UGC NET · Computer Science

Operating Systems Previous Year Questions

66 Operating Systems (Computer Science) questions asked in UGC NET previous year papers, each with the correct answer and explanation. Read them free on UTpaper, or attempt the full papers as timed online tests.

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Other Computer Science topics

Q1UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Which of the following is the correct value returned to the operating system upon the successful completion of a program?
  1. A0
  2. B1
  3. C− 1
  4. DProgram do not return a value.
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Correct answer: (A) 0

Explanation

0 is the identifier for a program to inform the operating system (OS) about proper running of a program. Each of the non-zero value signifies unique error doing running of a program.

Q2UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
KPA in CMM stands for
  1. AKey Process Area
  2. BKey Product Area
  3. CKey Principal Area
  4. DKey Performance Area
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Correct answer: (A) Key Process Area

Explanation

Key Process Area (KPA) used in CMM identifies a cluster of related activities that, when performed collectively, achieve a set of goals considered important.

Q3UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Consider a uniprocessor system where new processes arrive at an average of five processes per minute and each process needs an average of 6 seconds of service time. What will be the CPU utilization?
  1. A80%
  2. B50%
  3. C60%
  4. D30%
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Correct answer: (B) 50%

Explanation

New processes arrive at 5 per minute, i.e. one process every 60/5 = 12 seconds. Each process needs 6 seconds of service time. CPU utilization = (total service time required / total available time) × 100 = (6/12) × 100 = 50%. Hence the right option is (b).

Q4UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
A part Windows 2000 operating system that is not portable is
  1. ADevice Management
  2. BVirtual Memory Management
  3. CProcessor Management
  4. DUser Interface
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Correct answer: (B) Virtual Memory Management

Explanation

The design of virtual memory assumes that the underlying hardware supports virtual to physical mapping.

Q5UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Match the following with reference to Unix shell scripts : List-I (A) $? (B) $# (C) $0 (D) $* List-II (i) File name of the current script (ii) List of arguments (iii) The number of arguments (iv) Exit status of last command Codes : A B C D
  1. Aiii ii i iv
  2. Bii iii i iv
  3. Civ iii i ii
  4. Di iii i iv
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Correct answer: (C) iv iii i ii

Explanation

$? : exit status of the last command executed. $# : number of arguments supplied to a script. $0 : file name of the current script. $* : all the arguments (list of arguments). So A-iv, B-iii, C-i, D-ii, i.e. option (c).

Q6UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
The output generated by the LINUX command: $ seq 1 2 10 will be
  1. A1 2 10
  2. B1 2 3 4 5 6 7 8 9 10
  3. C1 3 5 7 9
  4. D1 5 10
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Correct answer: (C) 1 3 5 7 9

Explanation

In Linux the seq command prints a sequence of numbers from FIRST to LAST in steps of INCREMENT: seq FIRST INCREMENT LAST. If FIRST or INCREMENT is omitted, it defaults to 1. Here FIRST = 1, INCREMENT = 2 and LAST = 10, so the output is 1 3 5 7 9 (each number on its own line). Hence the answer is (c).

Q7UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
All the classes necessary for windows programming are available in the module :
  1. Awin.txt
  2. Bwin.main
  3. Cwin.std
  4. DMFC
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Correct answer: (A) win.txt

Explanation

Win·txt is a group of classes that is use for windows programming.

Q8UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Which one of the following is not a key process area in CMM level 5?
  1. ADefect prevention
  2. BProcess change management
  3. CSoftware product engineering
  4. DTechnology change management
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Correct answer: (C) Software product engineering

Explanation

CMM (Capability maturity model)– It is framework that describes an evolutionary improvement path for software organizations from an immature process to a mature, discipline one. The software process is at CMM level 5, if there is continuous process improvement, if there is quantitative feedback from the process and from piloting innovative ideas and technologies. Software processes are the CMM level 5, if there are process change management, detect prevention and technology change management, So software product engineering in not a key process area in CMM level 5

