UGC NET Computer Science System Software and Operating System Previous Year Questions (PYQs)
Practise 156 System Software and Operating System questions asked in UGC NET Computer Science from 2018–2026. Questions cover System software, Process management, Threads, Classical synchronisation problems, Deadlock and Memory management. Every question is shown with its options and the correct answer, free to read.
156 PYQs2018–2026Answers includedFree
Practise System Software and Operating System as timed sets
Focused 10-question sets with instant scoring, explanations, and weak-area analysis.
Which technology enables the creation of virtual servers in a cloud-computing environment?
ADNS server
BHypervisor
CFirewall
DLoad balancer
Answer:(B) Hypervisor
Explanation
A hypervisor abstracts physical hardware so that several isolated virtual machines can run on a host. DNS, firewalls, and load balancers serve different network or traffic-management roles.
In rate-monotonic scheduling, a task has a period of 80 ms and an execution time of 55 ms. What are its processor utilisation and task rate?
A0.69, 12.5 Hz
B0.58, 12.3 Hz
C0.48, 11.4 Hz
D0.38, 10.5 Hz
Answer:(A) 0.69, 12.5 Hz
Explanation
Utilisation is execution time divided by period: 55/80 = 0.6875, approximately 0.69. The task rate is 1/0.08 seconds = 12.5 Hz. The other rates do not correspond to an 80 ms period.
I. Converts high-level code to object (machine) code II. Used for code expansion III. Used to translate assembly-language code to object code IV. Used to allocate memory and relocate a program in memory
AA-IV, B-III, C-I, D-II
BA-II, B-III, C-I, D-IV
CA-II, B-III, C-IV, D-I
DA-IV, B-I, C-III, D-II
Answer:(C) A-II, B-III, C-IV, D-I
Explanation
A macro is used for code expansion, an assembler translates assembly language to object code, a loader allocates memory and relocates a program, and a compiler translates high-level code. Therefore A-II, B-III, C-IV, D-I is correct.
I. Number of tasks completed in a given time period II. Causes the current activity to be deferred III. Time elapsed before next activity starts IV. Time to complete a single task
AA-I, B-III, C-II, D-IV
BA-III, B-IV, C-I, D-II
CA-IV, B-II, C-III, D-I
DA-IV, B-III, C-I, D-II
Answer:(D) A-IV, B-III, C-I, D-II
Explanation
Latency is the time to complete one task, delay is the time before the next activity starts, throughput is the number of tasks completed per time period, and an interrupt defers the current activity. This gives A-IV, B-III, C-I, D-II.
For disk tracks numbered 0–49, the head starts at 15 and requests arrive for 4, 40, 11, 35, 7, and 14. Order these algorithms by total head movement in ascending order, assuming initial movement toward higher-numbered tracks when required.
A. FCFS
B. LOOK
C. SSTF
D. SCAN
AC, B, D, A
BB, A, C, D
CC, D, B, A
DD, B, C, A
Answer:(A) C, B, D, A
Explanation
SSTF travels 47 tracks, LOOK travels 61, SCAN travels 79, and FCFS travels 135. Thus the ascending order is SSTF, LOOK, SCAN, FCFS: C, B, D, A.
A. Disk scheduling B. Batch processing C. Time sharing D. Interrupt processing
I. Round robin II. SCAN III. LIFO IV. FIFO
AA-III, B-IV, C-II, D-I
BA-IV, B-III, C-II, D-I
CA-II, B-IV, C-I, D-III
DA-II, B-I, C-IV, D-III
Answer:(C) A-II, B-IV, C-I, D-III
Explanation
Disk scheduling uses SCAN, batch processing is handled in FIFO order, time sharing uses round robin, and interrupt processing uses LIFO nesting. Therefore the match is A-II, B-IV, C-I, D-III.
Which is not a characteristic of linked-list file allocation?
ANo space is lost to external disk fragmentation, apart from possible internal fragmentation in the final block
BReading files or random access is very fast
CThe data portion of a block is reduced because a pointer occupies some bytes
DThe disadvantages can be mitigated with a main-memory File Allocation Table (FAT)
Answer:(B) Reading files or random access is very fast
Explanation
Linked allocation follows pointers one block at a time, so direct/random access is slow. It avoids external fragmentation, reserves pointer space in each block, and FAT can retain the links in memory to improve traversal.
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 three frames, what is the hit ratio under optimal page replacement?
A11/20
B10/20
C9/20
D12/20
Answer:(A) 11/20
Explanation
Simulating OPT gives 11 hits in the 20 references. On each miss with full frames, OPT replaces the resident page whose next use is farthest in the future (or never occurs again), so the hit ratio is 11/20.
Which of the following is not an interprocess communication mechanism in UNIX?
APipes
BMessages
CShared memory
DSemaphore
Answer:(D) Semaphore
Explanation
Pipes, message queues, and shared memory transfer information between processes. A semaphore is principally a synchronization primitive: it coordinates access rather than carrying a message payload.
Consider the following set of processes, all of which arrive at time 0 in the order P1, P2, P3, P4 and P5, with the CPU burst time given in milliseconds. Based on the table, what is the average waiting time (in milliseconds) using priority scheduling? Assume a lower priority number indicates higher priority.
Process
Burst Time (ms)
Priority
P1
10
3
P2
1
1
P3
2
4
P4
1
5
P5
5
2
A8
B8.2
C8.3
D8.4
Answer:(B) 8.2
Explanation
In priority scheduling with a lower number meaning higher priority, the processes run in the order of their priority values rather than their arrival order.
Consider a disk queue with requests for I/O to blocks on cylinders 98, 183, 37, 122, 14, 124, 65, 67 in that order. The disk head is initially at cylinder 53.
Using SSTF scheduling, arrange the following cylinders in the order the head visits them.
A. 67
B. 37
C. 65
D. 98
E. 14
AA, B, C, D, E
BC, A, E, B, D
CC, A, B, E, D
DA, C, D, B, E
Answer:(C) C, A, B, E, D
Explanation
SSTF (Shortest Seek Time First) always moves the head to whichever pending request is nearest to its current position, not the original one.
Starting at 53, work through the queue {98, 183, 37, 122, 14, 124, 65, 67} step by step.
From 53, the nearest request is 65 (distance 12), so the head moves to 65.
From 65, the nearest remaining request is 67 (distance 2), so the head moves to 67.
From 67, the nearest remaining request is 37 (distance 30, versus 31 for 98), so the head moves to 37.
From 37, the nearest remaining request is 14 (distance 23), so the head moves to 14.
From 14, the nearest remaining request is 98 (distance 84), so the head moves to 98.
The head then continues on to 122, 124 and finally 183.
The full service order is 65, 67, 37, 14, 98, 122, 124, 183.
Picking out only the labelled cylinders in this order gives 65(C), 67(A), 37(B), 14(E), 98(D), which is C, A, B, E, D.
Assertion (A): Thrashing occurs when a process spends more time on paging than executing.
Reason (R): It happens when the degree of multiprogramming is too low and frames are insufficient.
ABoth A and R are correct and R is the correct explanation of A
BBoth A and R are correct but R is NOT the correct explanation of A
CA is correct but R is not correct
DA is not correct but R is correct
Answer:(C) A is correct but R is not correct
Explanation
Thrashing occurs when a system spends most of its time swapping pages in and out rather than executing processes, so Assertion A correctly describes the symptom.
Reason R claims this happens when the degree of multiprogramming is too low, but this is the opposite of the actual cause. Thrashing occurs when the degree of multiprogramming is increased too high, so that the combined working sets of all active processes exceed the number of available frames, forcing constant page faults.
Since A is true and R states an incorrect cause, the correct choice is that A is correct but R is not correct.
Match List-I (OS and distributed-systems terms) with List-II.
List-I
List-II
A. System Call
I. A process requests a function that is located in a different computer system.
B. Preemption
II. A request that a program makes to the operating system.
C. Real time application
III. The forced deallocation of the CPU from a process.
D. Remote procedure call
IV. A program that responds to activities in an external system within a maximum time determined by that external system.
AA-I, B-II, C-III, D-IV
BA-II, B-III, C-IV, D-I
CA-IV, B-III, C-II, D-I
DA-II, B-III, C-I, D-IV
Answer:(B) A-II, B-III, C-IV, D-I
Explanation
A system call is a request that a program makes to the operating system, so A pairs with II.
Preemption is the forced deallocation of the CPU from a running process, so B pairs with III.
A real-time application must respond to activities in an external system within a maximum time set by that external system, so C pairs with IV.
A remote procedure call lets a process invoke a function located on a different computer system, so D pairs with I.
A-II, B-III, C-IV, D-I is therefore the correct match.
Which of the following is correct about Real time operating system?
A. It permits creation of multiple processes within an application.
B. It does not permit priorities to be assigned to processes.
C. It permits a programmer to define interrupts and interrupt processing routines.
D. It does not use priority-driven or deadline-oriented scheduling.
E. It provides fault tolerance and graceful degradation capabilities.
AA, B & D Only
BB, C & D Only
CA, C & E Only
DB, D & E Only
Answer:(C) A, C & E Only
Explanation
A real-time operating system (RTOS) is built to guarantee timely, predictable responses to events, and checking each statement against that goal resolves the set.
A. It permits creation of multiple processes within an application. — correct, an RTOS supports multitasking so an application can split its work across several cooperating processes or tasks.
B. It does not permit priorities to be assigned to processes. — wrong, priority assignment is central to an RTOS since tasks with tighter deadlines must be able to preempt less urgent ones.
C. It permits a programmer to define interrupts and interrupt processing routines. — correct, RTOS kernels expose interrupt service routine hooks so time-critical events can be handled directly.
D. It does not use priority-driven or deadline-oriented scheduling. — wrong, priority-driven and deadline-oriented scheduling (for example rate-monotonic or earliest-deadline-first) are the defining scheduling techniques of an RTOS.
E. It provides fault tolerance and graceful degradation capabilities. — correct, RTOS designs for critical systems build in fault tolerance so the system degrades gracefully instead of failing outright.
What does a superblock object represent in the virtual file system of a Linux OS?
Aan open file
Ban entire file system
Cindividual directory entry
Dan individual file
Answer:(B) an entire file system
Explanation
The Linux Virtual File System (VFS) uses four principal object types, and the superblock is the one that represents a whole mounted file system, holding metadata such as its type, size, and status.
An open file is represented by a file object, an individual file's metadata is held in an inode object, and a directory entry is represented by a dentry object, each of which is distinct from the superblock.
Because the superblock stores information describing an entire mounted file system, an entire file system is the correct answer.
Using non-preemptive SJF and Round Robin with time quantum 10, which process finishes second last in each schedule?
Process
Burst time
Arrival time
P1
9
0
P2
30
0
P3
4
0
P4
8
2
P5
11
6
ASJF: P4; RR: P4
BSJF: P5; RR: P4
CSJF: P4; RR: P3
DSJF: P5; RR: P5
Answer:(D) SJF: P5; RR: P5
Explanation
For non-preemptive SJF the finishing order is P1, P3, P4, P5, P2, so P5 is second last. With RR(10), the completion order is P1, P3, P4, P5, P2; again P5 is second last.
BThe standard defines an API only for thread creation.
CIt specifies thread behaviour but not an implementation.
DSolaris, Linux, and macOS, except Tru64 UNIX, implement Pthreads.
Answer:(C) It specifies thread behaviour but not an implementation.
Explanation
Pthreads is an interface specification: it defines the observable thread API and behaviour, while operating systems may implement it differently. It covers much more than creation, including synchronization and thread management. The standard number in option A is not the POSIX Threads designation.
Arrange Linux interrupt-protection levels from increasing priority: user-mode programs, preemptible kernel system-service routines, bottom-half interrupt handlers, and top-half interrupt handlers.
AB → A → C → D
BC → B → A → D
CC → A → B → D
DA → D → B → C
Answer:(D) A → D → B → C
Explanation
User-mode work is the most preemptible and therefore lowest priority. Preemptible kernel service routines rank above it, bottom halves above ordinary kernel work, and top-half interrupt handlers are highest because they run immediately in interrupt context.
Which sequence gives the hierarchy of a layered file system from application level down to I/O control?
AA → B → D → C → E
BD → E → C → A → B
CE → A → B → C → D
DE → C → B → A → D
Answer:(C) E → A → B → C → D
Explanation
Application programs invoke the logical file system, which manages directories and metadata. It calls the file-organization module, then the basic file system, which finally relies on I/O control for device operations. The sequence is E, A, B, C, D.
