G.
Show answerHide answer
Correct answer: (D) P + 1, where P is the number of predicate nodes in the flow graph G.
Explanation
Cyclomatic complexity measures the logical complexity of a program (the number of independent paths). It can be computed as: (1) the number of regions of the flow graph; (2) V(G) = E – N + 2, where E is the number of edges and N the number of nodes; (3) V(G) = P + 1, where P is the number of predicate nodes in the flow graph G.
Q31UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Requirement Development, Organizational process focus, organizational training, risk management and integrated supplier management are process areas required to achieve maturity level
APerformed
BManaged
CDefined
DOptimized
Show answerHide answer
Correct answer: (C) Defined
Explanation
These process areas belong to maturity level 3 (Defined), which includes Requirements Development, Organizational Process Focus (OPF), Organizational Training (OT), Integrated Project Management (IPM) and Risk Management (RSKM).
Q32UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
The software _______ of a program of a computing system is the structure or structures of the system, which comprise software components, the externally visible properties of those components, and the relationships among them.
ADesign
BArchitecture
CProcess
DRequirement
Show answerHide answer
Correct answer: (B) Architecture
Explanation
Software architecture is the set of significant decisions about the organization of a software system: the structural elements and their interfaces, their behaviour and collaboration, and their composition into larger subsystems.
Paper III 19 questions
Q33UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
A computer based information system is needed :
(I) as it is difficult for administrative staff to process data
(II) due to rapid growth of information and communication technology
(III) due to growing size of organizations which need to process large volume of data
(IV) as timely and accurate decisions are to be taken
Which of the above statement (s) is/are true?
AI and II
BIII and IV
CII and III
DII and IV
Show answerHide answer
Correct answer: (B) III and IV
Explanation
A computer based information system is needed because of the growing size of organizations which need to process large volumes of data, and such data must support timely and accurate decisions. So III and IV are true.
Q34UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
Various storage devices used by an operating system can be arranged as follows in increasing order of accessing speed :
AMagnetic tapes → magnetic disks → optical disks → electronic disks → main memory → cache → registers
BMagnetic tapes → magnetic disks → electronic disks → optical disks → main memory → cache → registers
CMagnetic tapes → electronic disks → magnetic disks → optical disks → main memory → cache → registers
DMagnetic tapes → optical disks → magnetic disks → electronic disks → main memory → cache → registers
Show answerHide answer
Correct answer: (D) Magnetic tapes → optical disks → magnetic disks → electronic disks → main memory → cache → registers
Explanation
In increasing order of access speed: Magnetic tapes < Optical disks < Magnetic (hard) disks < Electronic disks (flash/SSD) < Main memory < Cache < Registers.
Q35UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
For switching from a CPU user mode to the supervisor mode following type of interrupt is most appropriate :
AInternal interrupts
BExternal interrupts
CSoftware interrupts
DNone of the above
Show answerHide answer
Correct answer: (C) Software interrupts
Explanation
Internal interrupts occur due to internal errors (like division by zero) and external interrupts come from hardware. Software interrupts are the most appropriate way to switch from user mode to supervisor (kernel) mode.
Q36UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
Any decision tree that sorts n elements has height
AΩ(n)
BΩ(lg n)
CΩ(n lg n)
DΩ(n²)
Show answerHide answer
Correct answer: (C) Ω(n lg n)
Explanation
For any comparison-based sort the decision tree has at least n! leaves, so its height is at least lg(n!) = Ω(n lg n).
Q37UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
Dijkstra algorithm, which solves the single-source shortest-paths problem, is a ................, and the Floyd-Warshall algorithm, which finds shortest paths between all pairs of vertices, is a............
Dijkstra's algorithm for the single-source shortest-paths problem is a greedy algorithm, and the Floyd-Warshall algorithm for all-pairs shortest paths is a dynamic programming algorithm.
Q38UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
We can show that the clique problem is NP-hard by proving that :
ACLIQUE ≤ₚ 3–CNF_SAT
BCLIQUE ≤ₚ VERTEX_COVER
CCLIQUE ≤ₚ SUBSET_SUM
DNone of the above
Show answerHide answer
Correct answer: (D) None of the above
Explanation
To show CLIQUE is NP-hard, a known NP-complete problem must be reduced to CLIQUE (e.g. 3-CNF-SAT ≤ₚ CLIQUE), not the other way round. Reducing CLIQUE to 3-CNF-SAT, VERTEX_COVER or SUBSET_SUM only shows CLIQUE is no harder than them. So the answer is None of the above.
Q39UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
Converting a primitive type data into its corresponding wrapper class object instance is called
ABoxing
BWrapping
CInstantiation
DAutoboxing
Show answerHide answer
Correct answer: (D) Autoboxing
Explanation
Autoboxing is the automatic conversion that the Java compiler makes from primitive types to their corresponding object wrapper classes.
Q40UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
Consider the following statements S1, S2 and S3 :
S1 : In call–by–value, anything that is passed into a function call is unchanged in the caller's scope when the function returns.
S2 : In call-by-reference, a function receives implicit reference to a variable used as argument.
