Operating System

51. Which of the following methods are used to pass any number of parameters to the operating system through system calls?
System calls के माध्यम से operating system को किसी भी संख्या में parameters pass करने के लिए कौन-से methods उपयोग किए जाते हैं?
(UGC NET C.S. 4 December 2019 (Paper-II))
A. Registers
B. Block or table in main memory
C. Stack
D. Block in main memory and stack

Correct Answer: D

Explanation (EN): Any number of parameters can be passed using a block in main memory or a stack.

Explanation (HI): किसी भी संख्या में parameters, main memory के block या stack की मदद से pass किए जा सकते हैं।

52. Database systems that store each relation in a separate operating system file may use the operating system's authorization scheme, instead of defining a special scheme themselves. In this case, which of the following is false?
यदि database system प्रत्येक relation को अलग operating system file में store करे और OS authorization scheme का उपयोग करे, तो निम्न में से कौन-सा कथन गलत है?
(UGC NET C.S. June-2018 (Paper-II))
A. The administrator enjoys more control on the grant option
B. It is difficult to differentiate among the update, delete and insert authorizations
C. Cannot store more than one relation in a file
D. Operations on the database are speeded up as the authorization procedure is carried out at the operating system level

Correct Answer: D

Explanation (EN): Using the operating system's authorization scheme does not necessarily speed up database operations in the way stated.

Explanation (HI): OS authorization scheme का उपयोग करने से यह आवश्यक नहीं कि database operations stated तरीके से faster हो जाएँ, इसलिए यह कथन false है।

53. Which of the following statements are true? (a) External Fragmentation exists when there is enough total memory space to satisfy a request but the available space is contiguous. (b) Memory Fragmentation can be internal as well as external. (c) One solution to external Fragmentation is compaction.
निम्न कथनों में कौन-से सही हैं? (a) External fragmentation तब होती है जब कुल memory पर्याप्त हो और available space contiguous हो। (b) Memory fragmentation internal भी हो सकती है और external भी। (c) External fragmentation का एक समाधान compaction है।
(UGC NET C.S. June-2018 (Paper-II))
A. (a) and (b) only
B. (a) and (c) only
C. (b) and (c) only
D. (a), (b) and (c)

Correct Answer: C

Explanation (EN): Statement (a) is false because external fragmentation occurs when total free memory is enough but it is not contiguous. Statements (b) and (c) are true.

Explanation (HI): Statement (a) गलत है क्योंकि external fragmentation तब होती है जब कुल free memory पर्याप्त हो लेकिन contiguous न हो। Statements (b) और (c) सही हैं।

54. Page information in memory is also called as Page Table. The essential contents in each entry of a page table is/are ______.
Page information in memory को Page Table कहते हैं। Page table की प्रत्येक entry का essential content क्या होता है?
(UGC NET C.S. June-2018 (Paper-II))
A. Page Access information
B. Virtual Page number
C. Page Frame number
D. Both virtual page number and Page Frame Number

Correct Answer: C

Explanation (EN): The essential content of a page table entry is the page frame number.

Explanation (HI): Page table entry का essential content page frame number होता है।

55. In which of the following scheduling criteria, context switching will never take place?
निम्न में से किस scheduling criterion में context switching नहीं होगा?
(UGC NET C.S. June-2018 (Paper-II))
A. ROUND ROBIN
B. Preemptive SJF
C. Non-preemptive SJF
D. Preemptive priority

Correct Answer: C

Explanation (EN): In non-preemptive SJF, once a process starts execution, it runs until completion, so preemptive context switches do not occur.

Explanation (HI): Non-preemptive SJF में process एक बार शुरू होने पर पूरा होने तक चलता है, इसलिए preemptive context switching नहीं होती।

56. Which of the following statements are CORRECT? (A) A process always check state of currently executing process to enter critical section. (B) Spin locks uses busy waiting. (C) Periodically testing a variable until some value appear is known as busy waiting. (D) Critical region is a part of program, where shared memory is kept. (E) Printer daemon, continuously checks to see if there are any file to be printed.
निम्न कथनों में कौन-से सही हैं? (A) कोई process critical section में प्रवेश के लिए हमेशा currently executing process की state check करता है। (B) Spin locks busy waiting का उपयोग करते हैं। (C) किसी variable को बार-बार test करना जब तक कोई value न मिल जाए, busy waiting कहलाता है। (D) Critical region program का वह भाग है जहाँ shared memory रखी जाती है। (E) Printer daemon लगातार check करता है कि print करने के लिए कोई file है या नहीं।
(UGC NET C.S. December-2023)
A. (A) and (B) Only
B. (B) and (C) Only
C. (B) and (D) Only
D. (B) and (E) Only

Correct Answer: C

Explanation (EN): According to the given source, the correct option is (B) and (D) only.

