Operating Systems Interview Questions — Set 1

25 OS interview staples — processes vs threads, scheduling, deadlock, paging and virtual memory — each with an interview-ready explanation.

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Practice sheet: all 25 questions

Prefer reading first? Every question from this practice set, with the answer and a worked explanation. (Solving them in the practice set above earns XP — up to ≈150.)

Q1. What is a process?
- a. A program stored on disk
- b. A program in execution
- c. A compiled binary file
- d. A CPU scheduling algorithm
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Answer: B. A program in execution

A process is a program in execution — the program is passive code on disk; the process is that code loaded in memory with a program counter, registers, stack and heap. This one-line distinction opens most OS interviews.
Q2. Which structure does the OS use to store the state of each process (PID, program counter, registers, open files)?
- a. Page table
- b. Process Control Block (PCB)
- c. File Allocation Table
- d. Interrupt vector table
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Answer: B. Process Control Block (PCB)

The PCB is the kernel's per-process record: PID, process state, program counter, CPU registers, scheduling info, memory pointers and open-file handles. On a context switch, the CPU state is saved into and restored from the PCB.
Q3. What happens during a context switch?
- a. The CPU state of the running process is saved and another process's state is loaded
- b. A process is permanently terminated
- c. Memory is compacted to remove fragmentation
- d. The CPU switches from user code to a device driver
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Answer: A. The CPU state of the running process is saved and another process's state is loaded

A context switch saves the running process's registers and program counter into its PCB and loads the next process's saved state. It is pure overhead — the CPU does no useful user work during the switch — which is why too-frequent switching hurts throughput.
Q4. Threads of the same process share all of the following EXCEPT:
- a. Code section
- b. Heap memory
- c. Open files
- d. Stack and registers
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Answer: D. Stack and registers

Threads share the process's code, data, heap and open files, but each thread keeps its own stack and register set — that's what lets them execute independently. This shared/private split is the most-asked threads question in interviews.
Q5. Which of these CPU scheduling algorithms is non-preemptive?
- a. Round Robin
- b. Shortest Remaining Time First
- c. First Come First Served (FCFS)
- d. Preemptive Priority
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Answer: C. First Come First Served (FCFS)

FCFS runs each process to completion in arrival order — the CPU is never snatched away, so it's non-preemptive. Round Robin preempts on the time quantum, and SRTF/preemptive priority preempt whenever a better candidate arrives.
Q6. Which scheduling algorithm provably minimises average waiting time (when burst times are known)?
- a. First Come First Served
- b. Shortest Job First
- c. Round Robin
- d. Priority scheduling
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Answer: B. Shortest Job First

SJF is optimal for average waiting time: running short jobs first makes every queued process wait less. Its catch — burst times aren't known in advance in practice, and long jobs can starve — is the standard follow-up question.
Q7. Round Robin scheduling is characterised by:
- a. Running the longest job first
- b. Giving each process a fixed time quantum in turn
- c. Running processes strictly by priority
- d. Never preempting a running process
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Answer: B. Giving each process a fixed time quantum in turn

Round Robin cycles through the ready queue giving each process one time quantum — great for fairness and responsiveness in time-sharing systems. Quantum size is the trade-off: too small wastes time on context switches, too large degrades into FCFS.
Q8. Processes P1, P2, P3 arrive together in that order with burst times 24, 3 and 3 ms. Under FCFS, what is the average waiting time in ms?

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Answer: 17

Waiting times: P1 = 0, P2 = 24, P3 = 27 → average (0 + 24 + 27)/3 = 17 ms. This is the classic convoy effect — one long job in front inflates everyone's wait; scheduling P2 and P3 first would drop the average to 3 ms.
Q9. Which of the following is NOT one of the four necessary conditions for deadlock?
- a. Mutual exclusion
- b. Hold and wait
- c. Preemption
- d. Circular wait
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Answer: C. Preemption

The four Coffman conditions are mutual exclusion, hold-and-wait, NO preemption and circular wait. Preemption being possible actually breaks deadlock — the condition is its absence, and this inversion is exactly what the question tests.
Q10. The Banker's algorithm is used for:
- a. Deadlock detection
- b. Deadlock avoidance
- c. Deadlock recovery
- d. CPU scheduling
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Answer: B. Deadlock avoidance

The Banker's algorithm avoids deadlock by granting a resource request only if the resulting state is still "safe" — i.e. some order exists in which every process can finish. Avoidance decides before allocation; detection finds cycles after the fact.
Q11. A binary semaphore initialised to 1 is functionally most similar to:
- a. A mutex lock
- b. A condition variable
- c. A spin counter
- d. A message queue
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Answer: A. A mutex lock

A binary semaphore takes values 0/1 and enforces one-at-a-time access, just like a mutex. The interview-grade nuance: a mutex has an owner (only the locker may unlock), while any process may signal a semaphore.
Q12. A race condition occurs when:
- a. Two processes deadlock on the same resource
- b. The result depends on the unpredictable order in which processes access shared data
- c. The CPU runs faster than the memory bus
- d. A process exceeds its time quantum
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Answer: B. The result depends on the unpredictable order in which processes access shared data

