IBPS IT Officer MCQ – 11 (Operating System)


The IBPS Specialist Officer (SO) Exam 2016 for recruitment of specialist officer vacancies in public sector banks has been announced. So we D2G Team is here with IBPS SO Material.

operating system

5-1 What does Belady’s Anomaly related to?
A. Page Replacement Algorithm
B. Memory Management Algorithm
C. Deadlock Prevention Algorithm
D. Disk Scheduling Algorithm

5-2 What are the two types of Semaphore?
A. Digital Semaphores and Binary Semaphores
B. Analog Semaphores and Octal Semaphores
C. Counting Semaphores and Binary Semaphores
D. Critical Semaphores and System Semaphores

5-3 What is dispatch latency?
A. The time taken by the dispatcher to stop one process and start another
B. The time taken by the processor to write a file into disk
C. The whole time taken by all processor
D. None of Above

5-4 Which of the following is not process states?
A. New
B. Running
C. Ready
D. Finished

5-5 What are the requirements for the solution to critical section problem?
A. Mutual Exclusion
B. Progress
C. Bounded Waiting
D. All of Above

5-6 Which of the following is the allocation method of a disk space?
A. Contiguous allocation
B. Linked allocation
C. Indexed allocation
D. All of the Above

5-7 What is the method of handling deadlocks?
A. Use a protocol to ensure that the system will never enter a deadlock state.
B. Allow the system to enter the deadlock state and then recover.
C. Pretend that deadlocks never occur in the system.
D. All of the Above

5-8 What do you mean by Memory Compaction?
A. Combine multiple equal memory holes into one big hole
B. Combine multiple small memory holes into one big hole
C. Divide big memory hole into small holes
D. Divide memory hole by 2

5-9 What is Thrashing?
A. A high paging activity is called thrashing.
B. A high executing activity is called thrashing
C. A extremely long process is called thrashing
D. A extremely long virtual memory is called thrashing

5-10 What hole will allocates in “WorstFit” algorithm of memory management?
A. It allocates the smaller hole than required memory hole
B. It allocates the smallest hole from the available memory holes
C. It allocates the largest hole from the available memory holes
D. It allocates the exact same size memory hole

1 – A / 2 – C / 3 – A / 4 – D / 5 – D / 6 – D / 7 – D / 8 – B / 9 – A / 10 – C

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