Operating Systems Quiz - AAT

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  • 1/20 Questions

    The possible solution to recover a deadlock is

    • Abort all deadlocked processes
    • Resource Preemption
    • Abort process by process
    • All the above
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About This Quiz

This Operating Systems QUIZ - AAT assesses understanding of process synchronization, focusing on race conditions, critical section problems, semaphore operations, and the Readers-Writers problem. It is designed to enhance practical knowledge of operating system functionalities crucial for IT professionals.

Operating Systems Quiz - AAT - Quiz

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  • 2. 

    An Access Matrix represents

    • Set of Objects

    • Set of Domains

    • Set of Access rights

    • All the above

    Correct Answer
    A. All the above
    Explanation
    An Access Matrix represents all the above options, which include a set of objects, a set of domains, and a set of access rights. An Access Matrix is a security model that defines the permissions and access control for different entities in a system. It consists of rows representing objects or resources, columns representing domains or subjects, and cells representing the access rights or permissions that domains have over objects. This matrix helps in managing and enforcing security policies by specifying who can access what resources and what actions they can perform.

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  • 3. 

    A semaphore in synchronization process is used for

    • Mutual Exclusion

    • Synchronization

    • Resource allocation

    • All the above

    Correct Answer
    A. All the above
    Explanation
    A semaphore in synchronization process is used for all the above purposes. Mutual Exclusion ensures that only one process can access a shared resource at a time, preventing conflicts and ensuring data integrity. Synchronization allows processes to coordinate their actions and ensure proper sequencing, avoiding race conditions. Resource allocation ensures that processes are allocated resources in a fair and controlled manner, preventing resource starvation. Therefore, a semaphore can be used for all of these purposes.

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  • 4. 

    The following operation on a file does allow only erasing the contents of a file but not its attributes.

    • Deleting a file

    • Renaming a file

    • Removing entry from directory

    • Truncating a file

    Correct Answer
    A. Truncating a file
    Explanation
    Truncating a file refers to the operation of erasing the contents of a file while preserving its attributes such as file name, size, and permissions. Unlike deleting a file, renaming a file, or removing an entry from a directory, truncating specifically focuses on removing the data within the file without affecting its attributes. Therefore, truncating a file is the correct answer as it allows only erasing the contents of a file without altering its attributes.

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  • 5. 

    When several processes access and manipulate the same data simultaneously and the result depends on particular order in which the access takes place is called

    • Synchronization

    • Concurrency

    • Race Condition

    • Inconsistency

    Correct Answer
    A. Race Condition
    Explanation
    A race condition occurs when multiple processes or threads access and manipulate the same data simultaneously, and the final result depends on the order in which the access takes place. In other words, the outcome of the program becomes unpredictable and inconsistent due to the non-deterministic nature of the concurrent execution. This can lead to bugs, data corruption, or unexpected behavior in the program.

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  • 6. 

    Which of the following is not the necessary condition for a deadlock to occur?

    • Hold and Wait

    • Preemption

    • Mutual Exclusion

    • Circular Wait

    Correct Answer
    A. Preemption
    Explanation
    Preemption is not a necessary condition for a deadlock to occur. Deadlock can occur even without preemption. Preemption refers to the act of forcibly removing a resource from a process and allocating it to another process. While preemption can help prevent or resolve deadlocks, it is not a required condition for deadlock to happen. Deadlock can occur due to hold and wait, mutual exclusion, and circular wait, where processes are waiting for resources in a circular chain.

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  • 7. 

    A vertical set of all tracks at same position in a disk pack is considered as

    • Tracks

    • Cylinder

    • Contiguous tracks

    • Non contiguous tracks

    Correct Answer
    A. Cylinder
    Explanation
    A vertical set of all tracks at the same position in a disk pack is called a cylinder. This means that all the tracks that are aligned vertically across the different platters in the disk pack and at the same radius or position are considered as a cylinder. It is a way to organize and access data on a disk, as it allows for efficient sequential access of data across multiple platters.

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  • 8. 

