As you want here I am giving UGC-NET exam Multiple Choice Questions related to Operating System topics:

Dijkstra's banking algorithm in an operating system, solves the problem of
A.

deadlock avoidance
B.

deadlock recovery
C.

mutual exclusion
D.

context switching



2:

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

increases
B.

decreases
C.

remains constant
D.

varies irregularly



3:

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

before the CPU time slice expires
B.

to allow starving processes to run
C.

when it requests I/O
D.

none of these



4:

Consider a set of 5 processes whose arrival time. CPU time needed and the priority are given below

Process Priority Arrival Time (in ms) CPU Time Needed (in ms) Priority
P1 0 10 5
P2 0 5 2
P3 2 3 1
P4 5 20 4
P5 10 2 3


smaller the number, higher the priority.

If the CPU scheduling policy FCFS, the average waiting time will be
A.

12.8 ms
B.

8 ms
C.

6 ms
D.

none of these



5:

Consider a set of 5 processes whose arrival time. CPU time needed and the priority are given below

Process Priority Arrival Time (in ms) CPU Time Needed (in ms) Priority
P1 0 10 5
P2 0 5 2
P3 2 3 1
P4 5 20 4
P5 10 2 3


smaller the number, higher the priority.

If the CPU scheduling policy is SJF, the average waiting time (without pre-emption) will be
A.

16 ms
B.

12.8
C.

6.8 ms
D.

none of these

Consider a set of 5 processes whose arrival time. CPU time needed and the priority are given below

Process Priority Arrival Time (in ms) CPU Time Needed (in ms) Priority
P1 0 10 5
P2 0 5 2
P3 2 3 1
P4 5 20 4
P5 10 2 3


smaller the number, higher the priority.

If the CPU scheduling policy is SJF with pre-emption, the average waiting time will be
A.

8 ms
B.

14 ms
C.

5.6 ms
D.

none of these



7:

Consider a set of 5 processes whose arrival time. CPU time needed and the priority are given below
Process Priority Arrival Time (in ms) CPU Time Needed (in ms) Priority
P1 0 10 5
P2 0 5 2
P3 2 3 1
P4 5 20 4
P5 10 2 3

smaller the number, higher the priority.

If the CPU scheduling policy is priority scheduling without pre-emption, the average waiting time will be
A.

12.8 ms
B.

11.8 ms
C.

10.8 ms
D.

none of these



8:

Consider a set of 5 processes whose arrival time. CPU time needed and the priority are given below

Process Priority Arrival Time (in ms) CPU Time Needed (in ms) Priority
P1 0 10 5
P2 0 5 2
P3 2 3 1
P4 5 20 4
P5 10 2 3


smaller the number, higher the priority.

If the CPU scheduling policy is priority sche duling with pre-emption, the average waiting time will be
A.

19 ms
B.

7.6 ms
C.

6.8 ms
D.

none of these



9:

Cascading termination refers to termination of all child processes before the parent terminates
A.

normally
B.

abnormally
C.

normally or abnormally
D.

none of these