Arrange virtualization steps in a cloud environment: install a hypervisor, create virtual machines, allocate their resources, and run isolated workloads.
AB, C, A, D
BA, B, C, D
CA, C, B, D
DB, C, D, A
Answer:(B) A, B, C, D
Explanation
A hypervisor must be installed on the physical server before it can create VMs. Resources are then allocated to those VMs, which can subsequently run isolated workloads. The order is A, B, C, D.
Once a process is executing, which events could not occur: issuing I/O then joining the ready queue; creating a child and waiting for it; time-slice expiry then joining the waiting queue; or interrupt-driven removal then joining the waiting queue?
AB only
BD only
CA and C only
DA, C, and D only
Answer:(D) A, C, and D only
Explanation
An I/O request blocks the process, sending it to waiting rather than ready. Expiry of a time slice preempts it to ready, not waiting. An interrupt can also preempt it to ready; it does not by itself make the process wait. Waiting for a child is a valid blocking event, so A, C, and D are the impossible descriptions.
Which statement about spinlock semaphores is correct?
AThey are busy-waiting semaphores.
BThey are not useful when locks are held briefly.
CWaiting for one may require multiple context switches.
DThey are often employed on uniprocessor systems.
Answer:(A) They are busy-waiting semaphores.
Explanation
A spinlock repeatedly tests a lock while busy waiting. It is useful for very short critical sections because it can avoid sleep-and-wakeup overhead; waiting does not itself cause context switches, and spinning is generally unsuitable on a uniprocessor because the lock holder needs CPU time to run.
Match clustered page table, hierarchical page table, segmentation, and inverted page table with their characteristics.
List I
List II
A. Clustered Page Table
I. Generally inappropriate for 64-bit architectures
B. Hierarchical Page Table
II. One entry for each real page/frame
C. Segmentation
III. Useful for sparse address spaces
D. Inverted Page Table
IV. Supports a user view of the system
AA-III, B-IV, C-II, D-I
BA-III, B-I, C-IV, D-II
CA-II, B-III, C-I, D-IV
DA-IV, B-II, C-I, D-III
Answer:(B) A-III, B-I, C-IV, D-II
Explanation
Clustered page tables are useful for sparse address spaces. Large hierarchical page tables are generally inappropriate for 64-bit address spaces, segmentation supports the user's logical view, and an inverted page table has one entry per physical frame. The mapping is A-III, B-I, C-IV, D-II.
Using FCFS disk scheduling, what is the total head movement for request queue 45, 81, 185, 33, 175, 99, 150, 77 with initial head location 58?
A536
B594
C647
D700
Answer:(C) 647
Explanation
FCFS visits requests in the given order rather than choosing a nearby request. The movements are 13, 36, 104, 152, 142, 76, 51, and 73 cylinders. Their sum is 647 cylinders; the smaller alternatives would require reordering the queue, which FCFS never does.
For C-SCAN moving right from initial head location 58, what is the total head movement for queue 45, 81, 185, 33, 175, 99, 150, 77?
A327
B307
C525
D385
Answer:(D) 385
Explanation
The head services 77, 81, 99, 150, 175, and 185 while moving right, continues to the end cylinder 199, wraps to 0, then services 33 and 45. The movement is 127 to cylinder 185, then 14 to 199, 199 for the wrap, and 45 after wrapping: 385 cylinders.
For C-LOOK moving right from initial head location 58, what is the second request processed for queue 45, 81, 185, 33, 175, 99, 150, 77?
A81
B45
C77
D33
Answer:(A) 81
Explanation
C-LOOK services requests in the current direction only as far as the last requested cylinder. Moving right from 58, it processes 77 first and 81 second, so the second request is 81.
I. bit-interleaved parity II. disk mirroring III. block-interleaved parity IV. ECC organisation
AA-II, B-I, C-IV, D-III
BA-II, B-IV, C-I, D-III
CA-III, B-IV, C-I, D-II
DA-IV, B-III, C-II, D-I
Answer:(B) A-II, B-IV, C-I, D-III
Explanation
RAID 1 is disk mirroring, RAID 2 uses ECC, RAID 3 uses bit-interleaved parity, and RAID 4 uses block-interleaved parity. Therefore option B supplies every correct pairing. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
If virtual memory is 32 MB and physical memory is 4 MB with page size 2 KB, calculate the number of frames available in physical memory.
A1024
B2048
C3072
D4096
Answer:(B) 2048
Explanation
The number of frames equals physical-memory size divided by page size: 4 MB ÷ 2 KB = 4096 KB ÷ 2 KB = 2048 frames. Virtual-memory size does not alter this frame count. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
Which is the correct sequence for an interrupt service routine?
A. Save contents of processor registers
B. Turn the interrupt facility on
C. Service the device whose flag is set
D. Check which flag is set
E. Restore contents of processor registers
AA, B, C, D, E
BD, B, C, A, E
CD, C, B, E, A
DA, D, C, E, B
Answer:(B) D, B, C, A, E
Explanation
The official final key selects D, B, C, A, E. The question’s terse wording makes the precise enable/restore ordering ambiguous across processor designs, so this answer follows the authoritative key.
Which of the following does not interrupt a running process?
ADevice
BTimer
CScheduler
DPower failure
Answer:(C) Scheduler
Explanation
Devices, timers and power failures cause interrupts. A scheduler is operating-system software that makes scheduling decisions; it is not itself an interrupt source. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
The displayed disk-scheduling graph processes requests 28, 36, 45, 84, 125, 170 and 172, then wraps to the low end. Which algorithm does it represent?
ASSTF
BSCAN
CC-SCAN
DC-LOOK
Answer:(D) C-LOOK
Explanation
The head services requests in one direction only up to the last pending request, then jumps to the lowest pending request without travelling to the physical end of the disk. That behaviour is C-LOOK.
Given the displayed process burst times and priorities, which is a value of average waiting time under priority scheduling?
A9.2
B6.8
C7.6
D8.2
Answer:(D) 8.2
Explanation
Priority scheduling orders ready processes by their priority number before summing waiting times. Applying the source table’s priority order gives the keyed average waiting time of 8.2. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
The entire set of parameters, including the return address, stored for a procedure invocation is called
Astack frame
Bstack base
Cstack limit
Dstack record
Answer:(A) stack frame
Explanation
A stack frame (activation record) holds a procedure call’s parameters, return address, local data and saved execution context. Base and limit refer to stack bounds, not a single invocation. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
Consider the following processes, with arrival and burst times in milliseconds. What is the average waiting time using non-preemptive shortest-job-first scheduling?
Process
Arrival time
Burst time
P₁
0
5
P₂
2
3
P₃
2
2
P₄
5
3
P₅
6
1
A3.0 ms
B2.1 ms
C4.6 ms
D3.2 ms
Answer:(D) 3.2 ms
Explanation
The official final key selects 3.2 ms. The result is obtained by scheduling the shortest available process whenever the processor becomes free and averaging the resulting waiting times. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
Calculate the average response time with non-preemptive priority scheduling. Assume all processes arrive at time 0 and lower numbers mean higher priority.
Process
Burst time
Priority
P₁
5
3
P₂
3
1
P₃
2
5
P₄
3
4
P₅
1
2
A5.6 ms
B5.2 ms
C4.7 ms
D3.8 ms
Answer:(A) 5.6 ms
Explanation
The official final key selects 5.6 ms. Order the ready processes by ascending priority number, calculate each start time (response time), and average them. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
For the given arrival-time and burst-time process set, find the average weighted turnaround time using Highest Response Ratio Next (HRN).
Process
Arrival time
Burst time
P₁
0
5
P₂
2
3
P₃
2
2
P₄
5
3
P₅
6
1
A3.2 ms
B2.834 ms
C1.632 ms
D6.721 ms
Answer:(B) 2.834 ms
Explanation
The official final key selects 2.834 ms. HRN chooses the process with the largest (waiting time + service time) / service time ratio at each decision point, then averages turnaround time divided by burst time.
When shortest-job-first scheduling causes a process to face partial starvation and wait longer to execute, which algorithm addresses this problem?
ARound Robin
BPriority
CHighest Response Ratio Next
DLeast Completed Next
Answer:(A) Round Robin
Explanation
Round Robin gives every ready process a time quantum, preventing the indefinite postponement that can occur under shortest-job-first scheduling. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
For the given arrival-time and burst-time process set, what is the average turnaround time using first-come, first-served scheduling?
Process
Arrival time
Burst time
P₁
0
5
P₂
2
3
P₃
2
2
P₄
5
3
P₅
6
1
A5.3 ms
B6.4 ms
C7.0 ms
D8.2 ms
Answer:(C) 7.0 ms
Explanation
The official final key selects 7.0 ms. FCFS executes processes in arrival order; turnaround time is completion time minus arrival time, and the required value is their average. The correct option follows from the stated definition or calculation; the other options omit a necessary condition or describe a different concept.
Which of the following statements are TRUE
about mutual exclusion in concurrent
programming ?
A. Mutual exclusion ensures that only one
process can be in a critical section at any given
time.
B. Mutual exclusion are designed to prevent
conflicts and ensure that only one process can
access shared resources at a time.
C. Mutual exclusion can use various algorithms
to ensure that processes do not enter the critical
section simultaneously.
D. Mutual exclusion allows multiple processes to
access the critical section simultaneously to
improve performance.
A(A), (B), (C) Only
B(B), (C), (D) Only
C(B), (D), (A) Only
D(A), (C), (D) Only
Answer:(A) (A), (B), (C) Only
Explanation
Statements (A), (B) and (C) all restate the same guarantee — at most one process inside the critical section at a time, enforced by locks or algorithms such as Peterson’s. (D) contradicts the definition outright, since allowing simultaneous access is exactly what mutual exclusion prevents. Only (A), (B) and (C), so option A is correct.
Which of the following statements are TRUE
about Privileged Instructions ?
A. It can only be executed by the Operating
System kernel and not by user applications.
B. It is designed to perform operations that can
directly affect the hardware or system state such
as I/O operations or changing memory
management setting.
C. User applications can execute privileged
instructions if they have to correct permissions,
set by the Operating System.
D. It usually executed in user mode to ensure
the safety and security of the system.
A(A) and (B) Only
B(A), (B) and (C) Only
C(B) and (C) Only
D(B), (C) and (D) Only
Answer:(A) (A) and (B) Only
Explanation
(A) is true — privileged instructions execute only in kernel (supervisor) mode. (B) is true — they cover exactly the operations touching hardware or system state, such as I/O and memory-management setup. (C) is false: no permission setting lets a user-mode application execute one directly; it must trap into the kernel. (D) is false: they run in kernel mode, not user mode. Only (A) and (B) — option A.
Which of the following statements are TRUE
about Process Control Block (PCB) ?
A. The PCB contains information about the
process state, such as whether it is running,
waiting or terminated
B. The PCB includes the program code and data
segments of the process
C. The PCB stores the process's memory
management information, such as page tables
and segment tables
D. The PCB is used to track process scheduling
information and CPU registers for process
execution
A(A), (B) and (C) Only
B(B), (C) and (D) Only
C(A), (C) and (D) Only
D(A) and (B) Only
Answer:(C) (A), (C) and (D) Only
Explanation
The process control block records the process state (A), its memory-management data such as page and segment tables (C), and its scheduling information together with the saved CPU registers (D). The program’s code and data segments live in the process’s address space, not in the PCB, so (B) is false. Only (A), (C) and (D) — option C.
Which of the following is NOT a
requirement for Banker's algorithm to
grant a resource request ?
The Banker’s Algorithm is a critical deadlock avoidance method in operating systems, designed to facilitate resource allocation without causing deadlock. It operates by maintaining information about the maximum resources each process may require, the currently allocated resources and the available resources in the system. The algorithm checks each resource request to determine if granting it would leave the system in a safe state, meaning that there is always a sequence in which all processes can complete their execution without getting stuck due to resource unavailability. Each process must specify its maximum demand for each resource type before it starts execution. When a process requests additional resources, the algorithm checks if granting the request will keep the system in a safe state. If so, the resources are allocated; otherwise the process must wait until its request can be safely fulfilled.
AThe requested resources must be available.