    A semaphore variable S has a value 12. What is the value of S if the synchronization problem performs 4-P (wait) operations and 6-V (signal) operations

    • 16

    • 14

    • 18

    • 15

    Correct Answer
    A. 14
    Explanation
    The semaphore variable S has an initial value of 12. For every P (wait) operation, the value of S is decremented by 1. For every V (signal) operation, the value of S is incremented by 1. In this case, there are 4 P operations and 6 V operations. Therefore, the value of S would be 12 - 4 + 6 = 14.

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  • 9. 

    The shortest seek distance among the tracks is achieved using which of the following disk scheduling algorithm.

    • FCFS

    • SCAN

    • LOOK

    • SSTF

    Correct Answer
    A. SSTF
    Explanation
    SSTF (Shortest Seek Time First) is the disk scheduling algorithm that achieves the shortest seek distance among the tracks. This algorithm selects the request with the shortest seek time from the current head position, resulting in reduced seek time and improved overall disk performance. By choosing the nearest track first, SSTF minimizes the movement of the disk arm, leading to faster data retrieval and improved system efficiency.

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  • 10. 

    In which of the following directory structure Master File Directory (MFD) is maintained for every User File Directory (UFD)

    • Single Level

    • Two Level

    • Tree Structured

    • Acyclic Graph

    Correct Answer
    A. Two Level
    Explanation
    In a two-level directory structure, the Master File Directory (MFD) is maintained for every User File Directory (UFD). This means that each user has their own UFD, which contains a list of files and directories specific to that user. The MFD keeps track of all the UFDs and their corresponding user information. This two-level structure allows for better organization and management of files and directories, as each user has their own separate UFD.

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  • 11. 

    In Readers – Writers Problem for process synchronization which of the following is/are true during read. I   Any number of reader processes can enter into critical section II  Any number of writer processes can enter into critical section III First reader has to block the writer process IV Last Reader has to resume the writer process

    • I , II and III only

    • I, II and IV only

    • I, III and IV only

    •  II , III and IV only

    Correct Answer
    A. I, III and IV only
    Explanation
    In the Readers-Writers problem, only one writer can enter the critical section at a time to maintain data consistency. Therefore, statement II is false. However, any number of reader processes can enter the critical section simultaneously as they do not modify the shared data. The first reader process has to block the writer process to prevent any concurrent writes while the last reader process has to resume the writer process to allow it to enter the critical section. Therefore, statements I, III, and IV are true.

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  • 12. 

    For deadlock avoidance  the false statement is I  A safe sequence keeps always the system in safe state II Every unsafe state is deadlock state III    A deadlock state is always unsafe state.

    • I

    • II

    • III

    • None of the above

    Correct Answer
    A. II
    Explanation
    The false statement for deadlock avoidance is II. Every unsafe state is not necessarily a deadlock state. An unsafe state is a state where a system may lead to a deadlock, but it does not guarantee that a deadlock will occur. A deadlock state is a specific type of unsafe state where processes are unable to proceed because they are waiting for resources held by other processes, resulting in a circular dependency.

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  • 13. 

    The protocol which is not used in deadlock prevention?

    • No Hold

    • No Wait

    • Shareable Resources

    • No Preemption

    Correct Answer
    A. No Preemption
    Explanation
    No Preemption is not used in deadlock prevention because it allows a process to hold on to a resource indefinitely, even if other processes are waiting for that resource. In deadlock prevention, resources should be preempted or taken away from a process if necessary to avoid deadlock situations. However, No Preemption does not allow this, making it ineffective in preventing deadlocks.

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  • 14. 

    Which of the following condition does not corresponds to critical section problem?

    • Consistency

    • Bounded Waiting

    • Progress

    • Mutual Exclusion

    Correct Answer
    A. Consistency
    Explanation
    Consistency is not a condition that corresponds to the critical section problem. The critical section problem focuses on ensuring mutual exclusion, which means that only one process can access the critical section at a time. Bounded waiting refers to the condition where a process waiting to enter the critical section is guaranteed to eventually enter it. Progress ensures that if no process is in the critical section and some processes want to enter it, then only those processes that do not violate other conditions can enter. Consistency, on the other hand, refers to the property of a system where all replicas or copies of data are always in the same state.

