Operating System

71. Match the following with respect to Input/Output management: A. Device controller B. Device driver C. Interrupt handler D. Kernel I/O subsystem
Input/Output management के संबंध में निम्न का मिलान कीजिए: A. Device controller, B. Device driver, C. Interrupt handler, D. Kernel I/O subsystem
(UGC NET C.S. June-2017 (Paper-II))
A. iii, iv, i, ii
B. ii, i, iv, iii
C. iv, i, ii, iii
D. i, iii, iv, ii

Correct Answer: D

Explanation (EN): According to the given source: Device controller → extracts information and stores it in data buffer, Device driver → performs data transfer, Interrupt handler → processing of I/O request, Kernel I/O subsystem → I/O scheduling.

Explanation (HI): दिए गए source के अनुसार: Device controller → information extract करके data buffer में store करता है, Device driver → data transfer करता है, Interrupt handler → I/O request process करता है, Kernel I/O subsystem → I/O scheduling करता है।

72. Which of the following scheduling algorithms may cause starvation? A. First-come-first-served B. Round Robin C. Priority D. Shortest process next E. Shortest remaining time first
निम्न scheduling algorithms में से किनसे starvation हो सकती है? A. FCFS B. Round Robin C. Priority D. Shortest process next E. Shortest remaining time first
(UGC NET C.S. June-2017 (Paper-II))
A. A, C and E
B. C, D and E
C. B, D and E
D. B, C and D

Correct Answer: B

Explanation (EN): Priority, shortest process next and shortest remaining time first may cause starvation.

Explanation (HI): Priority, shortest process next और shortest remaining time first starvation का कारण बन सकते हैं।

73. Distributed operating system consist of:
Distributed operating system किससे मिलकर बना होता है?
(UGC NET C.S. June-2017 (Paper-II))
A. Loosely coupled O.S. software on a loosely coupled hardware
B. Loosely coupled O.S. software on a tightly coupled hardware
C. Tightly coupled O.S. software on a loosely coupled hardware
D. Tightly coupled O.S. software on a tightly coupled hardware

Correct Answer: C

Explanation (EN): A distributed OS runs on loosely coupled hardware but provides a tightly coupled software view.

Explanation (HI): Distributed OS loosely coupled hardware पर चलता है लेकिन tightly coupled software view प्रदान करता है।

74. A memory management system has 64 pages with 512 bytes page size. Physical memory consists of 32 page frames. Number of bits required in logical and physical address are respectively:
एक memory management system में 64 pages हैं और page size 512 bytes है। Physical memory में 32 page frames हैं। Logical और physical address के लिए क्रमशः कितने bits चाहिए?
(UGC NET C.S. June-2017 (Paper-III))
A. 14 and 15
B. 14 and 29
C. 15 and 14
D. 16 and 32

Correct Answer: C

Explanation (EN): Logical address bits = log2(64×512) = log2(2^6×2^9) = 15. Physical address bits = log2(32×512) = log2(2^5×2^9) = 14.

Explanation (HI): Logical address bits = log2(64×512) = 15 और physical address bits = log2(32×512) = 14।

75. Some of the criteria for calculation of priority of a process are: A. Processor utilization by an individual process B. Weight assigned to a user or group of users C. Processor utilization by a user or group of processes. In fair share scheduler, priority is calculated based on:
किसी process की priority calculate करने के criteria हैं: A. किसी individual process द्वारा processor utilization B. किसी user या group को assigned weight C. किसी user या group of processes द्वारा processor utilization। Fair share scheduler में priority किस आधार पर calculate होती है?
(UGC NET C.S. June-2017 (Paper-III))
A. Only (A) and (B)
B. Only (A) and (C)
C. (A), (B), and (C)
D. Only (B) and (C)

Correct Answer: C

Explanation (EN): Fair-share scheduling considers all these criteria.

Explanation (HI): Fair-share scheduling में ये सभी criteria consider किए जाते हैं।

76. One of the disadvantages of user level threads compared to Kernel level threads is
Kernel level threads की तुलना में user level threads का एक disadvantage क्या है?
(UGC NET C.S. June-2017 (Paper-III))
A. If a user level thread of a process executes a system call, all threads in that process are blocked
B. Scheduling is application dependent
C. Thread switching doesn't require kernel mode privileges
D. The library procedures invoked for thread management in user level threads are local procedures

Correct Answer: A

Explanation (EN): In user level threads, a blocking system call by one thread can block all threads of the process.

Explanation (HI): User level threads में यदि एक thread blocking system call करे, तो process के सभी threads block हो सकते हैं।

77. Which statement is not correct about "init" process in Unix?
Unix के 'init' process के बारे में कौन-सा कथन सही नहीं है?
(UGC NET C.S. June-2017 (Paper-III))
A. It is generally the parent of the login shell
B. It has PID 1
C. It is the first process in the system
D. Init forks and execs a 'getty' process at every port connected to a terminal

Correct Answer: C

Explanation (EN): According to the given source, saying init is the first process in the system is not correct.

Explanation (HI): दिए गए source के अनुसार यह कथन सही नहीं है कि init system का first process है।

78. The Bounded buffer problem is also known as_____
Bounded buffer problem को और किस नाम से जाना जाता है?
(UGC NET C.S. November-2017 (Paper-III))
A. Producer-consumer problem
B. Reader-writer problem
C. Dining philosophers problem
D. Both (a) and (b)

Correct Answer: A

Explanation (EN): The bounded buffer problem is also called the producer-consumer problem.

Explanation (HI): Bounded buffer problem को producer-consumer problem भी कहा जाता है।

79. Consider a file currently consisting of 50 blocks. Assume that the file control block and the index block is already in memory. If a block is added at the end (and the block information to be added is stored in memory), then how many disk I/O operations are required for indexed (single-level) allocation strategy?
एक file में वर्तमान में 50 blocks हैं। मान लें कि file control block और index block पहले से memory में हैं। यदि अंत में एक block जोड़ा जाए और जोड़े जाने वाले block की information भी memory में हो, तो indexed (single-level) allocation strategy में कितनी disk I/O operations आवश्यक होंगी?
(UGC NET C.S. August-2016 (Paper-III))
A. 1
B. 101
C. 27
D. 0

Correct Answer: A

Explanation (EN): Since both the file control block and index block are already in memory, only one disk I/O is needed to write the new block information.

Explanation (HI): जब file control block और index block पहले से memory में हों, तो नए block की information लिखने के लिए केवल एक disk I/O चाहिए।

80. 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?
एक operating system paged virtual memory को support करता है। Central processor का cycle time 1 microsecond है। Current page के अलावा किसी अन्य page को access करने में 1 microsecond extra लगता है। Pages में 1000 words हैं, और paging device एक drum है जो 3000 revolutions per minute पर घूमता है तथा 10 लाख words per second transfer करता है। कुल instructions में से 1% instructions current page के अलावा किसी दूसरे page को access करती हैं। इन instructions में से 80% ऐसे page को access करती हैं जो पहले से memory में है, और जब नया page चाहिए होता है तब replaced page 50% मामलों में modified होता है। यदि system केवल एक process चला रहा है और drum transfer के दौरान processor idle रहता है, तो effective access time क्या होगा?
(UGC NET C.S. August-2016 (Paper-III))
A. 30 microseconds
B. 34 microseconds
C. 60 microseconds
D. 68 microseconds

Correct Answer: B

Explanation (EN): Using the given probabilities and timing values, the effective access time works out to 34 microseconds.

Explanation (HI): दिए गए probabilities और timing values को उपयोग करने पर effective access time 34 microseconds आता है।

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