Explanation (HI): दिए गए source के अनुसार सही उत्तर केवल (B) और (D) है।

57. In a multiuser operating system, 20 request are made to use a particular resource per hour, on an average. The probability that no request is made in 45 minutes is
एक multiuser operating system में किसी resource के लिए औसतन 20 requests per hour आती हैं। 45 minutes में कोई request न आने की probability क्या होगी?
(UGC NET C.S. June-2023)
A. e^-15
B. e^-5
C. 1 - e^-5
D. 1 - e^-10

Correct Answer: A

Explanation (EN): Using Poisson distribution with rate 20 per hour, in 45 minutes the mean becomes 15, so P(0) = e^-15.

Explanation (HI): Poisson distribution में rate 20 per hour है, इसलिए 45 minutes के लिए mean 15 होगा और P(0) = e^-15।

58. Given below are two statements: one is labelled as Assertion A and the other is labelled as Reason R. Assertion A: A process involves a library function to create a thread. Reason R: The threads make system calls to convey their resource and I/O requirement to the Kernel. In the light of the above statements choose the correct answer from the options given below.
निम्न दो statements में Assertion A और Reason R दिए गए हैं। Assertion A: कोई process thread create करने के लिए library function का उपयोग करता है। Reason R: Threads अपनी resource और I/O requirements को kernel तक पहुँचाने के लिए system calls करती हैं। सही उत्तर चुनिए।
(UGC NET C.S. June-2023)
A. Both A and R are true and R is the correct explanation of A
B. Both A and R are true but R is NOT the correct explanation of A
C. A is true but R is false
D. A is false but R is true

Correct Answer: B

Explanation (EN): Both statements are true, but the reason does not correctly explain why a library function is used to create a thread.

Explanation (HI): दोनों statements सही हैं, लेकिन reason यह नहीं बताता कि thread create करने के लिए library function क्यों उपयोग होता है।

59. An OS follows Round-Robin Scheduling with time quantum of 4 ms. Assuming that the CPU is free now and there are 20 processes waiting in the ready queue, the maximum amount of time that a process waits before getting into the CPU is ______.
एक OS Round-Robin scheduling को 4 ms time quantum के साथ follow करता है। यदि CPU अभी free है और ready queue में 20 processes waiting में हैं, तो किसी process को CPU मिलने से पहले maximum waiting time कितना होगा?
(UGC NET C.S. December-2022)
A. 80 ms
B. 76 ms
C. 84 ms
D. None of the above

Correct Answer: B

Explanation (EN): The last process waits for 19 other processes × 4 ms = 76 ms.

Explanation (HI): Last process को 19 अन्य processes के 4 ms each के हिसाब से wait करना होगा, अर्थात 76 ms।

60. Consider an operating system capable of loading and executing a single sequential user process at a time. The disk head scheduling algorithm used is first come first served (FCFS). If FCFS is replaced by shortest seek time first (SSTF) and the vendor claims 50% better benchmark results. What is the expected improvement in the I/O performance of user programs?
एक operating system एक समय में केवल single sequential user process को load और execute कर सकता है। Disk head scheduling algorithm FCFS है। यदि FCFS को SSTF से replace किया जाए, तो user programs की I/O performance में expected improvement कितना होगा?
(UGC NET C.S. December-2022)
A. 50%
B. 100%
C. 25%
D. 0%

Correct Answer: D

Explanation (EN): With a single sequential user process, disk requests occur in sequence, so changing the disk scheduling algorithm gives no improvement.

Explanation (HI): जब केवल single sequential user process हो, तो disk requests sequence में आती हैं, इसलिए disk scheduling algorithm बदलने से improvement नहीं होगा।

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