Q9UGC NET Computer Science, June 2014 (Paper II & III, Selected Questions)
Software testing is
  1. AThe process of establishing that errors are not present.
  2. BThe process of establishing confidence that a program does what it is supposed to do.
  3. CThe process of executing a program to show that it is working as per specifications.
  4. DThe process of executing a program with the intent of finding errors
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Correct answer: (D) The process of executing a program with the intent of finding errors

Explanation

Software testing is a process of executing a program or application with the intent of finding errors (the software bugs) and the main objective of software testing is to find any bug present in the software program. Besides, through the process of testing. One can determine the strengths and weaknesses of the program in other criteria such as path testing, dependency testing etc.

Q10UGC 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?
  1. AI, III and IV
  2. BI and II
  3. CIII and IV
  4. DI and III
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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.

Q11UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
_____are applied throughout the software process.
  1. AFramework activities
  2. BUmbrella activities
  3. CPlanning activities
  4. DConstruction activities
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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.

Q12UGC 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
  1. APerformed
  2. BManaged
  3. CDefined
  4. DOptimized
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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).

Q13UGC 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.
  1. ADesign
  2. BArchitecture
  3. CProcess
  4. DRequirement
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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.

Q14UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
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?
  1. AI and II
  2. BIII and IV
  3. CII and III
  4. DII and IV
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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.

Q15UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Various storage devices used by an operating system can be arranged as follows in increasing order of accessing speed :
  1. AMagnetic tapes → magnetic disks → optical disks → electronic disks → main memory → cache → registers
  2. BMagnetic tapes → magnetic disks → electronic disks → optical disks → main memory → cache → registers
  3. CMagnetic tapes → electronic disks → magnetic disks → optical disks → main memory → cache → registers
  4. DMagnetic tapes → optical disks → magnetic disks → electronic disks → main memory → cache → registers
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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.

Q16UGC NET Computer Science, December 2014 (Paper II & III, Selected Questions)
Which of the following versions of window O.S. contain built-in partition manager which allows us to shrink and expand pre-defined drives?
  1. AWindows vista
  2. BWindow 2000
  3. CWindow NT
  4. DWindow 98
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Correct answer: (A) Windows vista

Explanation

Windows Vista's built-in disk management (Logical Disk Manager) supports shrinking and expanding volumes.

Q17UGC NET Computer Science, June 2015 (Paper II & III, Selected Questions)
Which process model is also called as classic life cycle model?
  1. AWaterfall model
  2. BRAD model
  3. CPrototyping model
  4. DIncremental model
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Correct answer: (A) Waterfall model

Explanation

The waterfall model was first process model to be introduce. • It is also referred as linear sequential life cycle model • It is very simple to understand and use. In this each phase must be complicated fully before the next phase can begin. • In this model the testing stands only after the development is complete • In this model phases do not overlap

Q18UGC NET Computer Science, June 2015 (Paper II & III, Selected Questions)
Which of the following statements is not true for multi level feedback queue processor scheduling algorithm?
  1. AQueues have different priorities
  2. BEach queue may have different scheduling algorithm
  3. CProcesses are permanently assigned to a queue
  4. DThis algorithm can be configured to match a specific system under design
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Correct answer: (C) Processes are permanently assigned to a queue

Explanation

Processes are not permanently assigned to a queue they can vary between queues as their priorities change

Q19UGC NET Computer Science, June 2015 (Paper II & III, Selected Questions)
What is the most appropriate function of Memory management Unit (MMU)?
  1. AIt is an associative memory to store TLB
  2. BIt is a technique of supporting multiprogramming by creating dynamic partitions
  3. CIt is a chip to map virtual address to physical address
  4. DIt is an algorithm to allocate and deallocate main memory to a process
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Correct answer: (C) It is a chip to map virtual address to physical address