BThe system must be in a safe state after granting the request
CThe request must not exceed the maximum resources the process can request.
DThe process must be the only one requesting resources.
Answer:(D) The process must be the only one requesting resources.
Explanation
Banker’s algorithm grants a request only when the resources are available, when the request stays within the process’s previously declared maximum, and when the resulting state is safe. It never requires a process to be the only one requesting resources — several processes hold and request concurrently. Option D is therefore the statement that is NOT a requirement.
Which data structure does the Banker's
Algorithm use to maintain the state of
available, maximum and allocated
resources ?
The Banker’s Algorithm is a critical deadlock avoidance method in operating systems, designed to facilitate resource allocation without causing deadlock. It operates by maintaining information about the maximum resources each process may require, the currently allocated resources and the available resources in the system. The algorithm checks each resource request to determine if granting it would leave the system in a safe state, meaning that there is always a sequence in which all processes can complete their execution without getting stuck due to resource unavailability. Each process must specify its maximum demand for each resource type before it starts execution. When a process requests additional resources, the algorithm checks if granting the request will keep the system in a safe state. If so, the resources are allocated; otherwise the process must wait until its request can be safely fulfilled.
APriority Queue
BHash table
CWait-for-Graph
DMatrices and Vectors
Answer:(D) Matrices and Vectors
Explanation
The algorithm tracks Available as a vector, and Maximum, Allocation and Need as matrices indexed by process and by resource type. Priority queues, hash tables and wait-for graphs appear elsewhere in deadlock handling but not in this algorithm. Option D is correct.
What is the significance of the Banker's
algorithm in terms of resource
management?
The Banker’s Algorithm is a critical deadlock avoidance method in operating systems, designed to facilitate resource allocation without causing deadlock. It operates by maintaining information about the maximum resources each process may require, the currently allocated resources and the available resources in the system. The algorithm checks each resource request to determine if granting it would leave the system in a safe state, meaning that there is always a sequence in which all processes can complete their execution without getting stuck due to resource unavailability. Each process must specify its maximum demand for each resource type before it starts execution. When a process requests additional resources, the algorithm checks if granting the request will keep the system in a safe state. If so, the resources are allocated; otherwise the process must wait until its request can be safely fulfilled.
AIt enures that all processes can finish their execution without deadlock.
BIt eliminates the need for processes to request resources.
CIt accelerates the execution of critical sections in processes.
DIt minimizes the number of context switches between processes.
Answer:(A) It enures that all processes can finish their execution without deadlock.
Explanation
Its significance is that by admitting only requests which leave the system in a safe state, it guarantees a sequence exists in which every process can run to completion — deadlock is avoided rather than merely detected afterwards. It does not remove the need to request resources, accelerate critical sections, or reduce context switches. Option A is correct.
What is the primary goal of the Banker's
Algorithm?
The Banker’s Algorithm is a critical deadlock avoidance method in operating systems, designed to facilitate resource allocation without causing deadlock. It operates by maintaining information about the maximum resources each process may require, the currently allocated resources and the available resources in the system. The algorithm checks each resource request to determine if granting it would leave the system in a safe state, meaning that there is always a sequence in which all processes can complete their execution without getting stuck due to resource unavailability. Each process must specify its maximum demand for each resource type before it starts execution. When a process requests additional resources, the algorithm checks if granting the request will keep the system in a safe state. If so, the resources are allocated; otherwise the process must wait until its request can be safely fulfilled.
ATo allocate resources optimally
BTo prevent processes from requesting resources
CTo detect and recover from deadlocks
DTo maximise CPU utilization
Answer:(A) To allocate resources optimally
Explanation
Banker’s algorithm is a deadlock-avoidance method: its goal is to allocate resources in the way that keeps the system in a safe state. Option C describes deadlock detection and recovery, which is a different strategy, while options B and D describe aims the algorithm does not pursue. Option A is correct.
What information is used to determine if
a resource request can be granted ?
The Banker’s Algorithm is a critical deadlock avoidance method in operating systems, designed to facilitate resource allocation without causing deadlock. It operates by maintaining information about the maximum resources each process may require, the currently allocated resources and the available resources in the system. The algorithm checks each resource request to determine if granting it would leave the system in a safe state, meaning that there is always a sequence in which all processes can complete their execution without getting stuck due to resource unavailability. Each process must specify its maximum demand for each resource type before it starts execution. When a process requests additional resources, the algorithm checks if granting the request will keep the system in a safe state. If so, the resources are allocated; otherwise the process must wait until its request can be safely fulfilled.
AAvailable resources and current allocation of each process
BCPU utilization of each process
CNumber of processes waiting for resources
DArrival time of each process
Answer:(A) Available resources and current allocation of each process
Explanation
The safety check compares the resources currently available against what each process still needs, which follows from its declared maximum and its current allocation. CPU utilisation, the length of the waiting queue and arrival times play no part in the decision. Option A is correct.
What is the output of the following program ? # include <stdio.h> # define SQR(x) (x*x) int main () {int a, b=3; a=SQR(b+2); printf("%d",a); return 0; }
A25
B11
CGarbage value
D24
Answer:(B) 11
Explanation
The key idea is stated in the problem: What is the output of the following program ? # include <stdio.h> # define SQR(x) (x*x) int main () {int a, b=3; a=SQR(b+2); printf("%d",a); return 0; }. The correct answer is B: 11. It follows the governing definition, algorithm, or calculation in this system software and operating system item; the other choices fail to satisfy one or more stated conditions.
The head of a moving head disk with 200 tracks, numbered 0 to 199, has just finished a request at track 125, and currently serving a request at track 143. The queue of requests is given in the FIFO order as 86, 147, 91, 177, 94, 150, 102, 175, 130. What will be the total number of head movements required to satisfy these requests for SCAN algorithm ?
A259 cylinders
B169 cylinders
C154 cylinders
D264 cylinders
Answer:(B) 169 cylinders
Explanation
The key idea is stated in the problem: The head of a moving head disk with 200 tracks, numbered 0 to 199, has just finished a request at track 125, and currently serving a request at track 143. The queue of requests is given in the FIFO order as 86, 147, 91, 177, 94, 150, 102, 175, 130. What will b. The correct answer is B: 169 cylinders. It follows the governing definition, algorithm, or calculation in this system software and operating system item; the other choices fail to satisfy one or more stated conditions.
Given as 4 GB (= 4.3 x 10° bytes) of virtual space and typical page size of 4 KB and each page table entry is 5 bytes. How many virtual pages would this imply ? What is the size of whole page table ?
A107500 and 20480 bytes
B215000 and 40960 bytes
C10750 and 10240 bytes
D43000 and 1024 bytes
Answer:(A) 107500 and 20480 bytes
Explanation
NTA's final answer key marks this question as dropped. The provisional option A is included only to satisfy the required import schema; do not score this item until an editor replaces it with an approved answer.
The key idea is stated in the problem: In Linux, where is the user password stored ?. The correct answer is C: /etc/passwd. It follows the governing definition, algorithm, or calculation in this system software and operating system item; the other choices fail to satisfy one or more stated conditions.
Indexed/ grouped allocation is useful as :
A. It supports both sequential and direct access.
B. Entire block is available for data.
C. It does not require lots of space for keeping pointers.
D. No external fragmentation. Choose the correct answer from the options given below :
A(A) Only
B(B) and (C) Only
C(B) Only
D(A), (B) and (D) Only
Answer:(D) (A), (B) and (D) Only
Explanation
The key idea is stated in the problem: Indexed/ grouped allocation is useful as :
A. It supports both sequential and direct access.
B. Entire block is available for data.
C. It does not require lots of space for keeping pointers.
D. No external fragmentation. Choose the correct answer from the . The correct answer is D: (A), (B) and (D) Only. It follows the governing definition, algorithm, or calculation in this system software and operating system item; the other choices fail to satisfy one or more stated conditions.
Which of the following statements are CORRECT ?
A. A process always check state of currently executing process to enter critical schema.
B. Spin locks uses busy waiting.
C. Periodically testing a variable until some value appear is known as busy waiting.
D. Critical region is a part of program, where shared memory is kept.
E. Printer daemon, continuously checks to see if there are any file to be printed. Choose the correct answer from the options given below :
A(A) and (B) Only
B(B) and (C) Only
C(B) and (D) Only
D(B) and (B) Only
Answer:(C) (B) and (D) Only
Explanation
The key idea is stated in the problem: Which of the following statements are CORRECT ?
A. A process always check state of currently executing process to enter critical schema.
B. Spin locks uses busy waiting.
C. Periodically testing a variable until some value appear is known as busy waiting.
D. The correct answer is C: (B) and (D) Only. It follows the governing definition, algorithm, or calculation in this system software and operating system item; the other choices fail to satisfy one or more stated conditions.
The steps for analysis and design of object oriented system.
A. Draw interaction diagrams
B. Draw state chart and object diagram
C. Draw use case and activity diagram
D. Draw component and deployment diagram
E. Draw class diagram Choose the correct answer from the options given below :
AE → B → A → C → D
BB → A → E → D → C
CE → C → B → D → A
DC → A → E → B → D
Answer:(D) C → A → E → B → D
Explanation
The key idea is stated in the problem: The steps for analysis and design of object oriented system.
A. Draw interaction diagrams
B. Draw state chart and object diagram
C. Draw use case and activity diagram
D. Draw component and deployment diagram
E. Draw class diagram Choose the correct answer. The correct answer is D: C → A → E → B → D. It follows the governing definition, algorithm, or calculation in this system software and operating system item; the other choices fail to satisfy one or more stated conditions.
Result of *{x.s=y.s¥ is : [s|s[ec] er]. | J 1 M S, | CA B | 2/]N 1] P, | AB M |r |s3]uH |v, |} ve £ 4 G | Null] Null [s[s [ec] Pr]o] J 1 M S, | CA B}| 2 | N | P, | AB
X (S, Si, C)
Y (S, P, D)
(J, 1, M)
(B, 2, N)
(R, 3, H)
(T, 4, G)
(J, S₁, CA)
(B, P₁, AB)
(R, D₁, DC)
(A, H₁, MD)
A(J, 1, M, S₁, CA); (B, 2, N, P₁, AB); (R, 3, H, D₁, DC); (T, 4, G, Null, Null)
B(J, 1, M, S₁, CA); (B, 2, N, P₁, AB); (R, 3, H, D₁, DC); (A, Null, Null, H₁, MD)
The key idea is stated in the problem: Result of *{x.s=y.s¥ is : [s|s[ec] er]. | J 1 M S, | CA B | 2/]N 1] P, | AB M |r |s3]uH |v, |} ve £ 4 G | Null] Null [s[s [ec] Pr]o] J 1 M S, | CA B}| 2 | N | P, | AB. The correct answer is B: (J, 1, M, S₁, CA); (B, 2, N, P₁, AB); (R, 3, H, D₁, DC); (A, Null, Null, H₁, MD). It follows the governing definition, algorithm, or calculation in this system software and operating system item; the other choices fail to satisfy one or more stated conditions.
I. A page is written to secondary memory only when it has been selected for replacement
B. Demand cleaning
II. A page that has not been referenced for the longest time is replaced
C. Long term scheduling
III. The decision to add to the number of processes that are partially or fully in main memory
D. Medium term scheduling
IV. The decision to add to the pool of processes to be executed
A(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
B(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
C(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
D(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Answer:(A) (A)-(II), (B)-(I), (C)-(IV), (D)-(III)
Explanation
The correct match is (A)-(II), (B)-(I), (C)-(IV), (D)-(III):
(A) LRU (Least Recently Used) matches (II): A virtual memory page replacement algorithm that evicts the page that has not been referenced for the longest period of time, leveraging temporal locality.
(B) Demand cleaning matches (I): In virtual memory cleaning policies, demand cleaning specifies that a modified (dirty) page is written to secondary storage only when it has been selected for replacement.
(C) Long term scheduling matches (IV): The long-term (or job) scheduler determines which programs are admitted into the system from the job pool to become active processes, controlling the degree of multiprogramming.
(D) Medium term scheduling matches (III): Medium-term scheduling is part of the swapping mechanism, deciding when to remove or bring back processes that are partially or fully in main memory to balance memory load.
Assertion (A): I/O protection is ensured by a hardware trap.
Reason (R): I/O interrupt is caused by conditions like I/O completion and device malfunction occurring within the I/O devices.
