Operating
systems exams - multiple choice questions (MCQ) and answers, Objective
type questions, OS exam questions, MCQ in memory management, process management, disk scheduling, process state transition, race condition
Operating
Systems MCQ questions and answers – Set 23
1.
A computer system has a 36-bit virtual address space with a page size of 8K,
and 4 bytes per page table entry. How many pages are in the virtual address
space?
a) 8,192
b) 4,096
c) 68,71,94,76,736
d) 83,88,608
|
Answer: (d) 83,88,608
A
36 bit address can address 2^36 bytes in a byte addressable machine. Since
the size of a page 8K bytes (2^13), the number of addressable pages is 2^36 /
>2^13 = 2^23 = 83,88,608.
|
2.
Which is true about process termination?
a) The waitpid operation
always blocks the calling process
b) A child process can
kill another child process of the same parent process
c) A child process that
exits before the parent process does a waitpid operation becomes a zombie
process
d) None of the above
|
Answer: (d) None of the
above
Process
termination is a technique in which a process is terminated and releases the
CPU after completing the execution. It may happen in the following
situations;
normal
conditions (exit() call), abnormal termination (due to programming error,
runtime error, i/o error etc.), parent may terminate child process, due to
memory requirement etc.
|
3.
In the process state transition diagram, the transition from the Ready state to
the Running state:
a) Indicates that a
process was preempted by another process
b) Indicates that a process
is chosen by CPU for execution
c) Indicates that a
process is done waiting for an I/O operation
d) Indicates that a
process has blocked on an I/O operation
|
Answer: (b) Indicates
that a process is chosen by CPU for execution
Ready
state - The task has completed preparations for running, but cannot run
because a task with higher precedence is running. In this state, the task is
able to run whenever it becomes the task with the highest precedence among
the tasks in READY state.
Running
state - A process moves into the running state when it is chosen for
execution. The process's instructions are executed by one of the CPUs (or
cores) of the system.
The
transition from ready state to running state happens if the process is chosen
for execution by the CPU.
|
4.
Disk scheduling algorithms attempt to reduce
a) Transfer time
b) Rotational latency
c) Seek time
d) All of the above
|
Answer: (c) Seek time
In
operating systems, seek time is very important. Since all device requests are
linked in queues, the seek time is increased causing the system to slow down.
Disk Scheduling Algorithms are used to reduce the total seek time of any
request.
Seek
time – Seek time is the time taken for a hard disk controller to locate a
specific piece of stored data. It measures the delay for the disk head to
reach the track.
Transfer
time – It is about how long the actual data read or write takes. It is the
time actually taken to transfer a span of data.
Rotational
latency - Rotational latency (or just latency) is the delay waiting for the
rotation of the disk to bring the required disk sector under the read-write
head. It accounts for the time to get to the right sector.
|
5.
Two threads in the same process share the same
a) Address space
b) Program counter
c) Stack
d) All of the above
|
Answer: (a) Address
space
In
a multi-threaded process, all of the process’ threads share the same memory
and open files. Within the shared memory, each thread gets its own stack.
Each thread has its own instruction pointer and registers.
Threads
allow for efficient sharing of information, since the address space is shared,
multiple threads can directly access the same memory.
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Related links:
OS Interview questions with answers
How disk scheduling algorithms reduce the seek time
GATE questions in system programming and operating systems
Why threads are good at information sharing
Operating systems interview questions for competitive exams
objective type OS questions for IIT JEE entrance exam
