Operating System

181. Which of the statement is not true in regards to virtual memory?
Virtual memory के संबंध में कौन-सा statement सही नहीं है?
(UGC NET Computer Science December 2022)
A. The main objective for using virtual memory is to increase the effective capacity of the memory system
B. Size of the program can be as large as the size of the secondary memory
C. Program and data are stored in the secondary memory
D. None of these

Correct Answer: D

Explanation (EN): All the given statements are considered true in the provided explanation, so 'None of these' is correct.

Explanation (HI): दिए गए explanation के अनुसार सभी statements सही हैं, इसलिए 'None of these' सही उत्तर है।

182. The memory size for n address lines and m data lines is given by
n address lines और m data lines के लिए memory size किससे दी जाती है?
(UGC NET Computer Science December 2022)
A. 2^m × n
B. m × n^2
C. 2^n × m
D. n × m^2

Correct Answer: C

Explanation (EN): Memory size = 2^n × m, where n is the number of address lines and m is the number of data lines.

Explanation (HI): Memory size = 2^n × m होती है, जहाँ n address lines की संख्या है और m data lines की संख्या है।

183. Dirty bit is used to show the
Dirty bit किसे दिखाने के लिए उपयोग होती है?
(UGC NET Computer Science December 2018 (Paper-II))
A. Page with low frequency occurrence
B. Wrong page
C. Page that is modified after being loaded into cache memory
D. Page with corrupted data

Correct Answer: C

Explanation (EN): A dirty bit indicates that the page or block has been modified after loading into memory/cache.

Explanation (HI): Dirty bit यह दिखाती है कि page या block memory/cache में load होने के बाद modify हुआ है।

184. Let a computer has 512 KB of main memory having one byte per word. If all the words of memory are operational then there are minimum _____ bits in _____.
मान लीजिए एक computer में 512 KB main memory है और प्रति word 1 byte है। यदि memory के सभी words operational हैं, तो minimum _____ bits _____ में होंगी।
(RPSC Computer Science 2016 (Paper-I))
A. 18, Data bus
B. 19, Data bus
C. 18, Address bus
D. 19, Address bus

Correct Answer: D

Explanation (EN): 512 KB = 2^19 bytes, so 19 address bits are needed. Hence the answer is 19 bits in the address bus.

Explanation (HI): 512 KB = 2^19 bytes, इसलिए 19 address bits चाहिए। अतः उत्तर है address bus में 19 bits।

185. What is compaction?
Compaction क्या है?
(BPSC TRE 3.0 22.07.2024 (6-10))
A. A technique for overcoming internal fragmentation
B. A paging technique
C. A technique for overcoming external fragmentation
D. More than one of the above

Correct Answer: C

Explanation (EN): Compaction is used to overcome external fragmentation by combining scattered free spaces into one contiguous block.

Explanation (HI): Compaction external fragmentation को दूर करने के लिए scattered free spaces को एक contiguous block में बदलती है।

186. In virtual memory paging, page size and frame size have which of the following relations?
Virtual memory paging में page size और frame size के बीच कौन-सा relation होता है?
(LT Grade Computer Science 2018)
A. Both are of equal size
B. Both are of unequal size
C. The page size is exactly half of the frame size
D. The page size is twice the frame size

Correct Answer: A

Explanation (EN): In paging, page size and frame size must be equal so that pages can fit exactly into frames.

Explanation (HI): Paging में page size और frame size समान होते हैं ताकि pages frames में ठीक-ठीक fit हो सकें।

187. Sequential, direct, random and associative are access methods and key characteristics of computer ___ system.
Sequential, direct, random और associative access methods computer ___ system की key characteristics हैं।
(KVS PGT Computer Science 2013)
A. stack
B. counter
C. memory
D. core

Correct Answer: C

Explanation (EN): These are standard access methods of a computer memory system.

Explanation (HI): ये computer memory system के standard access methods हैं।

188. _____ are used to access data on secondary, sequential-access stores, such as disks and tapes.
Disks और tapes जैसे secondary, sequential-access stores पर data access करने के लिए _____ का उपयोग किया जाता है।
(KVS PGT Computer Science 2013)
A. Sequences
B. Arrays
C. Records
D. Registers

Correct Answer: A

Explanation (EN): According to the given source, sequences are used for such sequential-access storage.

Explanation (HI): दिए गए source के अनुसार ऐसे sequential-access storage पर sequences का उपयोग होता है।

189. The mechanism that brings a page into memory only when it is needed is called ______.
वह mechanism जो किसी page को memory में तभी लाता है जब उसकी आवश्यकता होती है, क्या कहलाता है?
(KVS PGT Computer Science 2017)
A. Page replacement
B. Segmentation
C. Fragmentation
D. Demand paging

Correct Answer: D

Explanation (EN): Demand paging loads a page into memory only when it is actually required.

Explanation (HI): Demand paging किसी page को memory में तभी लाती है जब वास्तव में उसकी आवश्यकता होती है।

190. Which of the following techniques allows execution of programs larger than the size of physical memory?
निम्न में से कौन-सी technique physical memory के size से बड़े programs के execution की अनुमति देती है?
(KVS PGT Computer Science 2017)
A. Thrashing
B. DMA
C. Buffering
D. Demand paging

Correct Answer: D

Explanation (EN): Demand paging allows execution of programs larger than physical memory by loading pages only when needed.

Explanation (HI): Demand paging आवश्यकता अनुसार pages को load करके physical memory से बड़े programs को भी execute करने देती है।