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GATE Computer Science Engineering(CSE) 2027 Test: Introduction, Process,


MCQ Practice Test & Solutions: Test: Introduction, Process, Threads & CPU Scheduling- 2 (15 Questions)

You can prepare effectively for Computer Science Engineering (CSE) GATE Computer Science Engineering(CSE) 2027 Mock Test Series with this dedicated MCQ Practice Test (available with solutions) on the important topic of "Test: Introduction, Process, Threads & CPU Scheduling- 2". These 15 questions have been designed by the experts with the latest curriculum of Computer Science Engineering (CSE) 2026, to help you master the concept.

Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 45 minutes
  • - Number of Questions: 15

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Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 1

The only state transition that is initiated by the user process itself is

Detailed Solution: Question 1

The only state transition that is initiated by user process itself is block.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 2

Pre-emptive scheduling, is the strategy of temporarily suspending a running process

Detailed Solution: Question 2

In pre-emptive scheduling suspanding temporarily a running process is done before the CPU time slice expires.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 3

In Round Robin CPU scheduling, as the time quantum is increased, the average turn around time

Detailed Solution: Question 3

The Round Robin CPU scheduling technique will become insignificance if the time slice is very small or very large. When it is large it acts as FIFO which does not have a fixed determination over the turn around time. If processes with small burst arrived earlier turn around time will be less else it will be more.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 4

Suppose that a process spends a fraction p of its time in I/O wait state. With n processes in memory at once, the probability that all n processes are waiting for I/O is 

Detailed Solution: Question 4

Considering that a process spends a fraction p of its time in I/O wait state. As n processes are there is memory at once, then Probability that all n processes are waiting for I/O = p x p x p x ... p (n times) = pn
Note: Moreover, the CPU utilization is given by the formula CPU utilization = 1 - pn.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 5

In a multi-user operating system, 20 requests are made to use a particular resource per hour, on an average. The probability that no requests are made in 45 minutes is

Detailed Solution: Question 5

The arrival pattern is a Poission distribution.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 6

In which of the following scheduling policies doe context switching never take place?
1. Round-robin
2. Shortest job first(non pre-emptive)
3. Pre-emptive
4. First-cum-first-served

Detailed Solution: Question 6

Context switching takes place when a process is preempted (forcefully) and another process goes into the running state. In FIFO and SJF (non- preemptive) techniques, the processes finishes their execution then only their context is switched to other processes.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 7

Which of the following is the most suitable scheduling scheme in a real-time operating system?

Detailed Solution: Question 7

In preemptive scheduling the process with higher priority is executed first and then next higher process with least priority is executed after the higher priority processes are executed, if occured simultaneously. This is the ideal technique to be used in real time O.S. in which critical or higher priority processes are to be entertained first.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 8

Which of the following scheduling algorithms gives minimum average waiting time?

Detailed Solution: Question 8

SJF will gives minimum average waiting time. SJF ensures that all smaller processes will finish first. So, less processes will be waiting since more smaller processes will end up in less number of time. Use processes waiting means less waiting time overall which will decrease the average waiting time.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 9

A process executes the following code: 

The total number of child processes created is

Detailed Solution: Question 9

A fork () call creates the child processes for the following loop creates 2n - 1 child processes, excluding the parent process. And including parent it is 2n.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 10

While designing a kernel, an operating system designer must decide whether to support kernel- level or user-level threading. Which of the following statements is/are true?
1. Kernel-level threading may be preferable to user-level threading because storing information about user-level threads in the process control block would create a security risk.
2. User-level threading may be preferable to kernel-level threading because in user-level threading, if one thread blocks on I/O, the process can continue.

Detailed Solution: Question 10

Kernel level threading may be preferrable to user level threading because storing information about user level thread in a PCB would create a security risk i.e. with each access to the non-critical services. We are going to the domain where both critical and non-critical services are residing. Any harm in this domain may creates problems to the critical services. Hence Kernel level threading is preferable.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 11

Which of the following should be allowed only in Kernel mode?
1. Changing mapping from virtual to physical address.
2. Mask and unmask interrupts
3. Disabling all interrups 
4. Reading status of processor
5. Reading time of day

Detailed Solution: Question 11

Only critical services must resides in the kernel. All services mentioned except reading status of processors and reading time of the day are critical.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 12

Consider the following statements with respect to user-level threads and Kernel-supported threads
(i) Context switching is faster with Kernel- supported threads.
(ii) For user-level threads, a system call can block the entire process.
(iii) Kernel-supported threads can be scheduled independently.
(iv) User-level threads are transparent to the Kernel.
Which of the above statements are true?

Detailed Solution: Question 12

(i) is false. Context switching between user-level threads is generally faster because it is done entirely in user space and does not require a switch to kernel mode or invocation of the kernel scheduler.

(ii) is true. If a user-level thread makes a blocking system call, the kernel sees the whole process as a single schedulable entity and the blocking call can cause the entire process (and thus all its user-level threads) to block.

(iii) is true. Kernel-supported threads are visible to the kernel and can be scheduled independently by the kernel scheduler.

(iv) is true. The kernel is unaware of the presence of multiple user-level threads inside a process, so user-level threads are transparent to the Kernel.

Therefore, statements (ii)(iii) and (iv) are correct and option (A) is the right choice.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 13

Consider the following statements about user level threads and kernel level threads. Which one of the following statements is FALSE?

Detailed Solution: Question 13

  • User level threads are not recognised by operating system, hence do not need any hardware support. Since, user level threads are not recognised by operating system, no separate PCB (Process Control Block) information is to stored. Hence, context switching is faster.  
  • Since, Kernel level threads are like separate processes. So, blocking one thread will not block other threads.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 14

Which of the following algorithms favour CPU bound processes?
1. Round-robin
2. Rirst-come-first-served
3. Multilevel feedback queues

Detailed Solution: Question 14

Only FCFS and non-preempting algorithms favour CPU bound processes.

Test: Introduction, Process, Threads & CPU Scheduling- 2 - Question 15

Consider a system contains two types of processes CPU bound processes and I/O bound processes, what will be the sufficient condition when ready queue becomes empty.

Detailed Solution: Question 15

It all the processes are I/O bound, they will go or reside into BLOCKED state and hence READY QUEUE becomes empty.

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