S3 : In call-by-reference, caller is unable to see the modified variable used as argument
AS3 and S2 are true
BS3 and S1 are true
CS2 and S1 are true
DS1, S2, S3 are true
Show answerHide answer
Correct answer: (C) S2 and S1 are true
Explanation
S1 is true: in call-by-value, anything passed into a function is unchanged in the caller's scope when the function returns. S2 is true: in call-by-reference the function receives an implicit reference to the argument variable. S3 is false, because in call-by-reference the caller always sees the modified value of the argument.
Q41UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
Which methods are utilized to control the access to an object in multi-threaded programming?
AAsynchronized methods
BSynchronized methods
CSerialized methods
DNone of the above
Show answerHide answer
Correct answer: (B) Synchronized methods
Explanation
Synchronization is the capability to control the access of multiple threads to shared resources. Without it, one Java thread could modify a shared variable while another thread is using or updating the same variable; synchronized methods prevent this.
Q42UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
Which method is called first by an applet program?
Astart ( )
Brun ( )
Cinit ( )
Dbegin ( )
Show answerHide answer
Correct answer: (C) init ( )
Explanation
In the applet life cycle, init() is called first, followed by start(), paint(), stop() and destroy().
Q43UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
______rules used to limit the volume of log information that has to be handled and processed in the event of system failure involving the loss of volatile information:
AWrite-ahead log
BCheck-pointing
CLog buffer
DThomas
Show answerHide answer
Correct answer: (B) Check-pointing
Explanation
Check-pointing saves a snapshot of the system state periodically, so that after a failure recovery can restart from the last checkpoint; this limits the volume of log information that has to be processed.
Q44UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
Which of the following versions of window O.S. contain built-in partition manager which allows us to shrink and expand pre-defined drives?
AWindows vista
BWindow 2000
CWindow NT
DWindow 98
Show answerHide answer
Correct answer: (A) Windows vista
Explanation
Windows Vista's built-in disk management (Logical Disk Manager) supports shrinking and expanding volumes.
Q45UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
Which of the following is false ?
AEvery binary relation is never be in BCNF
BEvery BCNF relation is in 3NF
C1 NF, 2 NF, 3 NF and BCNF are based on functional dependencies
DMultivalued dependency (MVD) is a special case of join dependency (JD)
Show answerHide answer
Correct answer: (A) Every binary relation is never be in BCNF
Explanation
Every binary relation (a relation with only 2 attributes) is always in BCNF: if both attributes form the key it is in BCNF, and if one attribute is the key the other is determined by it. So statement (a) is false.
Q46UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
______ is a bit-oriented protocol for communication over point-to-point and multipoint links.
AStop-and-wait
BHDLC
CSliding window
DGo-back-N
Show answerHide answer
Correct answer: (B) HDLC
Explanation
HDLC (High-level Data Link Control) is a bit-oriented, code-transparent, synchronous data link layer protocol from ISO. It can be used for point-to-point and point-to-multipoint links.
Q47UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
Which of the following is correct?
I. Two phase locking is an optimistic protocol
II. Two phase locking is pessimistic protocol
III. Time stamping is an optimistic protocol
IV. Time stamping is pessimistic protocol
AI and III
BII and IV
CI and IV
DII and III
Show answerHide answer
Correct answer: (D) II and III
Explanation
Two phase locking is a pessimistic protocol (it assumes conflicts will happen) because it applies locks. Time stamping is an optimistic protocol because it uses no locking and handles conflicts later. So II and III are correct.
Q48UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
The behavior of the document elements in XML can be defined by :
A JavaScript program can define the behaviour of document elements by registering appropriate event handlers.
Q49UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
A technique used to approximate halftones without reducing spatial resolution is known as______
AHalf toning
BDithering
CError diffusion
DNone of the above
Show answerHide answer
Correct answer: (B) Dithering
Explanation
Dithering approximates halftones without reducing spatial resolution. The dither matrix is treated like a floor tile that is repeatedly positioned one copy next to another to cover the entire image.
Q50UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
Which of the following categories of language do not refer to animation language?
AGraphical languages
BGeneral-purpose languages
CLinear-list notations
DNone of the above
Show answerHide answer
Correct answer: (D) None of the above
Explanation
All three categories refer to animation languages: linear-list notations (e.g. 'B rotate palm'), general-purpose languages (e.g. ASAS), and graphical languages. So the answer is None of the above.
Q51UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions) · Paper III
Given two languages :
L1 = {(ab)ⁿ aᵏ | n > k, k ≥ 0}
L2 = {aⁿ bᵐ | n ≠ m}
Using pumping lemma for regular language, it can be shown that
AL₁ is regular and L₂ is not regular
BL₁ is not regular and L₂ is regular
CL₁ is regular and L₂ is regular
DL₁ is not regular and L₂ is not regular
Show answerHide answer
Correct answer: (D) L₁ is not regular and L₂ is not regular
Explanation
The pumping lemma for regular languages can only be used to prove that a language is not regular. Both L₁ (which needs to compare n and k) and L₂ (which needs to compare n and m) require unbounded counting, so neither is regular.