Explanation

Memory management unit (MMU), sometimes called paged memory management unit (PMMU), is a computer hardware unit having all memory references passing through itself, primarily performing the translation of virtual memory addresses to physical addresses. It is usually implemented as part of the central processing unit (CPU), but it also can be in the form of a separate integrated circuit. So we can say it is a chip to map virtual address to physical address

Q20UGC NET Computer Science, June 2015 (Paper II & III, Selected Questions)
WOW32 is a :
  1. AWin 32 API library for creating processes and threads
  2. BSpecial kind of file system to the NT name space
  3. CKernel - mode objects accessible through Win 32 API
  4. DSpecial execution environment used to run 16 bit Window applications on 32-bit machines
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Correct answer: (D) Special execution environment used to run 16 bit Window applications on 32-bit machines

Explanation

Wow 32 is a running environment used in windows to run 16 bit windows application on a 32 bit machine

Q21UGC NET Computer Science, June 2015 (Paper II & III, Selected Questions)
The Unix command : $ vi file1 file2
  1. AEdits file1 and stores the contents of file1 in file2
  2. BBoth files i.e. file1 and file2 can be edited using 'ex' command to travel between the files
  3. CBoth files can be edited using 'mv' command to move between the files
  4. DEdits file1 first, saves it and then edits file2
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Correct answer: (B) Both files i.e. file1 and file2 can be edited using 'ex' command to travel between the files

Explanation

You can edit more than one file at a time with vi. vi file1 file2 vi two (or more) files at the same time.

Q22UGC NET Computer Science, June 2015 (Paper II & III, Selected Questions)
A Unix file may be of the type :
  1. ARegular file
  2. BDirectory file
  3. CDevice file
  4. DAny one of the above
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Correct answer: (D) Any one of the above

Explanation

A Unix file may be of any of the type like regular file, directory file, or device file.

Q23UGC NET Computer Science, June 2015 (Paper II & III, Selected Questions)
A Design concept refinement is a :
  1. ATop-down approach
  2. BComplementary of abstraction concept
  3. CProcess of elaboration
  4. DAll of the above
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Correct answer: (D) All of the above

Explanation

Refinement is surly a top down approach as we are extracting a particular type like person entity to s/w engineer. Refinement, process of elaboration, start with the statement of function defined at the abstract level, decompose the statement function in a step wise fashion until programming language statements are reached.

Q24UGC NET Computer Science, December 2015 (Paper II & III, Selected Questions)
In Unix, the login prompt can be changed by changing the contents of the file _____
  1. AContrab
  2. BInit
  3. CGettydefs
  4. DInittab
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Correct answer: (C) Gettydefs

Explanation

The gettydefs file supplies the getty command with information about various terminal attributes, such as the default initialization and the login message. (i) There is nothing like contrab in Unix. (ii) The inittab file controls what happens whenever a Unix system is rebooted or forced to change run levels. (iii) init (short for initialization) is the first process started during booting of the computer system.

Q25UGC NET Computer Science, December 2015 (Paper II & III, Selected Questions)
Which of the following is not a software process model?
  1. APrototyping
  2. BIterative
  3. CTimeboxing
  4. DGlassboxing
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Correct answer: (D) Glassboxing

Explanation

Prototyping and iterative are well-known software process models. In the timeboxing model, development is done iteratively as in the iterative enhancement model, with each iteration done in a timebox of fixed duration. Glass boxing is a software testing approach, not a process model.

Q26UGC NET Computer Science, December 2015 (Paper II & III, Selected Questions)
In an operating system, indivisibility of operation means :
  1. AOperation is interruptible
  2. BRace – condition may occur
  3. CProcessor cannot be pre-empted
  4. DAll of the above
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Correct answer: (C) Processor cannot be pre-empted

Explanation

In an OS, an indivisible (atomic) operation is performed entirely or not performed at all. This means the processor cannot be pre-empted during it.

