Operating System

81. 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।

82. User level threads are threads that are visible to the programmer and are unknown to the kernel. The operating system kernel supports and manages kernel level threads. Three different types of models relate user and kernel level threads. Which of the following statements is/are true? (A) (i) The many to one model maps many user threads to one kernel thread (ii) The one-to-one model maps one user thread to one kernel thread (iii) The many-to-many model maps many user threads to smaller or equal kernel threads (B) (i) Many-to-one model maps many kernel threads to one user thread (ii) One-to-one model maps one kernel thread to one user thread (iii) Many-to-many model maps many kernel threads to smaller or equal user threads
User level threads programmer को दिखाई देते हैं और kernel को ज्ञात नहीं होते। OS kernel, kernel level threads को support और manage करता है। User और kernel level threads के संबंध में निम्न statements में कौन-सा सही है? (A) (i) Many-to-one model में कई user threads एक kernel thread से map होते हैं (ii) One-to-one model में एक user thread एक kernel thread से map होता है (iii) Many-to-many model में कई user threads छोटे या बराबर संख्या के kernel threads से map होते हैं। (B) (i) Many-to-one model में कई kernel threads एक user thread से map होते हैं (ii) One-to-one model में एक kernel thread एक user thread से map होता है (iii) Many-to-many model में कई kernel threads छोटे या बराबर user threads से map होते हैं।
(UGC NET C.S. November-2017 (Paper-III))
A. (A) is true; (B) is false
B. (A) is false; (B) is true
C. Both (A) and (B) are true
D. Both (A) and (B) are false

Correct Answer: A

Explanation (EN): Statement set A correctly describes many-to-one, one-to-one and many-to-many models. Statement set B is false.

Explanation (HI): Statement set A many-to-one, one-to-one और many-to-many models को सही बताता है। Statement set B गलत है।

83. In an operating system, indivisibility of operation means:
Operating system में indivisibility of operation का क्या अर्थ है?
(UGC NET C.S. December-2015 (Paper-III))
A. Operation is interruptible
B. Race condition may occur
C. Processor cannot be pre-empted
D. All of the above

Correct Answer: C

Explanation (EN): An indivisible or atomic operation must execute completely or not at all, so it cannot be pre-empted midway.

Explanation (HI): Indivisible या atomic operation या तो पूरी execute होगी या बिल्कुल नहीं होगी, इसलिए उसे बीच में pre-empt नहीं किया जा सकता।

84. Function of memory management unit is:
Memory Management Unit (MMU) का कार्य क्या है?
(UGC NET C.S. December-2015 (Paper-III))
A. Address translation
B. Memory allocation
C. Cache management
D. All of the above

Correct Answer: A

Explanation (EN): The primary function of MMU is translation of virtual addresses to physical addresses.

Explanation (HI): MMU का मुख्य कार्य virtual address को physical address में translate करना है।

85. Which of the following statements is not true for multi level feedback queue processor scheduling algorithm?
Multi Level Feedback Queue processor scheduling algorithm के लिए निम्न में से कौन-सा statement सही नहीं है?
(UGC NET C.S. June-2015 (Paper-III))
A. Queues have different priorities
B. Each queue may have different scheduling algorithm
C. Processes are permanently assigned to a queue
D. This algorithm can be configured to match a specific system under design

Correct Answer: C

Explanation (EN): In a multi level feedback queue, processes can move between queues; they are not permanently assigned.

Explanation (HI): Multi Level Feedback Queue में process queues के बीच move कर सकते हैं; वे किसी एक queue में permanently assigned नहीं रहते।

86. 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:
Paging scheme में यदि average process size x bytes हो, page size y bytes हो और प्रत्येक page table entry z bytes की हो, तो page table overhead और internal fragmentation loss को minimize करने वाली optimum page size क्या होगी?
(UGC NET C.S. December-2014 (Paper-II))
A. x/2
B. xz/2
C. √(2xz)
D. x/z^2

Correct Answer: C

Explanation (EN): Minimizing the sum of page table overhead and internal fragmentation leads to optimum page size y = √(2xz).

Explanation (HI): Page table overhead और internal fragmentation के कुल प्रभाव को minimize करने पर optimum page size y = √(2xz) मिलती है।

87. In a demand paging memory system, page table is held in registers. The time taken to service a page fault is 8 msec if an empty frame is available or if the replaced page is not modified, and it takes 20 msec 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?
Demand paging memory system में page table registers में held है। यदि empty frame उपलब्ध हो या replaced page modified न हो तो page fault service time 8 msec है, और यदि replaced page modified हो तो 20 msec है। यदि replaced page 70% समय modified होता है, तो page fault service करने का average access time क्या होगा?
(UGC NET C.S. December-2014 (Paper-II))
A. 11.6 msec
B. 16.4 msec
C. 28 msec
D. 14 msec

Correct Answer: B

Explanation (EN): Average time = 0.7 × 20 + 0.3 × 8 = 16.4 msec.

Explanation (HI): Average time = 0.7 × 20 + 0.3 × 8 = 16.4 msec.

88. A LRU page replacement is used with four page frames and eight pages. How many page faults will occur with the reference string 0172327103 if the four frames are initially empty?
चार page frames और आठ pages के साथ LRU page replacement उपयोग किया जाता है। यदि reference string 0172327103 है और सभी चार frames शुरुआत में empty हैं, तो कितने page faults होंगे?
(UGC NET C.S. June-2015 (Paper-II))
A. 6
B. 7
C. 5
D. 8

Correct Answer: B

Explanation (EN): Simulating LRU for the given reference string gives 7 page faults.

Explanation (HI): दिए गए reference string पर LRU simulate करने पर 7 page faults आते हैं।

89. Which of the following system call include 'wait for time' operation command in Operating System?
Operating System में निम्न में से किस system call category में 'wait for time' operation command शामिल होता है?
(DSSSB TGT C.S. 07.02.2024 (Shift-II))
A. Process Control
B. Device Management
C. File Management
D. Communications

Correct Answer: A

Explanation (EN): Waiting for a specific time is part of process control operations.

Explanation (HI): किसी process को निश्चित समय तक wait कराना process control operations का हिस्सा है।

90. Which of the following statement(s) is/are correct for context switching in Operating System? I. When a context switch occurs, the kernel saves the context of the old process in its PCB (Process Control Block). II. Context-switch time is pure overhead, because the system does no useful work while switching.
Context switching के बारे में निम्न statements में कौन-से सही हैं? I. Context switch होने पर kernel पुराने process का context उसके PCB में save करता है। II. Context-switch time pure overhead है क्योंकि switching के दौरान system कोई useful work नहीं करता।
(DSSSB TGT C.S. 07.02.2024 (Shift-II))
A. Neither I nor II
B. Only I
C. Only II
D. Both I and II

Correct Answer: D

Explanation (EN): Both statements are correct for context switching.

Explanation (HI): Context switching के संदर्भ में दोनों statements सही हैं।

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