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System calls, Processes, Threads, Inter‐process communication, Concurrency and synchronization. Deadlock. CPU scheduling. Memory management and Virtual memory. File systems. Disks is also under this

$$\scriptsize{\overset{{\large{\textbf{Mark Distribution in Previous GATE}}}}{\begin{array}{|c|c|c|c|c|c|c|c|}\hline
\textbf{Year}&\textbf{2024-1} & \textbf{2024-2} &\textbf{2023}&  \textbf{2022} & \textbf{2021-1}&\textbf{2021-2}&\textbf{Minimum}&\textbf{Average}&\textbf{Maximum}
\\\hline\textbf{1 Mark Count} &2&2&3& 2 &4&2&2&2.5&4
\\\hline\textbf{2 Marks Count} & 4&4&3&4 &1&3&1&3.16&4
\\\hline\textbf{Total Marks} & 10&10&9&10 &6&8&\bf{6}&\bf{8.83}&\bf{10}\\\hline
\end{array}}}$$

Recent questions in Operating System

#2961
593
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1 answers
2 votes
Why sjf is not a practical implementation ?
#2962
933
views
1 answers
1 votes
Consider a dynamic partitioning scheme.Show that on an avg. memory contains half as many holes as segments??i am not able to understand the concept of relation bw holes and segments in context to this question..
#2963
261
views
1 answers
1 votes
what is the meaning of "interrupts are asynchronous">
#2964
1.3k
views
1 answers
2 votes
In a demand paging memory system, page table is held in registers. The time taken to service a page fault is 8 m.sec. if an empty frame is ... maximum acceptable page-fault rate for an effective access time of no more than 200nanoseconds?
#2965
568
views
0 answers
0 votes
A process executes following segment of code:for(i = 1; i <= n; n++)fork();The number of new process created is 1. n2. ((n(n+1))/2)3. (2^n)-14. (3^n)-1
#2966
7.6k
views
0 answers
0 votes
Why paging suffer from internal fragmentation and why segmentation suffer from external fragmentation
#2967
229
views
0 answers
0 votes
var occupied, var blocked;Entry section:{ if(occupied) { blocked = blocked + 1; sleep(); } else ... .Please elaborate the correct working of this, for all the possible initial values of occupied and blocked.
#2968
1.4k
views
2 answers
2 votes
1. Progress Satisfied => Deadlock free2. Starvation free => Deadlock free3.Starvation freedom => Bounded waitPlease explain these implications if valid or invalid
#2969
464
views
2 answers
1 votes
Is user-level threads transparent to the kernel?
#2970
769
views
0 answers
1 votes
12.8 Suppose that a disk drive has 5,000 cylinders, numbered 0 to 4999. Thedrive is currently serving a request at cylinder 143, and the previousrequest was at ... disk-scheduling algorithms?a. FCFSb. SSTFc. SCANd. LOOKe. C-SCANf. C-LOOK
#2971
4.7k
views
6 answers
8 votes
If we have only one process in ready queue with burst time "m", then how many context switching will happen using round robing scheduling with time quantum ... that dispatching the process first time is not counted as a context switch.
#2972
341
views
0 answers
2 votes
Mode SwitchValidate the following statements:1. Kernel To User Mode Switch can be done by a Non Privileged Instruction. eg Return From Interrupt. True2. ... as exception. TruePlease validate these and help me understand if I am wrong.
#2973
4.7k
views
1 answers
3 votes
Consider the organization of a UNIX file as represented by the inodeAssume that there are 12 direct block pointers, and a singly, doubly, and triply indirectpointer ... disk accesses are required to access the byte in position 13,423,956?
#2974
422
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1 answers
0 votes
" In the case of LRU, ( and particularly the stack implementation thereof ), the top N pages of the stack will be the same for all frame set ... ?Please refer : https://www.cs.uic.edu/~jbell/CourseNotes/OperatingSystems/9_VirtualMemory.html
#2975
741
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1 answers
0 votes
Struct Semaphore{ int value; Queue type L;} Down (Semaphore S){ S.value = S.value -1 ... please correct me. I referred Tanenbaum and Galvin but didn't find any such explanation for Counting Semaphores)
#2976
871
views
1 answers
0 votes
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#2977
426
views
0 answers
0 votes
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#2978
726
views
1 answers
2 votes
Let the number of bits for Interrupt gate descriptor be 48, ie 32:16 (segment selector:offset). Each entry in the interrupt descriptor table is 5 Bytes. What will the total size of the IDT in bytes ?1. 12562. 10243. 10204. 1280
#2979
2.7k
views
1 answers
0 votes
Why logical address and physical address are same in case of compile and load time binding ?In complie time as i have read compiler will generate absoute address ... Logical address can be anything?Why it has ti be same as physical address?
#2980
1.5k
views
1 answers
2 votes
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