Q27UGC NET Computer Science, December 2015 (Paper II & III, Selected Questions)
The character set used in windows 2000 operating system is _____
  1. A8 bit ASCII
  2. BExtended ASCII
  3. C16 bit UNICODE
  4. D12 bit UNICODE
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Correct answer: (C) 16 bit UNICODE

Explanation

Windows 2000 uses the 16-bit UNICODE character set. Extended ASCII encodings are 8-bit or larger encodings that include the 7-bit ASCII characters.

Q28UGC NET Computer Science, December 2015 (Paper II & III, Selected Questions)
In Unix, the command to enable execution permission for file "mylife" by all is _____
  1. AChmod ugo + X myfile
  2. BChmod a + X myfile
  3. CChmod + X myfile
  4. DAll of the above
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Correct answer: (D) All of the above

Explanation

All three commands enable execute permission: X stands for execute and + adds the permission; ugo means user, group and others; a means all; with no class given, + applies it for all.

Q29UGC NET Computer Science, December 2015 (Paper II & III, Selected Questions)
What will be the output of the following Unix command? $rm chap0\[1 – 3\]
  1. ARemove file chap0[1 – 3]
  2. BRemove file chap01, chap02, chap03
  3. CRemove file chap\[1 – 3\]
  4. DNone of the above
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Correct answer: (A) Remove file chap0[1 – 3]

Explanation

rm chap* would remove all files beginning with chap. To remove a file whose name contains a special character, escape it with the backslash (\); this is called escaping or despecializing. So rm chap0\[1 – 3\] removes the file named chap0[1 – 3] itself.

Q30UGC NET Computer Science, December 2015 (Paper II & III, Selected Questions)
In Unix operating system, special files are used to :
  1. ABuffer data received in its input from where a process reads
  2. BProvide a mechanism to map physical device to file names
  3. CStore list of file names plus pointers associated with i-nodes
  4. DStore information entered by a user application program or utility program
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Correct answer: (B) Provide a mechanism to map physical device to file names

Explanation

In the Unix file system, six types of file are distinguished: special, named pipes, links, symbolic links, regular (ordinary) and directory. A special file contains no data, but provides a mechanism to map physical devices to file names. The file names are used to access peripheral devices, such as terminals and printers.

Q31UGC NET Computer Science, December 2015 (Paper II & III, Selected Questions)
Blind image deconvolution is ______
  1. ACombination of blur identification and image restoration
  2. BCombination of segmentation and classification
  3. CCombination of blur and non-blur image
  4. DNone of the above
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Correct answer: (A) Combination of blur identification and image restoration

Explanation

In image processing, blind deconvolution is a deconvolution technique that permits recovery of the target scene from a single or a set of blurred images in the presence of a poorly determined or unknown point spread function (PSF). It combines blur identification and image restoration.

Q32UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Which of the following is not an inherent application of stack?
  1. AImplementation of recursion
  2. BEvaluation of a postfix expression
  3. CJob scheduling
  4. DReverse a string
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Correct answer: (C) Job scheduling

Explanation

Application of stack(s) are: (i) Implementation of recursion. (ii) Evaluation of a postfix expression. (iii) Reverse a string. Job scheduling is commonly done via queue data structure.

Q33UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
If the Disk head is located initially at track 32, find the number of disk moves required with FCFS scheduling criteria if the disk queue of I/O blocks requests are: 98, 37, 14, 124, 65, 67
  1. A320
  2. B322
  3. C321
  4. D319
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Correct answer: (C) 321

Explanation

Given I/O requests: 98, 37, 14, 124, 65, 67, head at 32. Using FCFS scheduling, total disk moves = (98 – 32) + (98 – 37) + (37 – 14) + (124 – 14) + (124 – 65) + (67 – 65) = 66 + 61 + 23 + 110 + 59 + 2 = 321

Q34UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
A scheduling Algorithm assigns priority proportional to the waiting time of a process. Every process starts with priority zero (lowest priority). The scheduler reevaluates the process priority for every 'T' time units and decides next process to be scheduled. If the process have no I/O operations and all arrive at time zero, then the scheduler implements _______ criteria.
  1. APriority scheduling
  2. BRound Robin Scheduling
  3. CShortest Job First
  4. DFCFS
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Correct answer: (B) Round Robin Scheduling

Explanation

The scheduling algorithm works as round robin with time quantum T. After a process has executed for T units, its waiting time becomes least (lowest priority), and its turn comes again only after every other process has had a chance for T units.

