अपरेटिङ सिस्टम – Operating System MCQ
अपरेटिङ सिस्टम – कम्प्युटर र IT
Operating System · Computer & IT (Technical) · ४०९ प्रश्न
४०९ प्रश्न उपलब्ध। मोड छान्नुहोस्:
अपरेटिङ सिस्टम: प्रश्न र उत्तर
सही लागेको विकल्पमा थिच्नुहोस्। सही भए हरियो, गलत भए रातो देखिन्छ; ३ सेकेन्डपछि सही उत्तर र व्याख्या देखिन्छ।
सही: ० / ०
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The set of pages that a process has referenced in its most recent Δ (delta) page references is called its:
सही उत्तर: घ) Working set
The working-set model uses the pages referenced in a recent window to estimate how many frames a process needs and to prevent thrashing.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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For the reference string 1, 2, 3, 2, 1, 5, 2, 1, 6, 2, 5, 6, 3, 1, 3 with 3 frames (initially empty), how many page faults does LRU cause?
सही उत्तर: ख) 8
Faults occur on the references to 1, 2, 3, 5, 6, 5, 3 and 1, i.e. 8 page faults; all other references are hits.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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For the reference string 7, 0, 1, 2, 0, 3, 0, 4, 2, 3, 0, 3, 2, 1, 2, 0, 1, 7, 0, 1 with 3 frames (initially empty), how many page faults does the Optimal (OPT) algorithm cause?
सही उत्तर: क) 9
Replacing the page whose next use is farthest in the future gives 9 faults; LRU gives 12 and FIFO gives 15 for this string.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Copy-on-write, used after fork(), means that:
सही उत्तर: क) Parent and child initially share pages, and a page is copied only when one of them modifies it
Copy-on-write makes fork() fast because only pages that are actually modified ever get duplicated.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Demand paging means that:
सही उत्तर: घ) A page is brought into memory only when it is referenced
In demand paging a lazy swapper loads a page only when the process actually accesses it, causing a page fault the first time.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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Memory access time is 200 ns, the average page-fault service time is 8 ms and the page-fault rate is 0.001. The effective access time is approximately:
सही उत्तर: ग) 8.2 microseconds
EAT = (1 - 0.001) × 200 + 0.001 × 8,000,000 ns = 199.8 + 8000 = 8199.8 ns ≈ 8.2 µs.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In a paged system, the hardware register that points to the page table of the currently running process is the:
सही उत्तर: ख) Page-Table Base Register (PTBR)
The PTBR holds the starting address of the current process's page table; on a context switch only this register needs to be changed.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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A page table that has one entry for each physical frame of memory, rather than one for each virtual page, is called a(n):
सही उत्तर: क) Inverted page table
An inverted page table keeps one entry per physical frame (storing process ID and page number), greatly reducing table size.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In segmentation, if the offset in a logical address is greater than or equal to the segment's limit:
सही उत्तर: घ) A trap (addressing error) is raised to the operating system
The hardware compares the offset with the limit; an out-of-range offset causes a trap so the process cannot access memory outside its segment.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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In a segmentation system, segment 2 has base 4300 and limit 400. What is the physical address of the logical address (segment 2, offset 53)?
सही उत्तर: ग) 4353
Since offset 53 is less than the limit 400, physical address = base + offset = 4300 + 53 = 4353.
💬 टिप्पणी / गल्ती रिपोर्ट (०)
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