A race condition is when correctness depends on interleaving — e.g. two threads doing x = x + 1 concurrently can lose one update because read-modify-write isn't atomic. The fix is mutual exclusion around the critical section.
Q13. Which of the following is NOT a requirement of a correct critical-section solution?
- a. Mutual exclusion
- b. Progress
- c. Bounded waiting
- d. Maximum CPU throughput
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Answer: D. Maximum CPU throughput

The three classical requirements are mutual exclusion, progress and bounded waiting. Throughput is a performance goal, not a correctness condition — a solution can be slow and still correct.
Q14. Paging eliminates which kind of fragmentation?
- a. Internal fragmentation
- b. External fragmentation
- c. Both internal and external
- d. Neither
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Answer: B. External fragmentation

Paging removes external fragmentation because any free frame can hold any page — no contiguous allocation needed. It still suffers internal fragmentation: on average half the last page of each process is wasted.
Q15. A system uses 4 KB pages with 32-bit logical addresses. How many bits are used for the page offset?

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Answer: 12

The offset must address every byte inside a page: 4 KB = 2¹² bytes → 12 offset bits, leaving 20 bits for the page number (2²⁰ pages). Page-table sizing questions all start from this split.
Q16. The main benefit of virtual memory is that it:
- a. Makes RAM physically larger
- b. Lets processes run without loading all of their pages into RAM
- c. Eliminates the need for disk storage
- d. Removes the need for CPU caches
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Answer: B. Lets processes run without loading all of their pages into RAM

Virtual memory lets each process see a large private address space while only its actively-used pages occupy RAM — the rest stay on disk until demanded. Programs larger than physical memory can run, and more processes fit in memory at once.
Q17. Belady's anomaly — more page frames causing MORE page faults — can occur with which replacement algorithm?
- a. LRU
- b. Optimal
- c. FIFO
- d. LFU
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Answer: C. FIFO

FIFO can fault more with extra frames because it evicts by arrival time, not by usefulness. LRU and Optimal are "stack algorithms" — the pages held with k frames are always a subset of those held with k+1 — so they can never exhibit the anomaly.
Q18. The LRU page-replacement algorithm evicts the page that:
- a. Was loaded into memory first
- b. Has not been used for the longest time
- c. Will not be used for the longest time in future
- d. Has the smallest page number
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Answer: B. Has not been used for the longest time

LRU evicts the least recently USED page, betting that the recent past predicts the near future (locality of reference). Option c describes the Optimal algorithm — unimplementable in practice since it needs future knowledge, but the benchmark LRU approximates.
Q19. Thrashing is the situation where:
- a. The CPU overheats under load
- b. The system spends more time paging than executing useful work
- c. Two processes deadlock on I/O
- d. The disk becomes completely full
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Answer: B. The system spends more time paging than executing useful work

Thrashing happens when processes don't have enough frames for their working sets, so every few instructions trigger a page fault — CPU utilisation collapses while the disk churns. The cure is reducing multiprogramming, not adding more processes.
Q20. In demand paging, a page fault occurs when:
- a. A process accesses a page that is not currently in physical memory
- b. A page contains a division-by-zero error
- c. Two processes access the same page
- d. The page table becomes full
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Answer: A. A process accesses a page that is not currently in physical memory

A page fault is a hardware trap raised when the referenced page's valid bit is off — the page isn't in RAM. The OS then finds the page on disk, loads it into a free frame (evicting one if needed) and restarts the instruction. It's a normal mechanism, not an error.
Q21. Fixed-size memory partitioning primarily suffers from:
- a. External fragmentation
- b. Internal fragmentation
- c. Thrashing
- d. Starvation
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Answer: B. Internal fragmentation

With fixed partitions, a process rarely fills its partition exactly — the unused space inside the allocated block is internal fragmentation. Variable partitioning flips the problem: allocations fit exactly, but free memory splinters into unusable external holes.
Q22. In a Unix-style file system, which piece of information is NOT stored in a file's inode?
- a. File size
- b. Owner and permissions
- c. The file's name
- d. Pointers to data blocks
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Answer: C. The file's name

The inode holds metadata — size, owner, permissions, timestamps, block pointers — but NOT the name. Names live in directory entries that map name → inode number, which is exactly why one file can have several hard links (names).
Q23. What does fork() return in the child process?
- a. 0
- b. The child's PID
- c. The parent's PID
- d. -1
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Answer: A. 0

fork() returns 0 in the child and the child's PID in the parent (or −1 on failure) — one call, two returns, and the return value is how the code tells which process it's in. The child can fetch its parent's PID with getppid().
Q24. Which of the following scheduling algorithms are preemptive? (Select all that apply.)
- a. Round Robin
- b. Shortest Remaining Time First
- c. First Come First Served
- d. Non-preemptive Shortest Job First
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Answer: A. Round Robin · B. Shortest Remaining Time First

Round Robin preempts on every quantum expiry, and SRTF preempts whenever a shorter-remaining job arrives. FCFS and non-preemptive SJF let the running process finish — the CPU is only released voluntarily.
Q25. In which CPU mode does the kernel execute system-call code?
- a. User mode
- b. Kernel (supervisor) mode
- c. Safe mode
- d. Virtual mode
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Answer: B. Kernel (supervisor) mode

A system call traps from user mode into kernel mode, where privileged instructions and hardware access are allowed, then returns to user mode. This dual-mode design is the hardware foundation of OS protection — user code can never touch devices directly.