Q35UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Suppose there are four processes in execution with 12 instances of a Resource R in a system. The maximum need of each process and current allocation are given below: Process | Max. Need | Current Allocation P₁ | 8 | 3 P₂ | 9 | 4 P₃ | 5 | 2 P₄ | 3 | 1 With reference to current allocation, is system safe? If so, what is the safe sequence?
  1. ANo
  2. BYes, P₁ P₂ P₃ P₄
  3. CYes, P₄ P₃ P₁ P₂
  4. DYes, P₂ P₁ P₃ P₄
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Correct answer: (C) Yes, P₄ P₃ P₁ P₂

Explanation

Remaining need = Max need – Current allocation: P₁ = 5, P₂ = 5, P₃ = 3, P₄ = 2. Total allocated = 10, so available = 12 – 10 = 2. Satisfy P₄ (need 2), it releases 3 → available 4. Satisfy P₃ (need 3), releases 5 → available 7 (after P₃ completes). Then P₁ and P₂ can be satisfied similarly. Hence the safe sequence is P₄, P₃, P₁, P₂.

Q36UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Given the following statements: S1: Strategic value of data mining is time stamping. S2: Information collection is an expensive process in building an expert system. Which of the following options is correct?
  1. ABoth S1 and S2 are false.
  2. BBoth S1 and S2 are true.
  3. CS1 is true, S2 is false.
  4. DS1 is false, S2 is true.
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Correct answer: (B) Both S1 and S2 are true.

Explanation

• Strategic value of data mining is time stamping → correct • Information collection is an expensive process in building an expert system → correct • An expert system is a computer system that emulates the decision making ability of a human expert.

Q37UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
The ______ model is preferred for software development when the requirements are not clear.
  1. ARapid Application Development
  2. BRational Unified Process
  3. CEvolutionary Model
  4. DWaterfall Model
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Correct answer: (C) Evolutionary Model

Explanation

Evolutionary Model is an iterative and incremental approach to software development. It is preferred for software development when requirements are not clear.

Q38UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
What is the function of the following UNIX command? wc -l <a >b &
  1. AIt runs the word count program to count the number of lines in its input, a, writing the result to b, as a foreground process.
  2. BIt runs the word count program to count the number of lines in its input, a, writing the result to b, but does it in the background.
  3. CIt counts the errors during the execution of a process, a, and puts the result in process b.
  4. DIt copies the 'l' numbers of lines of program from file, a, and stores in file b.
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Correct answer: (B) It runs the word count program to count the number of lines in its input, a, writing the result to b, but does it in the background.

Explanation

wc -l <a >b &: wc is the word count program, -l counts lines, <a takes input from a, >b writes output to b, and & runs the command in the background. So it counts the number of lines in a, writes the result to b, and does it in the background.

Q39UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Which of the following statement is not correct with reference to cron daemon in UNIX O.S.?
  1. AThe cron daemon is the standard tool for running commands on a pre-determined schedule.
  2. BIt starts when the system boots and runs as long as the system is up.
  3. CCron reads configuration files that contain list of command lines and the times at which they invoked.
  4. DCronatab for individual users are not stored.
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Correct answer: (D) Cronatab for individual users are not stored.

Explanation

The cron daemon is the standard tool for running commands on a pre-determined schedule. It starts when the system boots and runs as long as the system is up. Crontabs for individual users are stored under /var/spool/cron; there is one crontab file per user. So statement (d) is not correct.