ABoth A and R are true and R is the correct explanation of A
BBoth A and R are true but R is NOT the correct explanation of A
CA is true but R is false
DA is false but R is true
Answer:(B) Both A and R are true but R is NOT the correct explanation of A
Explanation
I/O protection and I/O interrupts are related but distinct hardware mechanisms in operating systems.
Assertion A is true because I/O instructions are privileged, so any attempt by a user program to execute one directly causes a hardware trap into the operating system, which then validates and performs the operation on the program's behalf.
Reason R is also true as a general statement, because I/O interrupts are indeed signals raised by a device to report events such as completion of a transfer or a malfunction.
R does not explain A because the trap mechanism that enforces I/O protection is triggered by a user program's attempt to run a privileged instruction, not by the device-generated completion or malfunction interrupts that R describes. These are two separate mechanisms serving different purposes, so R is true but not the reason A holds.
Consider a disk system with 100 cylinders. The request sequence is 4, 34, 10, 7, 19, 73, 2, 15, 6, 20. If the head is initially at cylinder 50 and it takes 1 ms to move from one cylinder to the adjacent cylinder, then using shortest seek time first (SSTF) scheduling, the total seek time is
A119 ms
B120 ms
C142 ms
D146 ms
Answer:(A) 119 ms
Explanation
SSTF always services the pending request closest to the disk head's current position, so the seek path must be traced step by step from the given starting cylinder.
Starting at cylinder 50, SSTF visits requests in this order, each time picking the nearest remaining one: 50 → 34 → 20 → 19 → 15 → 10 → 7 → 6 → 4 → 2 → 73.
The seek distances at each step are 16, 14, 1, 4, 5, 3, 1, 2, 2, and finally 71, because cylinder 73 is farthest from every intermediate position and is only serviced last, after the head has moved all the way down to 2.
A. 119 ms — correct, matches the sum of every SSTF seek segment above.
B. 120 ms — off by one from a miscount in the segment distances.
C. 142 ms and D. 146 ms — these totals do not match the correct SSTF path traced above, and typically result from servicing the requests out of nearest-first order or misplacing the long final seek to cylinder 73.
A virtual memory system uses the first-in first-out (FIFO) page replacement policy and allocates a fixed number of frames to a process.
Assertion (A): Increasing the number of page frames allocated to a process sometimes increases the page fault rate.
Reason (R): Some programs do not exhibit locality of reference.
ABoth A and R are true and R is the correct explanation of A
BBoth A and R are true but R is NOT the correct explanation of A
CA is true but R is false
DA is false but R is true
Answer:(B) Both A and R are true but R is NOT the correct explanation of A
Explanation
This assertion describes Belady's anomaly, a specific behavior of the FIFO page replacement algorithm.
Assertion A is true because FIFO is not a stack algorithm, so for certain reference strings, adding more frames can paradoxically increase the number of page faults instead of reducing them, as first demonstrated by Belady.
Reason R is also a true statement on its own, since some programs genuinely access memory in patterns with weak or no locality of reference.
R does not explain A because Belady's anomaly is caused by a structural property of the FIFO algorithm itself, not by whether the running program has locality of reference. The anomaly can occur even for reference strings from programs with reasonably strong locality, so R is an unrelated true fact rather than the actual cause of A.
At a particular time of computation, the value of a counting semaphore is 7. Then 20 P operations and x V operations were completed on this semaphore. If the final value of the semaphore is 5, x will be
A15
B22
C18
D13
Answer:(C) 18
Explanation
Each P operation decrements a counting semaphore by 1 and each V operation increments it by 1, so the final value equals the initial value minus the number of P operations plus the number of V operations.
Final value = 7 − 20 + x = 5.
Solving gives x = 5 − 7 + 20 = 18.
A. 15, B. 22, and D. 13 — these do not satisfy 7 − 20 + x = 5 and come from arithmetic errors in isolating x.
Consider the following statements:
S1: The LRU page replacement algorithm suffers from Belady's anomaly.
S2: Shortest remaining time first (SRTF) scheduling may cause starvation.
S3: The stack is shared by all threads in a process.
AS1, S2 and S3 are true
BS1, S3 false and S2 is true
CS1, S2 are false and S3 is true
DS1, S2 and S3 are false
Answer:(B) S1, S3 false and S2 is true
Explanation
Each statement is an independent, checkable fact about a different operating system mechanism.
S1 is false. Belady's anomaly, where adding more frames increases page faults, is a known weakness of FIFO replacement, not LRU, since LRU is a stack algorithm and stack algorithms are proven never to exhibit this anomaly.
S2 is true. SRTF always picks the process with the least remaining burst time, so a steady stream of short jobs can keep pushing a longer job back indefinitely, causing starvation.
S3 is false. Each thread within a process maintains its own private stack for function calls and local variables, while the code, data, and heap segments are what the threads share.
Consider arrival time and burst time for three processes P0, P1, P2:
P0: arrival 0 ms, burst 7
P1: arrival 1 ms, burst 3
P2: arrival 2 ms, burst 7
Using pre-emptive shortest job first (SRTF) scheduling, scheduling carried out only at arrival or completion of a process. What is the average waiting time for the three processes?
A3 ms
B3.67 ms
C4.47 ms
D4 ms
Answer:(B) 3.67 ms
Explanation
Trace SRTF (preemptive shortest remaining time first), rechecking the shortest remaining burst only when a process arrives or completes.
t=0 to 1: only P0 is available, so it runs. Remaining burst: P0 = 6.
t=1: P1 arrives (burst 3). Remaining P0 = 6 versus P1 = 3, so P1 preempts and runs.
t=2: P2 arrives (burst 7). The running P1 now has remaining burst 3 − 1 = 2, which is still the shortest among P0(6), P1(2), P2(7), so P1 continues.
t=4: P1 finishes, having run from t=1 to t=4. Remaining: P0 = 6, P2 = 7. P0 is shorter, so P0 runs uninterrupted from t=4 to t=10.
t=10: P0 finishes. Only P2 remains, running from t=10 to t=17.
Completion times: P1 = 4, P0 = 10, P2 = 17.
Waiting time = (Completion − Arrival) − Burst:
P0: (10 − 0) − 7 = 3
P1: (4 − 1) − 3 = 0
P2: (17 − 2) − 7 = 8
Average waiting time = (3 + 0 + 8) / 3 = 11/3 ≈ 3.67 ms
Which of the following statement/s is/are correct with respect to virtual memory?
Statement I: Address translation is performed for every logical address used during the execution of a program
Statement II: A program can execute only when all of its components are loaded in the memory
ABoth Statement I and Statement II are correct
BBoth Statement I and Statement II are incorrect
CStatement I is correct but Statement II is incorrect
DStatement I is incorrect but Statement II is correct
Answer:(C) Statement I is correct but Statement II is incorrect
Explanation
Statement I is a basic fact about virtual memory: every logical (virtual) address a running program generates must pass through address translation — via the page table or TLB — before it can be used to access physical memory. This happens for every single memory reference, so Statement I is correct.
Statement II is false, and contradicts the whole purpose of virtual memory. Virtual memory, through demand paging, is specifically designed to let a program start and run while only the pages it currently needs are resident in memory, loading the rest on demand as page faults occur. A program does not need all of its components loaded at once to execute.
A. Both Statement I and Statement II are correct — Statement II contradicts the demand-paging principle behind virtual memory.
B. Both incorrect — Statement I is a correct, basic fact about address translation.
C. Statement I is correct but Statement II is incorrect — correct. Every logical address is translated, but a program does not need every component loaded to run.
D. Statement I is incorrect but Statement II is correct — reverses which statement is actually true.
Assertion (A): A process involves a library function to create a thread.
Reason (R): The threads make system calls to convey their resource and I/O requirement to the Kernel.
Choose the correct answer from the options given below.
ABoth A and R are true and R is the correct explanation of A
BBoth A and R are true but R is NOT the correct explanation of A
CA is true but R is false
DA is false but R is true
Answer:(B) Both A and R are true but R is NOT the correct explanation of A
Explanation
Assertion A is true. Creating a thread is typically done by calling a threading library function, such as pthread_create() in POSIX threads, rather than by writing a raw system call directly.
Reason R is also true on its own. Threads do rely on system calls to request resources and perform I/O from the kernel, since only the kernel can grant access to hardware and shared resources.
However, R does not explain A. R describes how threads communicate resource and I/O needs to the kernel during their execution, while A is about how a thread is created in the first place, a separate concern handled by the threading library's creation routine. So both statements are true, but R is not the reason A is true.
If table T is stored in row-major order, what is the main memory hit ratio produced by this access pattern?
A table T holds 17 rows (i = 0 to 16) and 1024 columns (j = 0 to 1023) of 32-bit floating-point values. The following program computes the average of each column:
for j = 0 to 1023:
temp = 0
for i = 0 to 16:
temp = temp + T[i][j]
print(temp / 17.0)
Memory is word-addressable, and each word holds one float, so every access to T[i][j] is one memory reference. temp is held in a processor register, so its updates cost no memory reference. Main memory is paged, with 16 page frames of 1024 words each, and the LRU page replacement policy is used.
A0
B1
C2
D3
Answer:(A) 0
Explanation
Row-major storage places T[i][j] at address i x 1024 + j, so an entire row of 1024 columns fills exactly one 1024-word page. Table T therefore spans 17 pages, one per row, but main memory holds only 16 page frames.
The outer loop fixes a column j and the inner loop walks i from 0 to 16, so each pass over one column touches one word from every one of the 17 row-pages, in the fixed order page0, page1, ..., page16. This is a cyclic reference pattern over 17 distinct pages with only 16 frames available.
Under LRU, this exact situation, a cyclic sequence of 17 pages with only 16 frames, is the classic worst case: whichever page LRU evicts to make room is always the very next page the loop is about to request. Simulating it shows every access after the first is a miss, so the steady-state hit ratio is 0, matching the plain-integer option 0 among the four given choices.
If table T is instead stored in column-major order, what is the main memory hit ratio produced by this access pattern?
A table T holds 17 rows (i = 0 to 16) and 1024 columns (j = 0 to 1023) of 32-bit floating-point values. The following program computes the average of each column:
for j = 0 to 1023:
temp = 0
for i = 0 to 16:
temp = temp + T[i][j]
print(temp / 17.0)
Memory is word-addressable, and each word holds one float, so every access to T[i][j] is one memory reference. temp is held in a processor register, so its updates cost no memory reference. Main memory is paged, with 16 page frames of 1024 words each, and the LRU page replacement policy is used.
A80%
B95.6%
C97.8%
D99.9%
Answer:(D) 99.9%
Explanation
Column-major storage places T[i][j] at address j x 17 + i, so the loop's access order (j outer, i inner) becomes a strictly increasing sequence of addresses, 0, 1, 2, ..., 17407, touching every word of T exactly once in order.
Since T has 17 x 1024 = 17408 words and each page holds 1024 words, T spans exactly 17 pages. Because the access order is purely sequential with no revisits, each page is loaded once (a miss on its first word) and then supplies hits for its remaining 1023 words, with no page ever needing to be reloaded.
If table T is stored in column-major order, how many page faults does this program encounter?
A table T holds 17 rows (i = 0 to 16) and 1024 columns (j = 0 to 1023) of 32-bit floating-point values. The following program computes the average of each column:
for j = 0 to 1023:
temp = 0
for i = 0 to 16:
temp = temp + T[i][j]
print(temp / 17.0)
Memory is word-addressable, and each word holds one float, so every access to T[i][j] is one memory reference. temp is held in a processor register, so its updates cost no memory reference. Main memory is paged, with 16 page frames of 1024 words each, and the LRU page replacement policy is used.
A14
B15
C16
D17
Answer:(D) 17
Explanation
With column-major storage, address(i,j) = j x 17 + i, so the program's access order is exactly sequential through T's 17408 words, one page after another with no revisits.
T occupies 17408 / 1024 = 17 pages exactly. A purely sequential single pass over 17 distinct pages causes exactly one page fault per page, the first time each page is touched, since no page needs to be reloaded once the scan moves past it.
If table T is stored in row-major order, how many page faults does this program encounter?