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  • 15. 

    Which of the following is not a technique for free space management?

    • Bit Vector

    • Linked List

    • Counting

    • Linear List

    Correct Answer
    A. Linear List
    Explanation
    The technique of linear list is not used for free space management. In free space management, the goal is to keep track of the available and allocated memory blocks. Techniques like bit vector, linked list, and counting are commonly used for this purpose. However, a linear list is not a specific technique used for free space management. It is a general term that refers to a list where elements are arranged sequentially in a linear manner.

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  • 16. 

    The file allocation method for which the directory structure uses a separate block for holding block addresses of available or free disk blocks.

    • Contiguous allocation

    • Linked allocation

    • Indexed allocation

    • Non contiguous allocation

    Correct Answer
    A. Indexed allocation
    Explanation
    Indexed allocation is the correct answer because it is a file allocation method where the directory structure uses a separate block called an index block to hold the block addresses of available or free disk blocks. This index block acts as a lookup table, allowing quick access to the data blocks associated with a file. Each file has its own index block, which contains pointers to the actual data blocks on the disk. This method provides efficient random access to files but requires additional overhead for maintaining the index blocks.

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  • 17. 

    Wait for Graph model is used for which of the following deadlock handling method.

    • Deadlock Prevention

    • Deadlock Avoidance

    • Deadlock detection

    • Deadlock Recovery

    Correct Answer
    A. Deadlock detection
    Explanation
    The Wait for Graph model is used for deadlock detection. In this model, a directed graph is constructed to represent the resource allocation and request relationships between processes. If a cycle is found in the graph, it indicates the presence of a deadlock. This model helps in identifying and resolving deadlocks by determining which processes are involved and which resources they are waiting for. It does not prevent or avoid deadlocks, but rather detects them after they have occurred.

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  • 18. 

    The module which translates logical block addresses of disk to physical block addresses of disk is performed  by

    • Basic File System

    • Logical File System

    • File Organization Module

    • I/O Controller

    Correct Answer
    A. File Organization Module
    Explanation
    The File Organization Module is responsible for translating logical block addresses of a disk to physical block addresses. This module determines how files are organized and stored on the disk, including the allocation of blocks and the mapping of logical addresses to physical addresses. It ensures that data is stored and retrieved efficiently, optimizing disk usage and access times.

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  • 19. 

    The method which is not used in the implementation of Access Matrix.

    • Global Table

    • Capability Lists for Domains

    • Access lists for domains

    • Lock Key mechanism

    Correct Answer
    A. Access lists for domains
    Explanation
    Access lists for domains are not used in the implementation of an Access Matrix. An Access Matrix is a security model that defines the rights and permissions of subjects (users) on objects (resources). It typically uses a Global Table to represent the subjects and objects, and a Capability List for each subject to specify their access rights. The Lock Key mechanism is used to control access to the objects. However, Access lists for domains are not part of the Access Matrix implementation and therefore not used.

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  • 20. 

    A computer has six tape drives, with n processes competing for them. Each process may need 2 drives. What i s the maximum value of “n” for the system to be deadlock free?

    • 3

    • 4

    • 5

    • 6

    Correct Answer
    A. 5
    Explanation
    The maximum value of n for the system to be deadlock-free in this scenario is 5. Let’s break it down:
    Given tape drives: 6
    Each process may need: 2 drives
    When we allocate 1 drive to each process, the total number of processes would be 6. However, in this case, there would definitely be a deadlock because every process holds 1 drive and waits for another drive held by a different process. To avoid deadlock, we need to reduce the number of processes by 1.
    Hence, the maximum value of n (number of processes) is 6 - 1 = 5. 

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  • Current Version
  • Feb 06, 2024
    Quiz Edited by
    ProProfs Editorial Team
  • Apr 06, 2020
    Quiz Created by
    A Ramesh
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