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Saturday, 28 March 2020

OS MCQ With answers justified set 4

Operating systems MCQ questions set 4 for competitive exams with answers explained



Operating systems MCQ Set - 4



1. Consider a disk with the following characteristics; Find the size of each sector of the given disk.
• Number of surfaces: 8 (= 23)
• Number of tracks / surface: 512 K (= 219)
• Number of bytes / track: 8 MB (= 223 bytes)
• Number of sectors / track: 8 K (= 213)
• On-disk cache: 16 MB (= 224 bytes)
a. 1 KB
b. 2 KB
c. 4 KB
d. 8 KB
Answer:
Sector capacity = No. of bytes per track * no. of tracks per sector
                         = 223 * 1/213 = 223/213 = 210 bytes/sector

2. Suppose you had a computer that supported virtual memory and had 32-bit virtual addresses and 4KB (212 byte) pages. If a process actually uses 1024 (210) pages of its virtual address space, how much space would be occupied by the page table for that process if a single-level page table was used? Assume each page table entry occupies 4 bytes.
a. 2 MB
b. 4 MB
c. 8 MB
d. None of the above
Answer:
Number of pages in the address space = 232/212 = 220 pages.
Therefore, a single level page table would have 220 entries, where each entry is 4 bytes. Therefore,
Space required by the page table = 4 * 220 bytes = 4 MB.
This is independent of how many pages are actually being used.

3. A CPU burst is
a. an example of priority inversion where a low priority process gets access to the CPU.
b. a temporary increase in the priority of a process to ensure that it gets the CPU.
c. an unexpected increase in a process' need for computation.
d. the period of time that a process uses the processor between I/O requests.
Answer:
A CPU burst is the period of time that a process is executing instructions before it gets to a waiting state on some event (usually I/O but also things like messages or semaphores).

4. Compared to a non-preemptive scheduler, a preemptive scheduler can move processes from the:
a. running to the blocked state.
b. ready to the running state.
c. blocked to the ready state.
d. running to the ready state.
Answer:
A preemptive scheduler can stop a process from running and have
another process run.

5. The main advantage of DMA is that it
a. Increases the speed of the CPU
b. Increases the speed of the data bus
c. Increases the performance of the system by allowing more things to happen at once
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