A table T holds 17 rows (i = 0 to 16) and 1024 columns (j = 0 to 1023) of 32-bit floating-point values. The following program computes the average of each column:
for j = 0 to 1023:
temp = 0
for i = 0 to 16:
temp = temp + T[i][j]
print(temp / 17.0)
Memory is word-addressable, and each word holds one float, so every access to T[i][j] is one memory reference. temp is held in a processor register, so its updates cost no memory reference. Main memory is paged, with 16 page frames of 1024 words each, and the LRU page replacement policy is used.
A16,402
B17,408
C18,208
D18,608
Answer:(B) 17,408
Explanation
With row-major storage, each row of 1024 columns is exactly one page, so T spans 17 row-pages while only 16 frames exist. The column-by-column access pattern cycles through all 17 pages every single column, a cyclic reference string of length 17 with only 16 frames available.
This is the classic LRU worst case for a cyclic pattern one longer than the number of frames: every access after the very first is a page fault, because LRU always evicts the page that is about to be needed next. So essentially all accesses to T become page faults.
Total memory references to T = 17 rows x 1024 columns = 17408, and since virtually every one of them faults, the number of page faults is 17408.
What is the ratio of page faults between the row-major and column-major arrangements of table T?
A table T holds 17 rows (i = 0 to 16) and 1024 columns (j = 0 to 1023) of 32-bit floating-point values. The following program computes the average of each column:
for j = 0 to 1023:
temp = 0
for i = 0 to 16:
temp = temp + T[i][j]
print(temp / 17.0)
Memory is word-addressable, and each word holds one float, so every access to T[i][j] is one memory reference. temp is held in a processor register, so its updates cost no memory reference. Main memory is paged, with 16 page frames of 1024 words each, and the LRU page replacement policy is used.
A1024:1
B1301:1
C1240:1
D9107:8
Answer:(A) 1024:1
Explanation
Row-major storage produces 17408 page faults, since the cyclic 17-page access pattern thrashes against only 16 frames on nearly every access.
Column-major storage produces 17 page faults, since it accesses T sequentially and each of its 17 pages is loaded exactly once.
Fault ratio (row-major : column-major) = 17408 : 17 = 1024 : 1, since 17408 / 17 = 1024 exactly.
Machine language and assembly language are two distinct members of the same family — low level languages — with machine language being pure binary opcodes and assembly language being its symbolic mnemonic form.
Since machine language is not itself assembly language, the accurate general classification is that it is a low level language.
In Linux Operating System, when ______ is invoked, it is passed a set of flags that determine how much sharing is to take place between the parent and child tasks.
Afork()
Bclone()
Cpthread()
Dthread()
Answer:(B) clone()
Explanation
Linux implements process and thread creation through a single system call, clone(), which takes explicit flag arguments such as CLONE_VM and CLONE_FS to control exactly which resources (address space, file descriptors, and signal handlers) the parent and child share.
fork() is implemented on top of clone() with a fixed set of flags, so it does not itself expose configurable sharing.
The main function of the microkernel is to provide a communication facility between the ______ program and the various ______ that are also running in user space.
AVirtual, Processes
BSystem, Processes
CClient, Services
DVirtual, Services
Answer:(C) Client, Services
Explanation
In microkernel design, the kernel is stripped down to minimal functionality, and traditional OS functions (file systems, device drivers, networking) run as separate user-space services.
The microkernel's main job is to relay messages between a client program and those user-space services, so it can be swapped or restarted independently of the core kernel.
Consider an operating system capable of loading and executing a single sequential user process at a time. The disk head scheduling algorithm used is first come first served (FCFS). If FCFS is replaced by shortest seek time first (SSTF) and the vendor claims 50% better benchmark results, what is the expected improvement in the I/O performance of user programs?
A50%
B100%
C25%
D0%
Answer:(D) 0%
Explanation
Disk-scheduling algorithms such as FCFS and SSTF only change behaviour when several I/O requests are already queued and can be reordered.
A system that loads and executes only a single sequential user process at a time issues disk requests one at a time, so there is essentially no backlog of pending requests for the scheduler to reorder. With nothing meaningful to reorder, switching from FCFS to SSTF has little to no effect on real user-program I/O time, even though an isolated benchmark built around queued requests can show a large improvement.
For the following set of processes scheduled using FCFS policy, determine the average waiting time. Assume that the processes arrived in the order P1, P2, P3, P4.
Process
Burst Time (ms)
P1
8
P2
15
P3
10
P4
7
Burst times for processes P1-P4
A8
B16
C32
D48
Answer:(B) 16
Explanation
Under FCFS each process waits for the sum of the burst times of every process ahead of it in arrival order.
Which of the statement is not true in regards to virtual memory?
AThe main objective for using virtual memory is to increase the effective capacity of the memory system
BSize of the program can be as large as the size of the secondary memory
CProgram and data are stored in the secondary memory
DNone of these
Answer:(D) None of these
Explanation
Virtual memory extends the effective size of usable memory by keeping the full program and data on secondary storage and paging in only what is currently needed.
This makes the first three statements all accurate descriptions of virtual memory, since it does increase effective capacity, does allow program size to approach secondary storage capacity, and does rely on secondary memory as the backing store.
Because none of the first three statements is actually untrue, the correct response to "which is not true" is that none of them, matching option D.
A. Shortest remaining time first scheduling may cause starvation
B. Preemptive scheduling may cause starvation
C. Round robin is better than FCFS in terms of response time
Choose the correct answer from the options given below:
AA only
BB, C only
CA, B only
DA, B, C
Answer:(D) A, B, C
Explanation
Check each scheduling claim against standard CPU scheduling behaviour.
A. SRTF may cause starvation — true. A steady stream of newly arriving short processes can keep pushing a long process to the back indefinitely.
B. Preemptive scheduling may cause starvation — true, since SRTF is itself a preemptive policy, and priority-based preemptive scheduling can starve low-priority processes the same way.
C. Round robin is better than FCFS for response time — true. Time-slicing lets every process get CPU time quickly, avoiding the long waits FCFS causes through the convoy effect.
In design protocol of critical section problem, each process must ask permission to enter critical section in ____ code, it then executes in the critical section, once it finishes executing in the critical section it enters the ____ code. The process then enters the ____ code.
The critical section protocol cycles through three named regions of code in a fixed order.
A process first requests permission in the entry section, runs its critical section code, then releases access in the exit section, and finally does unrelated work in the remainder section before looping back to the entry section again.
Match each operating-system concept to what it provides.
A-II, B-IV, C-I, D-III
Inter-process communication lets cooperating processes share work, which is the classic textbook benefit of computational speedup, matching II. Demand paging loads pages only when referenced, which is how virtual memory is implemented, matching IV. Banker's algorithm decides whether a resource request can be safely granted, matching resource allocation, I. The process control block stores a process's saved state, which is why it functions as a task control block, matching III.
Select the correct order of events after power is initialized on a system.
A. Bootstrap loader is loaded from the disk
B. Kernel is loaded onto the memory
C. Firmware ROM loads boot block
Choose the correct answer from the options given below:
AB, C, A
BC, A, B
CA, B, C
DA, C, B
Answer:(B) C, A, B
Explanation
Booting proceeds from the most permanent, hardwired code toward the fully loaded operating system.
Firmware in ROM runs first and loads the boot block, which is C. That boot block then loads the bootstrap loader from disk, which is A. The bootstrap loader finally loads the kernel into memory, which is B.
An OS follows round-robin scheduling with time quantum of 4ms. Assuming that the CPU is free now and there are 20 processes waiting in the ready queue, the maximum amount of time that a process waits before getting into the CPU is ________.
A80ms
B76ms
C84ms
DNone of the above
Answer:(B) 76ms
Explanation
In round robin, a process's worst-case wait happens when it sits at the very back of the queue and every other process runs its full time quantum first.
With 20 total processes, the unluckiest process must wait for the other 19 to each use their full quantum.
In the context of Binding which one of the following is incorrect?
AStatic binding is done at Compile Time
BAbsolute addresses are known at Load Time
CInternal functions are known at link time
DTarget of a function is known at runtime
Answer:(D) Target of a function is known at runtime
Explanation
Binding can happen at several distinct stages, and a function's target address is not always deferred to runtime.
Static binding fixes a function's target at compile time, and internal (statically linked) functions are resolved when the linker combines object modules at link time, while absolute memory addresses for a program's variables and code become fixed once the loader places it in memory at load time.
Option D generalizes runtime binding as if it applied to every function, which contradicts static and link-time binding described in the other three statements, making it the incorrect claim among the four.
Which RAID level is appropriate for storing the log file of a ticket booking system where a huge database update is required?
ALevel 0
BLevel 1
CLevel 2
DLevel 3
Answer:(B) Level 1
Explanation
A transaction log for a heavily updated database is written frequently and sequentially, and it must survive a disk failure since it is used to recover the database, so the RAID scheme needs both good write performance and redundancy.
RAID 1 (mirroring) writes each block to two disks with no parity computation overhead, giving fast small writes alongside full redundancy, which suits a frequently updated log file.
RAID 0 offers no redundancy at all, so a single disk failure would destroy the log needed for recovery, while RAID 2 and RAID 3 add parity-based redundancy at the cost of extra computation and a shared parity disk bottleneck that hurts frequent small-write workloads like transaction logging.
The classical protection model represents the state of a system as an access matrix, with rows for protection domains and columns for objects, where each entry lists the operations a domain may perform on that object.
This access matrix model is the standard abstraction used to describe and reason about access rights across domains and resources in an operating system.
"Owner matrix", "domain matrix", and "global matrix" are not the standard terminology for this abstraction, since a domain is only one axis of the matrix rather than the model's name itself.
Process in memory also contain the heap section. What is the use of heap section?
AContains temporary data
BDynamically allocate the memory during process run time
CContains Process priority and pointers to scheduling queues
DContains amount of CPU and real time use
Answer:(B) Dynamically allocate the memory during process run time
Explanation
The heap is the memory region a process uses for dynamic memory allocation requested during execution, such as through malloc in C or new in C++, and it grows and shrinks as the program allocates and frees memory at runtime.
This makes option B the correct description, since "temporary data" describes the stack (used for local variables and function calls), while process priority and CPU/real-time usage information are stored in the process control block, not the heap.
How many page faults occur for optimal algorithm for the following reference string, with four frames?
1, 2, 3, 4, 5, 3, 4, 1, 6, 7, 8, 7, 8
A8
B7
C6
D5
Answer:(A) 8
Explanation
Simulating the optimal (OPT) page replacement algorithm means that whenever a frame must be evicted, the page whose next use is farthest in the future (or never used again) is chosen.
References 1, 2, 3, 4 each fault to fill the four empty frames {1,2,3,4}
Reference 5 faults. Among {1,2,3,4}, page 2 is never referenced again, so it is evicted, giving {1,3,4,5}
References 3, 4, 1 are all hits since they are already in {1,3,4,5}
Reference 6 faults. None of {1,3,4,5} appear again in the remaining string, so any one is evicted, giving a frame set including 6
Reference 7 faults, since none of the current frames reappear before it. Reference 8 also faults, evicting a page that is not reused, so that 7 is kept for its later reuse
References 7 and 8 near the end are then both hits, giving a total of 8 faults across the sequence.
The degree of concurrency a threading model achieves depends on how many kernel threads can actually run in parallel on available processors.
In the many-to-one model, all user threads map to a single kernel thread, so only one thread can be scheduled on a CPU at a time, giving the least concurrency of the three.
The one-to-one model maps each user thread to its own kernel thread, letting the kernel schedule several threads on multiple processors truly in parallel, which is commonly presented as offering the greatest concurrency.
The many-to-many model sits between the two, multiplexing many user threads onto a smaller or equal pool of kernel threads, which allows some but not full parallel scheduling.
This ranking, one-to-one above many-to-many above many-to-one, matches option B. Note that some operating-systems texts treat many-to-many as offering concurrency comparable to one-to-one since it can also scale kernel threads to the processor count, so this specific ranking should be spot-checked against the exact course reference in use.
A. If the time quantum is extremely small, round robin can result in a large number of context switches.
B. The average turnaround time of a set of processes does not necessarily improve as the time-quantum size increases.
C. The average turnaround time can be improved if most processes finish their next CPU burst in a single time quantum.
Choose the correct answer from the options given below:
AA only
BA, B only
CB, C only
DA, B, C
Answer:(D) A, B, C
Explanation
All three statements describe well-established, independently verifiable properties of round robin scheduling.
A very small time quantum forces the scheduler to switch between processes far more often, directly increasing the number of context switches and their overhead, which makes statement A true.
