Operating System

171. Which of the following is a primary task of memory management system? I. It keeps track of which parts of the memory are currently being used. II. When space is available, it allocates memory to the programs that are next to be loaded.
Memory management system का primary task कौन-सा है? I. यह track करता है कि memory के कौन-कौन से भाग currently use हो रहे हैं। II. Space उपलब्ध होने पर यह अगली load होने वाली programs को memory allocate करता है।
(DSSSB TGT Computer Science 07.08.2021 (Shift-II))
A. Only I
B. Only II
C. Both I and II
D. Neither I nor II

Correct Answer: C

Explanation (EN): A memory management system must both track memory usage and allocate free memory to incoming programs.

Explanation (HI): Memory management system को memory usage track भी करना होता है और free memory को incoming programs को allocate भी करना होता है।

172. Memory buffer which stores the image definition is known as ______.
वह memory buffer जो image definition store करता है, क्या कहलाता है?
(DSSSB PGT Computer Science M/F 11.07.2021)
A. Pixel Buffer
B. Frame Buffer
C. Graphic Buffer
D. Picture Buffer

Correct Answer: B

Explanation (EN): A frame buffer stores the pixel data used to display an image on the screen.

Explanation (HI): Frame buffer में pixel data store होता है, जिसका उपयोग image display करने के लिए होता है।

173. A specific editor has 200 K of program text, 15 K of initial stack, 50 K of initialized data, and 70 K of bootstrap code. If five editors are started simultaneously, how much physical memory is needed if shared text is used?
एक specific editor में 200 K program text, 15 K initial stack, 50 K initialized data, और 70 K bootstrap code है। यदि पाँच editors एक साथ start किए जाएँ और shared text उपयोग किया जाए, तो कितनी physical memory चाहिए होगी?
(UGC NET Computer Science December 2014 (Paper-II))
A. 1135 K
B. 335 K
C. 1065 K
D. 320 K

Correct Answer: B

Explanation (EN): Using shared text, the required memory is 335 K according to the given solution.

Explanation (HI): दिए गए solution के अनुसार shared text उपयोग करने पर required memory 335 K होगी।

174. What is the most appropriate function of Memory Management Unit (MMU)?
Memory Management Unit (MMU) का सबसे उपयुक्त कार्य क्या है?
(UGC NET Computer Science June 2015 (Paper-III))
A. It is an associative memory to store TLB
B. It is a technique of supporting multiprogramming by creating dynamic partitions
C. It is a chip to map virtual address to physical address
D. It is an algorithm to allocate and deallocate main memory to a process

Correct Answer: C

Explanation (EN): The MMU is hardware that translates virtual addresses to physical addresses.

Explanation (HI): MMU एक hardware unit है जो virtual address को physical address में map करती है।

175. A dynamic RAM has refresh cycle of 32 times per msec. Each refresh operation requires 100 nsec and a memory cycle requires 250 nsec. What percentage of memory's total operating time is required for refreshes?
एक dynamic RAM का refresh cycle 32 times per msec है। प्रत्येक refresh operation में 100 nsec और एक memory cycle में 250 nsec लगते हैं। Memory के total operating time का कितने प्रतिशत refreshes में लगता है?
(UGC NET Computer Science December 2015 (Paper-III))
A. 0.64
B. 0.96
C. 2
D. 0.32

Correct Answer: D

Explanation (EN): The required percentage of operating time for refresh is 0.32%.

Explanation (HI): Refresh में लगने वाला operating time 0.32% है।

176. Consider a virtual page reference string 7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2, 1, 2, 0, 1, 7, 0, 1. Suppose a demand paged virtual memory system has 3 page frames. Which page replacement algorithm has minimum number of page faults?
Virtual page reference string 7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2, 1, 2, 0, 1, 7, 0, 1 पर विचार कीजिए। यदि demand paged virtual memory system में 3 page frames हों, तो किस page replacement algorithm में minimum page faults होंगे?
(UGC NET Computer Science November 2017 (Paper-III))
A. FIFO
B. LIFO
C. LRU
D. Optimal

Correct Answer: D

Explanation (EN): Among the listed algorithms, Optimal page replacement gives the minimum number of page faults.

Explanation (HI): दिए गए algorithms में Optimal page replacement सबसे कम page faults देता है।

177. Match List-I with List-II. List-I: A. Paging B. LRU (Least Recently Used) C. C-SCAN D. Virtual Memory. List-II: I. Evicts least recently used process II. Extends physical memory III. Logical to physical mapping IV. Circular disk access.
List-I और List-II का मिलान कीजिए। List-I: A. Paging B. LRU C. C-SCAN D. Virtual Memory. List-II: I. Evicts least recently used process II. Extends physical memory III. Logical to physical mapping IV. Circular disk access.
(UGC NET Computer Science Re-Exam June 2024)
A. A-III, B-IV, C-I, D-II
B. A-III, B-I, C-IV, D-II
C. A-I, B-III, C-IV, D-II
D. A-I, B-IV, C-III, D-II

Correct Answer: B

Explanation (EN): Paging maps logical to physical addresses, LRU evicts least recently used item, C-SCAN is circular disk access, and virtual memory extends physical memory.

Explanation (HI): Paging logical address को physical address से map करता है, LRU least recently used item को हटाता है, C-SCAN circular disk access है, और virtual memory physical memory को extend करती है।

178. 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।

179. 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।

180. 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।

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