Operating System

41. In a disk system with 100 cylinders, the request to access the cylinder occur in following sequence: 34, 4, 7, 19, 10, 83, 2, 6, 20, 15 Assuming that the head is currently at cylinder 50, then what is the time taken to satisfy all requests if it takes 1 second to move from one cylinder to another? (Shortest seek time first policy is used)
100 cylinders वाले disk system में cylinder access requests इस क्रम में आती हैं: 34, 4, 7, 19, 10, 83, 2, 6, 20, 15। यदि head वर्तमान में cylinder 50 पर है, तो Shortest Seek Time First policy के अनुसार सभी requests को पूरा करने में कितना समय लगेगा, यदि एक cylinder से दूसरे cylinder तक जाने में 1 second लगता है?
(DSSSB PGT C.S. M/F 11.07.2021)
A. 119 seconds
B. 129 seconds
C. 109 seconds
D. 139 seconds

Correct Answer: B

Explanation (EN): Using SSTF from cylinder 50, the head always moves to the nearest pending request. According to the given answer key, the total time required is 129 seconds.

Explanation (HI): SSTF में head हर बार सबसे निकट वाले pending request पर जाता है। दिए गए answer key के अनुसार कुल समय 129 seconds है।

42. A disk drive has 100 cylinders, numbered 0 to 99. Disk requests come to the disk driver for cylinders 12, 26, 24, 4, 42, 8 and 50 in that order. The driver is currently serving a request at cylinder 24. A seek takes 6 msec per cylinder moved. How much seek time is needed for shortest seek time first (SSTF) algorithm?
एक disk drive में 100 cylinders हैं, जिनकी numbering 0 से 99 तक है। Disk requests cylinders 12, 26, 24, 4, 42, 8 और 50 के लिए आती हैं। Driver वर्तमान में cylinder 24 पर request serve कर रहा है। यदि एक cylinder move करने में 6 msec लगते हैं, तो SSTF algorithm में कुल seek time कितना होगा?
(UGC NET C.S. June-2015 (Paper-II))
A. 0.984 sec
B. 0.396 sec
C. 0.738 sec
D. 0.42 sec

Correct Answer: D

Explanation (EN): Using SSTF, total head movement is 70 cylinders. Therefore total seek time = 70 × 6 msec = 420 msec = 0.42 sec.

Explanation (HI): SSTF के अनुसार कुल head movement 70 cylinders है। इसलिए कुल seek time = 70 × 6 msec = 420 msec = 0.42 sec होगा।

43. In direct file organization:-
Direct file organization में:
(RPSC C.S. 2016 (Paper-II))
A. A collision always results into an overflow
B. A collision never results into an overflow
C. A collision may or may not lead to an overflow
D. A collision never occurs

Correct Answer: C

Explanation (EN): In direct or hash file organization, when two records map to the same address, a collision occurs. Depending on the collision handling technique, it may or may not lead to overflow.

Explanation (HI): Direct या hash file organization में जब दो records एक ही address पर map करते हैं, तो collision होता है। Collision handling technique पर निर्भर करता है कि overflow होगा या नहीं।

44. Sector interleaving in disks is done by?
Disks में sector interleaving किसके द्वारा किया जाता है?
(RPSC C.S. 2016 (PAPER-I))
A. Disk Manufacturer
B. Disk Controller
C. Operating System
D. None of these

Correct Answer: C

Explanation (EN): Sector interleaving is managed to optimize disk access and, as per the given answer key, it is done by the operating system.

Explanation (HI): Sector interleaving disk access को optimize करने के लिए किया जाता है और दिए गए answer key के अनुसार यह operating system द्वारा किया जाता है।

45. In _____ allocation method for disk block allocation in a file system, insertion and deletion of blocks in a file is easy.
File system में disk block allocation की _____ method में file के blocks का insertion और deletion आसान होता है।
(UGC NET C.S. June-2015 (Paper-III))
A. Index
B. Linked
C. Contiguous
D. Bit Map

Correct Answer: B

Explanation (EN): In linked allocation, each block contains a pointer to the next block. So insertion and deletion are easy because only pointers need to be adjusted.