Q40UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
In Unix, files can be protected by assigning each one a 9-bit mode called right bits. Now, consider the following two statements : I. A mode of 641 (octal) means that the owner can read and write the file, other members of the owner's group can read it, and users can execute only. II. A mode of 100 (octal) allows the owner to execute the file, but prohibits all other access. Which of the following options is correct with reference to above statements?
  1. AOnly I is correct
  2. BOnly II is correct
  3. CBoth I and II are correct
  4. DBoth I and II are incorrect
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Correct answer: (C) Both I and II are correct

Explanation

In UNIX, file access permissions are given as three octal digits: the first for the owner, the second for the group, the third for others. Each digit is rwx: 7 = rwx, 6 = rw-, 5 = r-x, 4 = r--, 3 = -wx, 2 = -w-, 1 = --x, 0 = none. 641 → owner can read and write, group members can read, others can execute only. 100 → owner can execute; all other access is prohibited. So both statements are correct.

Q41UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
In UNIX, processes that have finished execution but have not yet had their status collected are known as ................
  1. ASleeping processes
  2. BStopped processes
  3. CZombie processes
  4. DOrphan processes
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Correct answer: (C) Zombie processes

Explanation

Zombie Process : The UNIX system assumes that if child process terminate, the parent collects some data about the child, As soon as this collection is made, the system releases information bits about the child process and removes its PID from the process table. The UNIX is forced to keep the child's Process ID (PID) and termination data in the process table indefinitely. Such a terminated process whose status is not collected called a zombie process.

Q42UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
In Unix operating system, when a process creates a new process using the fork () system call, which of the following state is shared between the parent process and child process?
  1. AHeap
  2. BStack
  3. CShared memory segments
  4. DBoth Heap and Stack
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Correct answer: (C) Shared memory segments

Explanation

Parent and child share only the share memory segments everything else (like stack, heap etc) is duplicated.

Q43UGC NET Computer Science, June 2016 (Paper II & III, Selected Questions)
Which of the following information about the UNIX file system is not correct?
  1. ASuper block contains the number of i-nodes, the number of disk blocks, and the start of the list of free disk blocks.
  2. BAn i-node contains accounting information as well as enough information to locate all the disk block that holds the file's data.
  3. CEach i-node is 256-bytes long.
  4. DAll the files and directories are stored in data blocks.
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Correct answer: (C) Each i-node is 256-bytes long.

Explanation

UNIX file system: Super block contains the number of i-nodes, the number of disk blocks, and the start of the list of free disk blocks.Correct An i-node contains accounting information as well as enough information to locate all the disk blocks that holds the file’s data.Correct All the files and directories are stored in data blocks.Correct Each i-node is 256-bytes long.Incorrect.It is 128 byte long.

Q44UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Consider a system with seven processes A through G and six resources R through W. Resource ownership is as follows : process A holds R and wants T process B holds nothing but wants T process C holds nothing but wants S process D holds U and wants S & T process E holds T and wants S process F holds W and wants S process G holds V and wants U Is the system deadlocked? If yes, ............... processes are deadlocked.
  1. ANo
  2. BYes, A, B, C
  3. CYes, D, E, G
  4. DYes, A, B, F
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Correct answer: (C) Yes, D, E, G

Explanation

Note: the original exam reads "process E holds T and wants V"; as printed (E wants S, which no process holds) there would be no cycle, but the key assumes the original wording. In the resource allocation graph there is a cycle D → T → E → V → G → U → D (D waits for T held by E, E waits for V held by G, G waits for U held by D). So the system is deadlocked and processes D, E and G are deadlocked. Answer is option (c).

Q45UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Discovery of cross sales opportunities is called as _____
  1. AAssociation
  2. BVisualization
  3. CCorrelation
  4. DSegmentation
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Correct answer: (A) Association

Explanation

Association rule learning is a method for discovering intersecting relations between variables in large data base. It is intended to identify strong rules disconnected in database using some measures of interestingness.