Increasing the quantum does not guarantee better turnaround time. Too large a quantum makes round robin behave like first-come-first-served, which can worsen turnaround for processes with short bursts stuck behind long ones, so improvement is not guaranteed, making statement B true.
When most processes complete their current CPU burst within a single quantum, they finish without being preempted and re-queued, cutting down the waiting caused by repeated cycling through the ready queue, which improves average turnaround, making statement C true.
Since all three hold, the correct option is A, B, C.
Consider the following set of processes that need to be scheduled on a single CPU. All the times are given in milliseconds.
Using the shortest remaining time first scheduling algorithm, the average process turnaround (in milliseconds) is ________.
Process Name
Arrival Time
Execution Time
A
0
6
B
3
2
C
5
4
D
7
6
E
10
3
Process arrival and execution times for SRTF scheduling
A7.2
B8.2
C6
D9
Answer:(A) 7.2
Explanation
Simulating SRTF (preemptive shortest-remaining-time-first) minute by minute tracks which process has the least remaining work at every arrival.
A (remaining 6) runs from 0 until B arrives at 3 with only 2 units of work, which is shorter than A's remaining 3, so B preempts and runs 3 to 5, completing at 5.
At 5, C arrives needing 4 units, but A's remaining 3 units is still shorter, so A resumes and runs until D arrives at 7, where A now has only 1 unit left versus D's fresh 6, so A finishes at 8.
C (remaining 4) then runs from 8, and even though E arrives at 10 needing 3 units, C's remaining 2 units at that point is shorter, so C finishes at 12.
E (3 units) then runs from 12 to 15, and finally D (remaining 6) runs from 15 to 21 with no more arrivals to preempt it.
Turnaround (completion − arrival): A = 8−0 = 8, B = 5−3 = 2, C = 12−5 = 7, D = 21−7 = 14, E = 15−10 = 5
Select the correct order of basic page replacement.
A. Page replacement algorithm is used in case there is no free frame
B. The process is continued by restarting the instruction caused by TRAP
C. Location of desired frame is found from disk
D. Desired page is brought into newly freed frame and page & frame tables are updated
AA, B, D, C
BC, D, A, B
CC, A, D, B
DA, D, B, C
Answer:(C) C, A, D, B
Explanation
The standard page-fault service routine first validates the faulting reference and finds where the needed page lives on disk before it worries about where in memory to put it.
Once the desired page's disk location is known (C), the OS checks for a free frame, and only if none exists does it invoke a page-replacement algorithm to select a victim frame to free up (A).
With a frame now available, the desired page is read in from disk into that frame, and the page and frame tables are updated to reflect the new mapping (D).
Finally, control returns to the interrupted process, and the instruction that originally caused the page-fault trap is restarted (B).
II. Probably will be used again soon and need to write out
C. (reference bit=1, modify bit=0)
III. Best page to replace
D. (reference bit=1, modify bit=1)
IV. Probably will be used again soon
AA-II, B-I, C-III, D-IV
BA-II, B-I, C-IV, D-III
CA-III, B-I, C-II, D-IV
DA-III, B-I, C-IV, D-II
Answer:(D) A-III, B-I, C-IV, D-II
Explanation
This classification follows the standard enhanced second-chance page-replacement scheme, which ranks pages by their reference and modify bits.
A page that is neither recently used nor modified (0,0) is the safest and cheapest to evict, since discarding it loses nothing and needs no disk write, giving A-III, the best page to replace.
A page that is not recently used but has been modified (0,1) is not needed again soon, but its changes must be saved first, giving B-I, must write out before replacement.
A page that is recently used but not modified (1,0) is likely to be accessed again soon, and since it is unmodified, it needs no write-back if evicted, giving C-IV, probably will be used again soon.
A page that is both recently used and modified (1,1) is likely to be needed again soon and would also require a write-back if evicted, giving D-II, probably will be used again soon and need to write out.
A disk has 300 cylinders (0 to 299). If the initial position of the read-write head is at cylinder 101, moving towards the higher end of the disk, calculate the total head movement following C_LOOK policy for the read request sequence:
123, 45, 197, 87, 250, 198, 280
A456
B450
C458
D445
Answer:(A) 456
Explanation
C-LOOK services requests only up to the highest pending request in the current direction, then jumps directly to the lowest pending request and continues in the same direction, without sweeping all the way to either end of the disk.
Sorted requests: 45, 87, 123, 197, 198, 250, 280
Starting at 101 and moving upward, the head services 123, 197, 198, 250, 280 in order, then jumps down to the lowest request, 45, and continues upward to 87.
Sequence of moves and distances: 101→123 (22), 123→197 (74), 197→198 (1), 198→250 (52), 250→280 (30), 280→45 (235, the jump), 45→87 (42)
Statement I: The main objective for using virtual memory is to increase the effective capacity of the memory system
Statement II: The size of virtual memory can be larger than the main memory
In the light of the above statements, choose the correct answer from the options given below
ABoth Statement I and Statement II are true.
BBoth Statement I and Statement II are false.
CStatement I is true but Statement II is false.
DStatement I is false but Statement II is true.
Answer:(A) Both Statement I and Statement II are true.
Explanation
Virtual memory exists specifically to let a system run programs whose combined address space exceeds the amount of physical RAM installed, effectively increasing the usable capacity of the memory system as seen by running processes. Statement I is true.
Because virtual memory is backed by disk-based storage in addition to RAM, its total addressable size can indeed exceed the size of main memory, which is the entire point of using it. Statement II is true.
Since both statements hold, the correct option is that both Statement I and Statement II are true.
If an operating system does not allow a child process to exist when the parent process has been terminated, this phenomenon is called as -
AThreading
BCascading termination
CZombie termination
DProcess killing
Answer:(B) Cascading termination
Explanation
When an operating system's policy forces every child process to be terminated as soon as its parent terminates, so that termination propagates down the process tree, this is called cascading termination. It is a deliberate OS design choice, not the same as a zombie process, which is a terminated process whose exit status has not yet been collected by its parent.
What is called Journalling in Linux operating system?
AProcess scheduling
BFile saving as transaction
CA type of thread
DAn editor
Answer:(B) File saving as transaction
Explanation
A journaling file system logs pending file system changes as transactions in a journal before committing them to the main structures. If the system crashes mid-write, the journal lets it replay or discard the incomplete transaction on recovery, keeping the file system consistent. This is why journaling is described as saving file changes as transactions.
Match List I with List II regarding types of interrupts.
List-I
List-II
A. Stack overflow
I. Software interrupt
B. Timer
II. Internal interrupt
C. Invalid opcode
III. External interrupt
D. Superior call
IV. Machine check interrupt
A(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
B(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
C(A)-(I), (B)-(II), (C)-(IV), (D)-(III)
D(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
Answer:(D) (A)-(II), (B)-(III), (C)-(IV), (D)-(I)
Explanation
Interrupts split into four classic categories.
Stack overflow is a program-execution error detected by the processor itself, an internal interrupt.
Timer expiry is signaled by external hardware, an external interrupt.
Invalid opcode is a fault the machine's hardware raises against a bad instruction, grouped here with machine-check style interrupts.
A supervisor call is a deliberately executed trap instruction used to request OS services, the definition of a software interrupt.
For the following page reference string 4, 3, 2, 1, 4, 3, 5, 4, 3, 2, 1, 5, the number of page faults that occur in Least Recently Used (LRU) page replacement algorithm with frame size 3 is
A6
B8
C10
D12
Answer:(C) 10
Explanation
Simulate LRU with 3 frames, evicting the least recently used page on every miss.
An activation record is the runtime frame created for each subroutine call, matching Subroutine Call.
The location counter is the running address pointer an assembler maintains while assembling code, matching Assembler.
A reference count is the technique reference-counting garbage collectors use to know when an object can be reclaimed, matching Garbage Collection.
Address relocation is performed by a linking loader when it places a program at its final load address, matching Linking Loader.
A-III (Activation record with Subroutine Call) and B-IV (Location counter with Assembler) are unambiguous. The printed key pairs C with Linking Loader and D with Garbage Collection, the reverse of the standard textbook definitions, where Reference count belongs with Garbage Collection and Address relocation belongs with Linking Loader. This item is kept for a human check against the source PDF and official key.
Match List I with List II regarding operating system concepts.
List-I
List-II
A. Least frequently used
I. Memory is distributed among processors
B. Critical Section
II. Page replacement policy in cache memory
C. Loosely coupled multiprocessor system
III. Program section that once begin must complete execution before another processor access the same shared resource
D. Distributed operating system organization
IV. O/S routines are distributed among available processors
A(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
B(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
C(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
D(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
Answer:(C) (A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Explanation
Least frequently used is a page replacement policy applied in cache memory.
A critical section is the program section that, once begun, must run to completion before another processor can access the same shared resource.
A loosely coupled multiprocessor system is one where memory is distributed among the processors rather than shared.
A distributed operating system organization is one where OS routines themselves are distributed among the available processors.
Match List I with List II regarding systems concepts.
List-I
List-II
A. Firmware
I. Number of logical records into physical blocks
B. Batch file
II. ASCII format
C. Packing
III. Resource allocation
D. Banker's Algorithm
IV. ROM
A(A)-(II), (B)-(I), (C)-(IV), (D)-(III)
B(A)-(II), (B)-(I), (C)-(III), (D)-(IV)
C(A)-(IV), (B)-(II), (C)-(I), (D)-(III)
D(A)-(IV), (B)-(I), (C)-(II), (D)-(III)
Answer:(C) (A)-(IV), (B)-(II), (C)-(I), (D)-(III)
Explanation
Firmware refers to software permanently held in ROM.
A batch file is a plain ASCII-format text file of commands.
Packing is the grouping of a number of logical records into physical blocks.
Banker's algorithm is a resource allocation technique used to avoid deadlock.
Match List I with List II regarding classic operating system algorithms.
List-I
List-II
A. Stack algorithm
I. Deadlock
B. Elevator algorithm
II. Disk scheduling
C. Priority scheduling algorithm
III. Page replacement
D. Havender's algorithm
IV. CPU scheduling
A(A)-(III), (B)-(II), (C)-(IV), (D)-(I)
B(A)-(II), (B)-(III), (C)-(IV), (D)-(I)
C(A)-(III), (B)-(II), (C)-(I), (D)-(IV)
D(A)-(II), (B)-(III), (C)-(I), (D)-(IV)
Answer:(A) (A)-(III), (B)-(II), (C)-(IV), (D)-(I)
Explanation
Stack algorithms, page replacement policies that satisfy the stack property such as LRU and OPT, belong to page replacement.
The elevator algorithm is another name for the SCAN disk-scheduling algorithm.
A priority scheduling algorithm assigns the CPU based on priority, a CPU scheduling technique.
Havender's algorithm imposes an ordering on resource requests to prevent deadlock.
Assertion (A): A load-and-go assembler avoids the overhead of writing the object program out and reading it back in.
Reason (R): This can be done with either one-pass or two pass assembler.
In the light of the above statements, choose the correct answer from the options given below:
ABoth (A) and (R) are true and (R) is the correct explanation of (A)
BBoth (A) and (R) are true but (R) is NOT the correct explanation of (A)
C(A) is true but (R) is false
D(A) is false but (R) is true
Answer:(C) (A) is true but (R) is false
Explanation
Assertion (A) is true, a load-and-go assembler places generated code directly into memory for immediate execution, which is exactly what lets it skip writing an object file to disk and reading it back in.
Reason (R) is false. Load-and-go is characteristically a one-pass assembler technique, since a one-pass design can resolve addresses and emit executable code in memory as it scans the source once. A conventional two-pass assembler instead builds a complete symbol table on the first pass and typically produces a separate object module rather than loading directly into memory, so claiming this works equally well with either design misstates how two-pass assemblers are normally used.
Memory access time is p nanoseconds and an extra q nanoseconds are needed to handle a page fault. If a page fault occurs once every 100 instructions, what is the effective memory access time?
Ap + q/100
B(p + q)/100
Cp + q
Dp/100 + q
Answer:(A) p + q/100
Explanation
Every access costs p. A page fault adds q, but that happens on only 1 in 100 accesses, contributing q/100 on average. Effective access time = p + q/100.
In a file-allocation system three schemes are used:
A. Contiguous allocation
B. Indexed allocation
C. Linked allocation
Which scheme(s) do NOT suffer from external fragmentation?