Explanation (HI): Linked allocation में प्रत्येक block अगले block का pointer रखता है। इसलिए insertion और deletion आसान होते हैं क्योंकि केवल pointers को बदलना पड़ता है।

46. ______ 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:
______ नियम system failure की स्थिति में, जब volatile information का loss हो जाए, log information की मात्रा को सीमित करने के लिए उपयोग किए जाते हैं।
(UGC NET C.S. December-2014 (Paper-III))
A. Write-ahead log
B. Check-pointing
C. Log buffer
D. Thomas

Correct Answer: B

Explanation (EN): Checkpointing saves the state of the system at intervals so that, after a failure, recovery can start from the latest checkpoint and less log information needs to be processed.

Explanation (HI): Checkpointing में system की state को समय-समय पर save किया जाता है ताकि failure के बाद recovery latest checkpoint से शुरू हो सके और कम log information process करनी पड़े।

47. K-mean clustering algorithm has clustered the given 8 observations into 3 clusters after Ist iteration as follows: C1 : {(3, 3), (5, 5), (7, 7)} C2 : {(0, 6), (6, 0), (3, 0)} C3 : {(8, 8), (4, 4)} What will be the Manhattan distance for observation (4, 4) from cluster centroid C1 in second iteration?
K-means clustering algorithm ने 8 observations को प्रथम iteration के बाद 3 clusters में बाँटा है: C1 : {(3, 3), (5, 5), (7, 7)} C2 : {(0, 6), (6, 0), (3, 0)} C3 : {(8, 8), (4, 4)}। दूसरे iteration में observation (4, 4) की cluster centroid C1 से Manhattan distance क्या होगी?
(UGC NET C.S. Exam-24 June-2019 (Paper-II))
A. 2
B. 2
C. 0
D. 18

Correct Answer: A

Explanation (EN): The centroid of C1 in the second iteration is (5,5). Manhattan distance from (4,4) to (5,5) is |5-4| + |5-4| = 2.

Explanation (HI): दूसरे iteration में C1 का centroid (5,5) होगा। (4,4) से (5,5) की Manhattan distance |5-4| + |5-4| = 2 है।

48. Which of the following statements is not true about disk-arm scheduling algorithms?
Disk-arm scheduling algorithms के बारे में निम्न में से कौन-सा statement true नहीं है?
(UGC NET C.S. June-2014 (Paper-III))
A. SSTF (shortest seek time first) algorithm increases performance of FCFS.
B. The number of requests for disk service are not influenced by file allocation method.
C. Caching the directories and index blocks in main memory can also help in reducing disk arm movements.
D. SCAN and C-SCAN algorithms are less likely to have a starvation problem.

Correct Answer: B

Explanation (EN): The number of requests for disk service can be influenced by the file allocation method. Hence option (b) is not true.

Explanation (HI): Disk service requests की संख्या file allocation method से प्रभावित हो सकती है। इसलिए option (b) सही नहीं है।

49. Which of the following File operation need not involve any actual I/O (Input/Output) and is also known as 'File Seek'?
निम्न में से कौन-सी file operation actual I/O को involve किए बिना भी हो सकती है और जिसे 'File Seek' भी कहा जाता है?
(DSSSB TGT C.S. 07.02.2024 (Shift-II))
A. Deleting a file
B. Repositioning within a file
C. Reading a file
D. Truncating a file

Correct Answer: B

Explanation (EN): Repositioning within a file changes the current file pointer position and may not require any actual disk I/O. This is known as file seek.

Explanation (HI): File के भीतर repositioning केवल current file pointer की position बदलती है और इसके लिए actual disk I/O आवश्यक नहीं भी हो सकता। इसे file seek कहा जाता है।

50. Which of the following statement is correct for a 'Volume Control block' in File-System Implementation?
File-System Implementation में 'Volume Control Block' के लिए निम्न में से कौन-सा statement सही है?
(DSSSB TGT C.S. 07.02.2024 (Shift-II))
A. It is used to organize the file
B. It has a unique identifier number to allow association with a directory entry
C. It contain the number of block in the partition
D. It can contain information needed by the system to boot an operating system from that volume

Correct Answer: C

Explanation (EN): A Volume Control Block stores metadata about a disk volume or partition, including information such as the number of blocks in the partition.

Explanation (HI): Volume Control Block disk volume या partition के बारे में metadata store करता है, जैसे partition में blocks की संख्या।