Operating System

151. Identify the circumstances under which pre-emptive CPU scheduling is used: (A) A process switches from Running state to Ready state (B) A process switches from Waiting state to Ready state (C) A process completes its execution (D) A process switches from Ready to Waiting state
बताइए कि किन परिस्थितियों में pre-emptive CPU scheduling उपयोग होती है: (A) Process Running से Ready state में जाए (B) Process Waiting से Ready state में जाए (C) Process execution complete करे (D) Process Ready से Waiting state में जाए
(UGC NET C.S. 4 December 2019 (Paper-II))
A. (A) and (B)
B. (A) and (D)
C. (C) and (D)
D. (A), (B), (C) only

Correct Answer: A

Explanation (EN): Pre-emptive scheduling is associated with situations like Running to Ready and Waiting to Ready.

Explanation (HI): Pre-emptive scheduling Running से Ready और Waiting से Ready जैसी situations में लागू होती है।

152. The problem of indefinite blockage of low-priority jobs in general priority scheduling algorithm can be solved using:
General priority scheduling algorithm में low-priority jobs की indefinite blockage की समस्या का समाधान किससे होता है?
(KVS PGT C.S. 2017)
A. Swapping
B. Dirty bit
C. Aging
D. Compaction

Correct Answer: C

Explanation (EN): Aging gradually increases priority of waiting jobs to avoid starvation.

Explanation (HI): Aging waiting jobs की priority धीरे-धीरे बढ़ाकर starvation को रोकता है।

153. A free list contains three memory areas of sizes 6 KB, 15 KB and 12 KB. The next three memory requests are for 10 KB, 2 KB and 14 KB. Which memory allocation strategy would be able to accommodate all the three requests?
एक free list में 6 KB, 15 KB और 12 KB के तीन memory areas हैं। अगली तीन requests 10 KB, 2 KB और 14 KB की हैं। कौन-सी memory allocation strategy तीनों requests को accommodate कर पाएगी?
(DSSSB PGT C.S. M/F 11.07.2021)
A. Worst Fit
B. Best Fit
C. Next Fit
D. First Fit

Correct Answer: B

Explanation (EN): Best Fit allocates the smallest suitable block each time and can accommodate all three requests here.

Explanation (HI): Best Fit प्रत्येक request के लिए सबसे छोटा suitable block चुनती है और यहाँ तीनों requests को accommodate कर सकती है।

154. If there are 64 segments, each of size 2KB, then how many bits should a logical address have?
यदि 64 segments हों और प्रत्येक का size 2KB हो, तो logical address में कितने bits होने चाहिए?
(DSSSB PGT C.S. (Male) 27.07.2018 (Shift-II))
A. 11 bits
B. 6 bits
C. 5 bits
D. 17 bits

Correct Answer: D

Explanation (EN): Segment number needs 6 bits and offset needs 11 bits, so total logical address bits = 17.

Explanation (HI): Segment number के लिए 6 bits और offset के लिए 11 bits चाहिए, इसलिए logical address = 17 bits।

155. When we apply Shortest Job First (SJF) scheduling policy/algorithm, Process (P2) faces partial starvation and waits for the longer time to execute. Which one of the following algorithms addresses this problem?
जब हम Shortest Job First (SJF) scheduling algorithm लागू करते हैं, तो Process P2 को partial starvation का सामना करना पड़ता है और उसे execute होने के लिए अधिक समय तक प्रतीक्षा करनी पड़ती है। निम्न में से कौन-सा algorithm इस समस्या को संबोधित करता है?
(UGC NET Computer Science December 2024)
A. Round Robin
B. Priority
C. Highest Response Ratio Next
D. Least Completed Next

Correct Answer: C

Explanation (EN): Highest Response Ratio Next (HRRN) addresses starvation by increasing a process's response ratio as its waiting time increases.

Explanation (HI): Highest Response Ratio Next (HRRN) waiting time बढ़ने पर process का response ratio बढ़ाता है, इसलिए starvation की समस्या कम हो जाती है।

156. What is the average turnaround time for these processes with First Come First Serve (FCFS) scheduling algorithm?
First Come First Serve (FCFS) scheduling algorithm के साथ इन processes का average turnaround time क्या है?
(UGC NET Computer Science December 2024)
A. 5.3 ms
B. 6.4 ms
C. 7.0 ms
D. 8.2 ms

Correct Answer: C

Explanation (EN): Using FCFS with the given arrival and burst times, the turnaround times are 5, 6, 8, 8, and 8 ms. Their average is 7.0 ms.

Explanation (HI): दिए गए arrival time और burst time के साथ FCFS लगाने पर turnaround times 5, 6, 8, 8, और 8 ms आते हैं। इनका औसत 7.0 ms है।

157. Which one of the following is not a condition for process state transition from running state to wait state?
निम्न में से कौन-सी running state से waiting state में process state transition की condition नहीं है?
(DSSSB PGT Computer Science M/F 11.07.2021)
A. If a process requests an Input/Output operation
B. If a higher priority process arrives
C. If a lower priority process arrives
D. If a process waits for some action performed by another process

Correct Answer: C

Explanation (EN): Arrival of a lower priority process does not force the currently running process to go into waiting state.

Explanation (HI): किसी lower priority process के आने से currently running process waiting state में नहीं जाता।

158. Consider three processes P1, P2, P3, which require 24, 3, and 3 units of CPU time respectively. For time quantum of 4 units and Round-Robin scheduling policy, the sequence in which the processes be executed for each time quantum is __________.
तीन processes P1, P2, P3 पर विचार कीजिए, जिन्हें क्रमशः 24, 3 और 3 units CPU time चाहिए। यदि time quantum 4 units हो और Round Robin scheduling policy लागू हो, तो execution sequence क्या होगा?
(NVS PGT Computer Science 15.12.2022 (Shift-II))
A. P1, P2, P3, P1, P1, P1, P1, P1
B. P1, P2, P3, P1, P1, P1, P1
C. P1, P2, P3, P1, P1, P1
D. P2, P3, P1, P1, P1

Correct Answer: A

Explanation (EN): P1 gets the first quantum, then P2 and P3 finish in one quantum each. The CPU then keeps returning to P1 until it completes.

Explanation (HI): पहले P1 को quantum मिलता है, फिर P2 और P3 एक-एक quantum में पूरे हो जाते हैं। उसके बाद CPU बार-बार P1 को मिलता है जब तक वह पूरा न हो जाए।

159. In which of the following state, the process is waiting for processor?
निम्न में से किस state में process processor की प्रतीक्षा करता है?
(KVS PGT Computer Science 2017)
A. Running
B. New
C. Ready
D. Waiting

Correct Answer: C

Explanation (EN): In the Ready state, the process is prepared for execution and waits for CPU allocation.

Explanation (HI): Ready state में process execution के लिए तैयार रहता है और CPU मिलने की प्रतीक्षा करता है।

160. If the time quantum size is 2 units and there is only one job of 14 time units in a ready queue, the round-robin scheduling algorithm will cause _____ context switches.
यदि time quantum 2 units है और ready queue में केवल 14 time units का एक ही job है, तो round-robin scheduling algorithm _____ context switches करेगा।
(KVS PGT Computer Science 2017)
A. 8
B. 5
C. 6
D. 7

Correct Answer: C

Explanation (EN): The 14-unit job is divided into 7 quanta of 2 units each. Excluding the initial dispatch and final completion, there are 6 context switches.

Explanation (HI): 14 units के job को 2-2 units के 7 quanta में बाँटा जाएगा। प्रारंभ और अंत को छोड़कर 6 context switches होंगे।

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