Operating System

1. In an operating system, a "buffer cache" is used to:
Operating system में buffer cache का उपयोग किसलिए किया जाता है?
(Bihar STET C.S. 12.09.2023 (Shift-I))
A. Store files in memory
B. Store copies of frequently used files in memory
C. Store the operating system kernel
D. Store input/output devices

Correct Answer: B

Explanation (EN): The buffer cache is a high-speed memory area where copies of frequently accessed files or data are kept. This reduces the number of disk accesses and improves system performance.

Explanation (HI): Buffer cache high-speed memory area होती है जहाँ frequently accessed files या data की copies रखी जाती हैं। इससे disk access कम होते हैं और system performance बढ़ती है।

2. _____ refers to the amount of time required to position the read-write head of a hard disk on appropriate sector.
Hard disk के read-write head को उचित sector पर लाने में लगने वाला समय क्या कहलाता है?
(KVS PGT C.S. 2017)
A. Load time
B. Seek time
C. Access time
D. Rotational latency

Correct Answer: B

Explanation (EN): Seek time is the time taken by the disk head to move from one track to another to reach the required sector.

Explanation (HI): Seek time वह समय है जो disk head को एक track से दूसरे track पर जाकर required sector तक पहुँचने में लगता है।

3. An operating system contains 3 user processes each requiring 2 units of resource R. The minimum number of units of R such that no deadlock will ever occur is:
एक operating system में 3 user processes हैं और प्रत्येक को resource R की 2 units चाहिए। R की न्यूनतम कितनी units हों कि deadlock कभी न हो?
(KVS PGT C.S. 2017)
A. 5
B. 2
C. 3
D. 4

Correct Answer: D

Explanation (EN): For n processes each requiring m units, the minimum number of resources to avoid deadlock is n(m−1)+1. Here it is 3(2−1)+1 = 4.

Explanation (HI): n processes और प्रत्येक को m units चाहिए हों, तो deadlock avoid करने के लिए minimum resources n(m−1)+1 होते हैं। यहाँ 3(2−1)+1 = 4।

4. The Kernel keeps track of the state of each executing program by using a data structure called _____.
Kernel प्रत्येक executing program की state को किस data structure की मदद से track करता है?
(KVS PGT C.S. 2017)
A. Process control block
B. User control block
C. File control block
D. Memory control block

Correct Answer: A

Explanation (EN): The kernel uses a Process Control Block (PCB) to store information about each process.

Explanation (HI): Kernel प्रत्येक process की information store करने के लिए Process Control Block (PCB) का उपयोग करता है।

5. To load a byte of data parallelly into a shift register with a synchronous load, there must be _____.
Synchronous load के साथ किसी shift register में एक byte data parallel load करने के लिए क्या आवश्यक है?
(KVS PGT C.S. 2018)
A. One clock pulse
B. One clock pulse for each 1 in the data
C. Eight clock pulses
D. One clock pulse for each 0 in the data

Correct Answer: A

Explanation (EN): In synchronous parallel load, the entire byte is loaded at once on a single clock pulse.

Explanation (HI): Synchronous parallel load में पूरा byte एक साथ केवल एक clock pulse पर load हो जाता है।

6. File operations that manipulate the 'open-count' maintained for each entry in open-file table include:
Open-file table में प्रत्येक entry के लिए रखे गए 'open-count' को manipulate करने वाले file operations कौन-से हैं?
(KVS PGT C.S. 2018)
A. Open, write
B. Read, write
C. Write, close
D. Open, close

Correct Answer: D

Explanation (EN): The open-count is incremented on open and decremented on close.

Explanation (HI): Open-count, file open करने पर बढ़ता है और close करने पर घटता है।

7. The process of removing deficiencies and loop holes in the data is called as _____.
Data में कमियाँ और loopholes दूर करने की प्रक्रिया क्या कहलाती है?
(KVS PGT C.S. 2018)
A. Data aggregation
B. Extraction of data
C. Compression of data
D. Cleaning of data

Correct Answer: D

Explanation (EN): Removing errors, inconsistencies and unwanted parts from data is called data cleaning.

Explanation (HI): Data से errors, inconsistencies और unwanted parts हटाने की प्रक्रिया data cleaning कहलाती है।

8. In a computer system, memory mapped access takes 100 nanoseconds when a page is found in TLB. In case the page is not in TLB, it takes 400 nanoseconds to access. Assuming a hit ratio of 80%, the effective access time is:
एक computer system में page TLB में मिलने पर memory mapped access 100 ns लेता है और TLB miss होने पर 400 ns लेता है। यदि hit ratio 80% है, तो effective access time क्या होगा?
(KVS PGT C.S. 2018)
A. 120 nanoseconds
B. 160 nanoseconds
C. 200 nanoseconds
D. 500 nanoseconds

Correct Answer: B

Explanation (EN): Effective access time = 0.8×100 + 0.2×400 = 160 nanoseconds.

Explanation (HI): Effective access time = 0.8×100 + 0.2×400 = 160 nanoseconds.

9. Both the first-fit and best-fit strategies for memory allocation suffer from
Memory allocation की first-fit और best-fit strategies किस समस्या से प्रभावित होती हैं?
(KVS PGT C.S. 2013)
A. External fragmentation
B. Internal fragmentation
C. 50-percent rule
D. Segmentation

Correct Answer: A

Explanation (EN): Both first-fit and best-fit leave small unusable holes in memory over time, causing external fragmentation.

Explanation (HI): First-fit और best-fit समय के साथ memory में छोटे unusable holes छोड़ती हैं, जिससे external fragmentation होती है।

10. The simplest, but most expensive, approach to introductory redundancy is duplicate to every disk. This technique is called
Redundancy के लिए सबसे सरल लेकिन महँगी तकनीक, जिसमें हर disk की duplicate copy रखी जाती है, क्या कहलाती है?
(KVS PGT C.S. 2013)
A. Swap space
B. Mirroring
C. Page slots
D. None of these

Correct Answer: B

Explanation (EN): Mirroring duplicates data on every disk to provide redundancy.

Explanation (HI): Mirroring में redundancy के लिए data की duplicate copy हर disk पर रखी जाती है।

12345678910111213141516171819202122Next