Q46UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
An Operating System (OS) crashes on the average once in 30 days, that is, the Mean Time Between Failures (MTBF) = 30 days. When this happens, it takes 10 minutes to recover the OS, that is, the Mean Time To Repair (MTTR) = 10 minutes. The availability of the OS with these reliability figures is approximately:
  1. A96.97%
  2. B97.97%
  3. C99.009%
  4. D99.97%
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Correct answer: (D) 99.97%

Explanation

Converting 30 days into minutes, the fraction of time the system is not available = 10/(30 × 24 × 60) × 100 ≈ 0.023%. Availability = 100 – 0.023 ≈ 99.97%.

Q47UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
An operating system supports a paged virtual memory, using a central processor with a cycle time of one microsecond. It costs an additional one microsecond to access a page other than the current one. Pages have 1000 words, and the paging device is a drum that rotates at 3000 revolutions per minute and transfers one million words per second. Further, one percent of all instructions executed accessed a page other than the current page. The instruction that accessed another page, 80% accessed a page already in memory and when a new page was required, the replaced page was modified 50% of the time. What is the effective access time on this system, assuming that the system is running only one process and the processor is idle during drum transfers?
  1. A30 microseconds
  2. B34 microseconds
  3. C60 microseconds
  4. D68 microseconds
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Correct answer: (B) 34 microseconds

Explanation

Effective access time = 0.99 × (1 µsec + 0.008 × 2 µsec) + 0.002 × (10000 µsec + 1000 µsec) + 0.001 × (10000 µsec + 1000 µsec) ≈ (0.99 + 0.016 + 22.0 + 11.0) µsec ≈ 34.0 µsec.

Q48UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Consider the following operations to be performed in Unix: "The pipe sorts all files in the current directory modified in the month of "June" by order of size and prints them to the terminal screen. The sort option skips ten fields then sorts the lines in numeric order." Which of the following Unix command will perform above set of operation?
  1. Als -l | grep "June" | sort +10n
  2. Bls -l | grep "June" | sort +10r
  3. Cls -l | grep -v "June" | sort +10n
  4. Dls -l | grep -n "June" | sort +10x
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Correct answer: (A) ls -l | grep "June" | sort +10n

Explanation

ls -l lists the files with details, grep "June" keeps the lines for files modified in June, and sort +10n skips ten fields and sorts the lines in numeric order. So the correct command is ls -l | grep "June" | sort +10n.

Q49UGC NET Computer Science, August 2016 (Paper II & III, Selected Questions)
Which of the following statement(s) is/are True regarding 'nice' command of UNIX? I. It is used to set or change the priority of a process. II. A process's nice value can be set at the time of creation. III. 'nice' takes a command line as an argument.
  1. AI, II only
  2. BII, III only
  3. CI, II, III
  4. DI, III only
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Correct answer: (C) I, II, III

Explanation

A process's nice value (priority) can be set at the time of creation with the nice command, which takes a command line as an argument, and adjusted later with the renice command. So I, II and III are all true.

Q50UGC NET Computer Science, June 2017 (Paper II & III, Selected Questions)
There are three processes P1, P2 and P3 sharing a semaphore for synchronizing a variable. Initial value of semaphore is one. Assume that negative value of semaphore tells us how many processes are waiting in queue. Processes access the semaphore in following order : (A) P2 needs to access (B) P1 needs to access (C) P3 needs to access (D) P2 exits critical section (E) P1 exits critical section The final value of semaphore will be :
  1. A0
  2. B1
  3. C–1
  4. D–2
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Correct answer: (A) 0

Explanation

Initial value S = 1. (1) P2 needs to access: S = 0 (no one waiting) (2) P1 needs to access: S = –1 (one process waiting) (3) P3 needs to access: S = –2 (two processes waiting) (4) P2 exits critical section: S = –1 (one process waiting) (5) P1 exits critical section: S = 0 (no process waiting)

How to use these UGC NET Computer Science questions

Every question here was asked in a real UGC NET paper; the paper and shift are shown on the card, and the number opens the question on its own page. Try each one before tapping Show answer, then attempt the full UGC NET papers as timed tests to check your speed under exam conditions.