Choose the correct answer from the options given below:
AA only
BB and C only
CA and B only
DC only
Answer:(B) B and C only
Explanation
Contiguous allocation needs a run of consecutive free blocks, so free space breaks into unusable gaps - external fragmentation. Indexed and linked allocation place file blocks anywhere on disk, so any free block is usable and there is no external fragmentation.
Three statements about the interrupt-handling mechanism:
A. The interrupt handler routine is not stored at a fixed address in memory.
B. CPU hardware has a dedicated wire called the interrupt-request line used for handling interrupts.
C. The interrupt vector contains the memory addresses of the specialized interrupt handlers.
Choose the correct answer from the options given below:
AA is TRUE only
BBoth B and C are TRUE only
CBoth A and B are TRUE only
DBoth A and C are TRUE only
Answer:(B) Both B and C are TRUE only
Explanation
The CPU has an interrupt-request line it senses each instruction cycle (B true), and the interrupt vector table holds the addresses of the interrupt service routines (C true). Statement A is false: handler routines occupy fixed, known addresses that the vector points to.
Match List-I (System Calls) with List-II (Description).
List-I
List-II
A. fork()
I. Sends a signal from one process to a particular process
B. exec()
II. Indicates termination of the current process
C. kill()
III. Loads the specified program into memory
D. exit()
IV. Creates a child process
AA-II, B-III, C-IV, D-I
BA-IV, B-III, C-I, D-II
CA-IV, B-I, C-II, D-III
DA-IV, B-III, C-II, D-I
Answer:(B) A-IV, B-III, C-I, D-II
Explanation
fork() creates a child process (IV). exec() replaces the current process image with a new program loaded into memory (III). kill() sends a signal to a specified process (I). exit() terminates the calling process (II).
The number of pages in the given virtual address space is:
Consider a machine with 16 GB main memory and a 32-bit virtual address space, with a page size of 4 KB. Frame size equals page size for this machine.
A2^10
B2^20
C2^30
D2^12
Answer:(B) 2^20
Explanation
The number of pages in a process address space equals the total virtual address space divided by the page size, because the virtual space is partitioned into equal fixed-size pages. A 32-bit virtual address covers 2^32 bytes, and each page is 4 KB, which is 2^12 bytes. Dividing, 2^32 divided by 2^12 leaves 2^20 pages.
The minimum number of bits needed for a physical address is:
Consider a machine with 16 GB main memory and a 32-bit virtual address space, with a page size of 4 KB. Frame size equals page size for this machine.
A28
B34
C24
D12
Answer:(B) 34
Explanation
A physical address must be wide enough to name every byte of installed main memory, so the number of address bits is the base-two logarithm of the memory size. Sixteen gigabytes equals 2^4 multiplied by 2^30 bytes, which is 2^34 bytes altogether. Naming that many byte locations therefore requires exactly 34 bits.
The size of the page table for this virtual address space, given a page-table entry of 2 bytes, is:
Consider a machine with 16 GB main memory and a 32-bit virtual address space, with a page size of 4 KB. Frame size equals page size for this machine.
A2 MB
B2 KB
C32 MB
D12 KB
Answer:(A) 2 MB
Explanation
A single-level page table stores one entry for every page in the virtual address space, so its size is the page count multiplied by the entry size. There are 2^20 pages, and each page-table entry occupies 2 bytes. Multiplying gives 2^20 times 2, which is 2^21 bytes, and that equals 2 megabytes.
If a process of size 34 KB runs on this machine, the internal fragmentation for the process is:
Consider a machine with 16 GB main memory and a 32-bit virtual address space, with a page size of 4 KB. Frame size equals page size for this machine.
A4 KB
BZero
C1 KB
D2 KB
Answer:(D) 2 KB
Explanation
Internal fragmentation is the wasted space inside the last frame allocated to a process, because frames are allocated whole even when the process does not fill them. A 34 KB process needs the ceiling of 34 divided by 4, which is 9 pages, occupying 36 KB of frame space. The final frame holds only 2 KB of useful data, so the internal fragmentation is 36 minus 34, that is 2 KB.
A machine has 3 page frames and the reference stream A, B, C, D, A, B, E, A, B, C, D, E, B, A, B. If P and Q are the page faults under FIFO and LRU respectively, what is (P, Q)?
A(11, 10)
B(12, 11)
C(10, 11)
D(11, 12)
Answer:(D) (11, 12)
Explanation
Simulating FIFO replaces the oldest loaded page and produces 11 faults. Simulating LRU instead evicts the least recently used page, which produces 12 faults for this sequence; replacement policies need not rank the same on every reference string.
A disk has 60 cylinders. Requests are 10, 22, 20, 2, 40, 6, 38; the head starts at 20. Movement by one cylinder takes 2 ms. Under SSTF, what is the total service time?
A240 milliseconds
B96 milliseconds
C120 milliseconds
D112 milliseconds
Answer:(C) 120 milliseconds
Explanation
SSTF serves 20, 22, 10, 6, 2, 38, 40, always choosing the nearest pending cylinder. The movement is 0 + 2 + 12 + 4 + 4 + 36 + 2 = 60 cylinders, and at 2 ms per cylinder the total is 120 ms; other totals result from a different service order.
A Linux filesystem has 2 KB blocks and 32-bit disk addresses. An i-node has 12 direct pointers, one single-indirect pointer, and one double-indirect pointer. Approximately what is the largest representable file?
A513 Kbytes
B513 Mbytes
C537 Mbytes
D537 Kbytes
Answer:(B) 513 Mbytes
Explanation
A 2 KB indirect block contains 2,048/4 = 512 addresses. The i-node addresses 12 + 512 + 512² blocks, about 262,668 blocks, which at 2 KB each is approximately 513 MB.
In a single-level page-table system with the page table in memory, TLB hit rate is 80%. TLB search takes 15 ns and a main-memory access takes 150 ns. What is the effective memory access time?
A185
B195
C205
D175
Answer:(B) 195
Explanation
On a TLB hit, time is 15 + 150 = 165 ns. On a miss, it is 15 + 2(150) = 315 ns because both the page table and the data are accessed. The weighted average is 0.8(165) + 0.2(315) = 195 ns; treating every reference as a hit would wrongly give 165 ns.
Assuming that the system call fork () never fails, consider the following C programs P1 and
P2 executed on a UNIX / Linux system:
/*P1+] j= P2+]
int main () { int main () {
fork (): fork ():
fork (): printf("Happy\n"):
fork (): fork()
printf("Happy\n"): printf("Happy\n"):
} fork( ):
printf("Happy\n");
}
Statement I: P1 displays "Happy" 8 times.
Statement II: P2 displays "Happy" 12 times.
In the light of the above statements, choose the correct answer from the options given below
ABoth Statement I and Statement II are true
BBoth Statement I and Statement II are false
CStatement I is correct but Statement II is false
DStatement I is incorrect but Statement II is true.
Answer:(C) Statement I is correct but Statement II is false
Explanation
P1 has three successive forks, creating 2³ = 8 processes, each printing once. In P2, prints occur after the first, second, and third forks, producing 2 + 4 + 8 = 14 prints, not 12; therefore only statement I is true.
A disk has the parameters shown in the table. If T is the capacity of one track and S is the capacity of one surface, what is (T,S)?
Disk parameter
Value
Sector size
512 bytes
Tracks per surface
2,000
Sectors per track
50
Platters
5 double-sided
Average seek time
10 ms
A(50 K, 50,000 K)
B(25 K, 25,000 K)
C(25 K, 50,000 K)
D(40 K, 36,000 K)
Answer:(C) (25 K, 50,000 K)
Explanation
A track holds 50 × 512 = 25,600 bytes, recorded as 25 K in the choices. A surface has 2,000 tracks, giving 2,000 × 25 K = 50,000 K. The 25,000 K option misses the factor from the number of tracks, while 50 K is only the sector-scale figure.
Using the disk parameters shown in the table, what is the disk capacity?
Disk parameter
Value
Sector size
512 bytes
Tracks per surface
2,000
Sectors per track
50
Platters
5 double-sided
Average seek time
10 ms
A25,000 K
B500,000 K
C250,000 K
D50,000 K
Answer:(B) 500,000 K
Explanation
Five double-sided platters supply ten recording surfaces. Each surface is 50,000 K, so the whole disk holds 10 × 50,000 K = 500,000 K. The other choices account for only one surface, one platter, or an incorrect partial total.
Using the disk parameters shown in the table, assess the statements.
Statement I: The disk has 2,000 cylinders.
Statement II: 51,200 bytes is not a valid block size for this disk.
Disk parameter
Value
Sector size
512 bytes
Tracks per surface
2,000
Sectors per track
50
Platters
5 double-sided
Average seek time
10 ms
ABoth statements are true.
BBoth statements are false.
CStatement I is true but Statement II is false.
DStatement I is false but Statement II is true.
Answer:(A) Both statements are true.
Explanation
A cylinder contains one track on every surface at the same radius, so the number of cylinders equals the 2,000 tracks per surface. A block is normally an integral number of sectors with the relevant disk block constraints; 51,200 bytes is not a valid block size in this setup.
If the disk rotates at 5,400 rpm, approximately what is the maximum rotational delay?
Disk parameter
Value
Sector size
512 bytes
Tracks per surface
2,000
Sectors per track
50
Platters
5 double-sided
Average seek time
10 ms
A0.011 seconds
B0.11 seconds
C0.0011 seconds
D1.1 seconds
Answer:(A) 0.011 seconds
Explanation
At 5,400 rpm, one revolution takes 60/5,400 ≈ 0.0111 seconds. The maximum rotational delay is one full revolution, so 0.011 seconds is the closest choice.
If one track can be transferred per revolution, what is the data-transfer rate for the disk in the table rotating at 5,400 rpm?
Disk parameter
Value
Sector size
512 bytes
Tracks per surface
2,000
Sectors per track
50
Platters
5 double-sided
Average seek time
10 ms
A2,850 KBytes/second
B4,500 KBytes/second
C5,700 KBytes/second
D2,250 KBytes/second
Answer:(D) 2,250 KBytes/second
Explanation
One track contains 50 sectors × 512 bytes = 25,600 bytes. At 5,400 rpm, the disk makes 90 revolutions each second, so one-track-per-revolution transfer gives 25,600 × 90 = 2,304,000 bytes/s, approximately 2,250 KBytes/s. The larger alternatives incorrectly use a different track capacity or rotation rate.
Consider a paging system where translation look aside buffer (TLB) a special type of associative memory is used with hit ratio of 80%.
Assume that memory reference takes 80 nanoseconds and reference time to TLB is 20 nanoseconds. What will be the effective memory access time given 80% hit ratio?
A110 nanoseconds
B116 nanoseconds
C200 nanoseconds
D100 nanoseconds
Answer:(B) 116 nanoseconds
Explanation
Calculate access times for hit and miss cases:
On a TLB hit (probability 0.80):
Access time = TLB time + Main memory access time = 20 + 80 = 100 ns
On a TLB miss (probability 0.20):
Access time = TLB time + Page table lookup time + Main memory access time = 20 + 80 + 80 = 180 ns
Which of the following CPU scheduling algorithms is/are supported by LINUX operating system?
ANon-preemptive priority scheduling
BPreemptive priority scheduling and time sharing CPU scheduling
CTime sharing scheduling only
DPriority scheduling only
Answer:(B) Preemptive priority scheduling and time sharing CPU scheduling
Explanation
The Linux operating system kernel implements both preemptive priority scheduling and time-sharing scheduling. Real-time tasks are scheduled using preemptive priority policies (SCHED_FIFO and SCHED_RR), while normal user processes are managed via the Completely Fair Scheduler (CFS) using time-sharing proportional resource allocation.
Identify the circumstances under which pre-emptive CPU scheduling is used:
A. A process switches from Running state to Ready state
B. A process switches from Waiting state to Ready state
C. A process completes its execution
D. A process switches from Ready to Waiting state
Choose the correct option:
AA and B only
BA and D only
CC and D only
DA, B, C only
Answer:(A) A and B only
Explanation
CPU scheduling algorithms make scheduling decisions under specific process state transitions:
Running to Ready (such as a timer interrupt in round-robin) — preemptive scheduling (A).
Waiting to Ready (such as I/O completion of a higher-priority task) — preemptive scheduling (B).
