Operating System Conceptsऑपरेटिंग सिस्टम अवधारणाएँ questions and answersप्रश्न और उत्तर

How operating system concepts worksऑपरेटिंग सिस्टम अवधारणाएँ को समझिए

A process has its own address space; threads of one process share it. Context switches between processes cost more than between threads.

Scheduling algorithms trade off waiting time and fairness: SJF minimises average waiting time, round robin gives every process regular turns.

Virtual memory lets a process use more memory than is physically present; too many page faults (thrashing) slow everything down.

Each trick says when to use it, why it works and when it fails. Formula tricks are checked by code over their whole range.हर ट्रिक बताती है कि उसे कब इस्तेमाल करें, वह क्यों काम करती है और कब नहीं। सूत्र वाली ट्रिक्स को कोड से उनकी पूरी रेंज पर जाँचा गया है।

Answers and full solutions are shown. Hide them to test yourself first.उत्तर और पूरे हल दिखाए गए हैं। पहले खुद को परखने के लिए इन्हें छिपाइए।

1. Example 1उदाहरण 1 · Operating System Concepts · Easyआसान

   Which of the following operations must be performed in kernel mode?

   1. Calling a C library function such as strcpy
   2. Updating the page table
   3. Executing a simple arithmetic expression in user code
   4. Reading a global variable from the process's data segment
   Answer:उत्तर: Updating the page table

   1. Identify operations that modify kernel data structures.
   2. Page tables are kernel‑only structures; updating them requires privilege.
   3. All other options are ordinary user‑mode actions.

2. Example 2उदाहरण 2 · Operating System Concepts · Mediumमध्यम

   Three processes arrive at time 0 with burst times 6 ms, 8 ms, and 4 ms. What is the average waiting time under non‑preemptive Shortest Job First (SJF) scheduling?

   1. 4.67 ms
   2. 5.33 ms
   3. 6.00 ms
   4. 3.33 ms
   Answer:उत्तर: 4.67 ms

   1. Sort the burst times ascending: 4, 6, 8.
   2. Waiting time for first (4) = 0; second (6) = 4; third (8) = 4+6 = 10.
   3. Average = (0 + 4 + 10) / 3 = 14/3 ≈ 4.67 ms.

3. Example 3उदाहरण 3 · Operating System Concepts · Hardकठिन

   A counting semaphore is initialized to 10. After 6 P (wait) operations and 4 V (signal) operations, what is the current value of the semaphore?

   1. 8
   2. 10
   3. 12
   4. 4
   Answer:उत्तर: 8

   1. Start with initial value: 10.
   2. P operation (wait) decrements the value: 10 - 6 = 4.
   3. V operation (signal) increments the value: 4 + 4 = 8.

4. Example 4उदाहरण 4 · Operating System Concepts · Mediumमध्यम

   Thrashing in a virtual memory system is best described as:

   1. The system spends most of its time swapping pages in and out rather than executing processes.
   2. The CPU utilization drops to zero because all processes are blocked on I/O.
   3. The page replacement algorithm selects the optimal page every time.
   4. The working set of each process fits entirely in physical memory.
   Answer:उत्तर: The system spends most of its time swapping pages in and out rather than executing processes.

   1. Recall thrashing = high paging activity, low CPU utilization.
   2. Match the description that mentions excessive swapping.

5. Example 5उदाहरण 5 · Operating System Concepts · Mediumमध्यम

   When a user program needs to read a file from the hard disk, which transition must occur?

   1. User mode to Kernel mode
   2. Kernel mode to User mode
   3. User mode to Hardware mode
   4. No mode change is required
   Answer:उत्तर: User mode to Kernel mode

   1. Identify that I/O is a privileged operation.
   2. User programs cannot access hardware directly.
   3. They must trigger a system call to switch to Kernel mode.

6. Example 6उदाहरण 6 · Operating System Concepts · Mediumमध्यम

   Thrashing in a virtual memory system is best described as:

   1. The system spends most of its time swapping pages in and out rather than executing processes.
   2. The CPU utilization drops to zero because all processes are blocked on I/O.
   3. The page replacement algorithm selects the optimal page every time.
   4. The working set of each process fits entirely in physical memory.
   Answer:उत्तर: The system spends most of its time swapping pages in and out rather than executing processes.

   1. Recall thrashing = high paging activity, low CPU utilization.
   2. Match the description that mentions excessive swapping.

Common mistakesआम गलतियाँ

• Thinking one deadlock condition is enough; all four must hold.
• Mixing up turnaround and waiting time.
• Forgetting that threads share heap memory.

Frequently asked questionsअक्सर पूछे जाने वाले प्रश्न