Running to Waiting or Process termination (completion) — non-preemptive (cooperative) scheduling.
A process cannot transition directly from Ready to Waiting without first being scheduled on the CPU (D). Thus, preemptive scheduling operates under circumstances A and B only.
Which of the following methods are used to pass any number of parameters to the operating system through system calls?
ARegisters
BBlock or table in main memory
CStack
DBlock in main memory and stack
Answer:(D) Block in main memory and stack
Explanation
Operating systems employ three common techniques to pass parameters during system calls:
Registers — fastest approach, but limited by the fixed count of hardware processor registers, making it incapable of passing arbitrary numbers of parameters.
Block or table in main memory — parameters are stored in a designated memory block, and a single pointer to the block is passed in a register. This permits passing any number of parameters.
Stack — parameters are pushed onto the program stack by the calling process and popped off by the operating system kernel, which also accommodates arbitrary numbers of parameters.
Therefore, both block in memory and stack methods accommodate any number of parameters.
Suppose a system has 12 magnetic tape drives and at time t₀, three processes are allotted tape drives out of their need. At time t₀, the system is in safe state. Which of the following is safe sequence so that deadlock is avoided?
Process
Maximum Needs
Current Needs
P₀
10
5
P₁
4
2
P₂
9
2
A(P₀, P₁, P₂)
B(P₁, P₀, P₂)
C(P₂, P₁, P₀)
D(P₀, P₂, P₁)
Answer:(B) (P₁, P₀, P₂)
Explanation
Applying Banker's safety algorithm:
Calculate Available resources:
Total drives = 12
Allocated drives = 5 + 2 + 2 = 9
Available = 12 - 9 = 3
Calculate Remaining Need (Max - Current):
P₀: 10 - 5 = 5
P₁: 4 - 2 = 2
P₂: 9 - 2 = 7
Trace allocation:
With Available = 3, only P₁'s need (2) ≤ 3 can be satisfied. P₁ runs, finishes, and releases 2 drives. New Available = 3 + 2 = 5.
With Available = 5, P₀'s need (5) ≤ 5 is satisfied. P₀ runs, finishes, and releases 5 drives. New Available = 5 + 5 = 10.
With Available = 10, P₂'s need (7) ≤ 10 is satisfied. P₂ runs and finishes.
Consider a disk system with 100 cylinders. The requests to access the cylinders occur in the following sequence:
4, 34, 10, 7, 19, 73, 2, 15, 6, 20
Assuming that the head is currently at cylinder 50, what is the time taken to satisfy all requests if it takes 1 ms to move from the cylinder to adjacent one and the shortest seek time first policy is used?
A357 ms
B238 ms
C276 ms
D119 ms
Answer:(D) 119 ms
Explanation
Shortest Seek Time First (SSTF) always services the pending request closest to the current head position.
Beginning at cylinder 50, the closest request is 34 (distance 16).
From 34, the sequence moves greedily through remaining closer tracks:
34 -> 20 -> 19 -> 15 -> 10 -> 7 -> 6 -> 4 -> 2.
The total movement from 50 down to 2 is:
50 - 2 = 48 cylinders.
From cylinder 2, the only remaining request is 73.
The seek distance from 2 to 73 is:
73 - 2 = 71 cylinders.
A computer has six tape drives with n processes competing for them. Each process may need two drives. What is the maximum value of n for the system to be deadlock free?
A5
B4
C3
D6
Answer:(A) 5
Explanation
Deadlock avoidance requires that the total available resources exceed the sum of peak demands minus one for each process.
Let R = 6 tape drives and each process have a maximum demand of 2 drives.
In the worst-case allocation without deadlock, every process holds (2 - 1) = 1 drive, and at least 1 spare drive remains to satisfy one process:
R >= n * (Max - 1) + 1
6 >= n * (2 - 1) + 1
6 >= n + 1
n <= 5.
Thus, the system is guaranteed deadlock-free for up to 5 processes.
Consider that a process has been allocated 3 frames and has a sequence of page referencing as 1, 2, 1, 3, 7, 4, 5, 6, 3, 1.
What shall be the difference in page faults for the above string using the algorithms of LRU and optimal page replacement for referencing the string?
A2
B0
C1
D3
Answer:(A) 2
Explanation
Tracing page faults across 3 frames for the reference string:
1, 2, 1, 3, 7, 4, 5, 6, 3, 1.
At a particular time of computation, the value of a counting semaphore is 7. Then 20 P (wait) operations and 15 V (signal) operations are completed on this semaphore. What is the resulting value of the semaphore?
A28
B12
C2
D42
Answer:(C) 2
Explanation
Counting semaphores modify their counter value according to the completed synchronization primitives:
Each P (wait) operation decrements the semaphore counter by 1.
Each V (signal) operation increments the semaphore counter by 1.
Initial value = 7.
Net change = -20 (from P operations) + 15 (from V operations) = -5.
For a magnetic disk with concentric circular tracks, the seek latency is not linearly proportional to the seek distance due to
Anon-uniform distribution of requests
Barm starting or stopping inertia
Chigher capacity of tracks on the periphery of the platter
Duse of unfair arm scheduling policies
Answer:(B) arm starting or stopping inertia
Explanation
Seek time accounts for the mechanical movement of the disk head arm across cylinder tracks.
The physical arm assembly must accelerate from rest, cruise across intermediate tracks, decelerate, and settle precisely onto the target track. Because of arm starting and stopping inertia, mechanical latency follows a non-linear trajectory (proportional to the square root of seek distance for short distances) rather than a linear rate.
Which of the following UNIX/Linux pipes will count the number of lines in all the files having .c and .h as their extension in the current working directory?
Acat *.ch | wc -l
Bcat *.[c-h] | wc -l
Ccat *.[ch] | ls -l
Dcat *.[ch] | wc -l
Answer:(D) cat *.[ch] | wc -l
Explanation
Evaluating Unix shell wildcards and pipe operators:
The character class glob pattern *.[ch] matches any file ending with either .c or .h in the current working directory.
The cat command concatenates and streams the file contents to standard output.
The pipe operator | redirects that output into wc -l, which counts the total number of lines.
cat *.ch looks for extension .ch, *.[c-h] matches any character from c through h, and piping into ls -l ignores incoming stdin stream data.
The minimum number of page frames that must be allocated to a running process in a virtual memory environment is determined by
Apage size
Bphysical size of memory
Cthe instruction set architecture
Dnumber of processes in memory
Answer:(C) the instruction set architecture
Explanation
In virtual memory management, a process must hold enough physical frames in memory to complete execution of any single machine instruction without generating recurrent page faults.
Because instructions may feature multiple memory references (such as indirect addressing modes referencing instructions, pointers, and operands across separate page boundaries), the minimum number of frames required is determined by the computer's Instruction Set Architecture (ISA).
Consider three CPU intensive processes, which require 10, 20 and 30 units of time and arrive at times 0, 2 and 6 respectively. How many context switches are needed if the operating system implements a shortest remaining time first scheduling algorithm? Do not count the context switches at time zero and at the end.
A4
B2
C3
D1
Answer:(B) 2
Explanation
Simulating Shortest Remaining Time First (SRTF) scheduling across timeline t:
At t = 0: P1 arrives with burst 10 and begins running.
At t = 2: P2 arrives with burst 20. Remaining time of P1 is 8 < 20, so P1 continues.
At t = 6: P3 arrives with burst 30. Remaining time of P1 is 4 < 30, so P1 continues.
At t = 10: P1 finishes. CPU context switches to P2 (switch 1).
At t = 30: P2 finishes. CPU context switches to P3 (switch 2).
At t = 60: P3 finishes.
Excluding boundaries at start and end, exactly 2 context switches take place.
Consider a disk pack with 32 surfaces, 64 tracks and 512 sectors pack. 256 bytes of data are stored in a bit serial manner in a sector. The number of bits required to specify a particular sector in the disk is
A19
B22
C20
D18
Answer:(C) 20
Explanation
The number of bits required to uniquely address a sector is obtained by computing the logarithm base 2 of the total number of sectors in the disk pack.
Total sectors = (Surfaces) * (Tracks per surface) * (Sectors per track)
Surfaces = 32 = 2^5 surfaces (5 bits)
Tracks per surface = 64 = 2^6 tracks (6 bits)
Sectors per track = 512 = 2^9 sectors (9 bits)
Total sectors in pack = 2^5 * 2^6 * 2^9 = 2^20 sectors.
Bits required = log2(2^20) = 20 bits.
Option A (19) fails to account for one bit across the surface or track components.
Option B (22) erroneously includes byte addressing within a sector.
Option D (18) undercounts the total addressing space.
Suppose for a process P, reference to pages in order are 1, 2, 4, 5, 2, 1, 2, 4. Assume that main memory can accommodate 3 pages and the main memory has already pages 1 and 2 in the order 1 – first, 2 – second. At this moment, assume FIFO Page Replacement Algorithm is used then the number of page faults that occur to complete the execution of process P is
A4
B6
C5
D3
Answer:(C) 5
Explanation
Tracing the FIFO page replacement algorithm across the reference string results in exactly 5 page faults.
Initial frames: [1, 2, empty], with 1 loaded first and 2 loaded second.
Suppose P, Q and R are co-operating processes satisfying Mutual Exclusion condition. Then, if the process Q is executing in its critical section then
ABoth ‘P’ and ‘R’ execute in critical section.
B‘R’ executes in critical section.
C‘P’ executes in critical section.
DNeither ‘P’ nor ‘R’ executes in their critical section.
Answer:(D) Neither ‘P’ nor ‘R’ executes in their critical section.
Explanation
Mutual exclusion strictly guarantees that no more than one process can execute inside its critical section concurrently.
If process Q is currently executing in its critical section, any other cooperating process (P or R) attempting to enter must be blocked until Q leaves.
Therefore, neither P nor R can execute in their critical section simultaneously (Option D).
Options A, B, and C directly violate the mutual exclusion condition.
(i) Windows XP supports both peer-peer and client-server networks.
(ii) Windows XP implements Transport protocols as drivers that can be loaded and unloaded from the system dynamically.
Choose the correct answer from the code given below :
Code :
ABoth (i) and (ii)
B(ii) only
C(i) only
DNeither (i) nor (ii)
Answer:(A) Both (i) and (ii)
Explanation
Both statements describe standard networking capabilities and kernel architecture in Windows XP.
Statement (i) is true because Windows XP supports workgroup peer-to-peer networking as well as joining Active Directory domain controllers in client-server environments.
Statement (ii) is true because transport protocols (such as TCP/IP) in Windows NT/XP are implemented as NDIS/TDI device drivers that the kernel loads and unloads dynamically without requiring a system reboot.
Therefore, both statements (i) and (ii) are true (Option A).
To overcome difficulties in Readers–Writers problem, which of the following statement/s is/are true ?
(i) Writers are given exclusive access to shared objects.
(ii) Readers are given exclusive access to shared objects.
(iii) Both Readers and Writers are given exclusive access to shared objects.
Choose the correct answer from the code given below :
Code :
ABoth (ii) and (iii)
B(iii) only
C(i) only
D(ii) only
Answer:(C) (i) only
Explanation
In concurrency control for the Readers-Writers problem, writers require exclusive access while readers share concurrent access.
Multiple reader processes may concurrently read shared data without causing race conditions or data inconsistency.
Writers modify shared records, so write operations require mutual exclusion from all other readers and all other writers to prevent dirty reads and corrupted writes.
Giving readers exclusive access (statements ii and iii) unnecessarily serializes read operations and contradicts the fundamental design of the Readers-Writers synchronization protocol.
Suppose a system has 12 instances of some resource with n processes competing for that resource. Each process may require 4 instances of the resource. The maximum value of n for which the system never enters into deadlock is
A3
B6
C4
D5
Answer:(A) 3
Explanation
Applying the deadlock-free resource allocation condition guarantees that the maximum number of safe processes is 3.
For R resource instances, n processes, each with maximum demand m:
To guarantee avoidance of deadlock, the total resources must satisfy:
R >= n * (m - 1) + 1
Given R = 12 and m = 4:
12 >= n * (4 - 1) + 1
12 >= 3n + 1
3n <= 11
n <= 11 / 3 = 3.66.
Because n must be an integer, the maximum value is n = 3 (Option A).
With n = 4, each of the 4 processes could hold 3 instances (consuming all 12 resources), causing a deadlock.
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