Operating-Systems

Question 1
Consider the following threads, T 1 , T 2 , and T 3 executing on a single processor, synchronized using three binary semaphore variables, S 1 , S 2 , and S 3 , operated upon using standard wait() and signal(). The threads can be context switched in any order and at any time. Which initialization of the semaphores would print the sequence BCABCABCA….?
A
S 1 = 1; S 2 = 1; S 3 = 1
B
S 1 = 1; S 2 = 1; S 3 = 0
C
S 1 = 1; S 2 = 0; S 3 = 0
D
S 1 = 0; S 2 = 1; S 3 = 1
Question 1 Explanation: 
Explanation: Since B is to be printed first semaphore S1 has to be set to 1, so that the thread enters its critical section and prints B. Then C needs to be printed, this happens when Thread 1 is blocked on semaphore s3 (blocked, s3=0). Then Thread1 invokes Thread 3 to print A.
Question 2
Which of the following statements is/are TRUE with respect to deadlocks?
A
Circular wait is a necessary condition for the formation of deadlock.
B
In a system where each resource has more than one instance, a cycle in its wait-for graph indicates the presence of a deadlock.
C
If the current allocation of resources to processes leads the system to unsafe state, then deadlock will necessarily occur.
D
In the resource-allocation graph of a system, if every edge is an assignment edge, then the system is not in deadlock state.
Question 3
Consider four processes P, Q, R, and S scheduled on a CPU as per round robin algorithm with a time quantum of 4 units. The processes arrive in the order P, Q, R, S, all at time t = 0. There is exactly one context switch from S to Q, exactly one context switch from R to Q, and exactly two context switches from Q to R. There is no context switch from S to P. Switching to a ready process after the termination of another process is also considered a context switch. Which one of the following is NOT possible as CPU burst time (in time units) of these processes?
A
P = 4, Q = 10, R = 6, S = 2
B
P = 2, Q = 9, R = 5, S = 1
C
P = 4, Q = 12, R = 5, S = 4
D
P = 3, Q = 7, R = 7, S = 3
Question 3 Explanation: 
The Gantt chart can be assumed as :
The combinations of given context switches are possible for options A,B and C. However, we can observe that option D does not follow the context switch. Hence, D is the answer.
Question 4
Consider two files systems A and B , that use contiguous allocation and linked allocation, respectively. A file of size 100 blocks is already stored in A and also in B. Now, consider inserting a new block in the middle of the file (between 50 th and 51 st block), whose data is already available in the memory. Assume that there are enough free blocks at the end of the file and that the file control blocks are already in memory. Let the number of disk accesses required to insert a block in the middle of the file in A and B are n A and n B , respectively, then the value of n A + n B is_________.
A
102
Question 4 Explanation: 
Assume that there are enough free blocks at the end of the file and that the file control blocks are already in memory.
For linked allocation we have to trace the first 51 blocks, which requires 51 disk accesses. Similarly for contiguous allocation we have to move the block number 50-100 after insertion of a new block, which takes 51 accesses. Hence, the answer is 51+51 = 102.
Question 5
Consider a demand paging system with four page frames (initially empty) and LRU page replacement policy. For the following page reference string
7, 2, 7, 3, 2, 5,3, 4, 6, 7, 7,1, 5, 6,1
the page fault rate, defined as the ratio of number of page faults to the number of memory accesses (rounded off to one decimal place) is_________.
A
0.60
Question 5 Explanation: 
We have a total of 9 page faults using LRU page replacement algorithm with on demand paging. Out of 15 references, 9 page faults implies the page fault rate is 9/16 = 0.60.
Question 6

A memory page containing a heavily used variable that was initialized very early and is in constant use is removed when

A
LRU page replacement algorithm is used
B
FIFO page replacement algorithm is used
C
LFU page replacement algorithm is used
D
None of the above
Question 6 Explanation: 
In FIFO, whichever comes first that can be removed first. If the variable was initialized very early, it is in set of first pages. So it was removed.
In LRU which can eliminate (or) removed which is least recently used.
In LFU the frequency of the page is more. So it is in constant use so cannot be replaced.
Question 7

Consider the following heap (figure) in which blank regions are not in use and hatched region are in use.

The sequence of requests for blocks of size 300, 25, 125, 50 can be satisfied if we use

A
either first fit or best fit policy (any one)
B
first fit but not best fit policy
C
best fit but first fit policy
D
None of the above
Question 7 Explanation: 
In first fit, block request will be satisfied from the first free block that fits it.
So, request for 300 will be satisfied by 350 size block reducing the free size to 50.
Request for 25, satisfied by 125 size block, reducing it to 125.
Request for 125 satisfied by the 125 size block.
And request for 50 satisfied by 50 size block.
So, all requests can be satisfied.
In best fit strategy, a block request is satisfied by the smallest block in which can fit it. So, request for 300 will be satisfied by 350 size block reducing the free size to 50.
Request for 25, satisfied by 50 size block as its the smallest size that fits 25, reducing it to 25.
Request for 125, satisfied by 150 size block, reducing it to 25.
Now, request for 50 cannot be satisfied as the two 25 size blocks are not contiguous.
Question 8

Consider the resource allocation graph given in the figure.

(a) Find if the system is in a deadlock state. (b) Otherwise, find a safe sequence.

A
Theory Explanation.
Question 9

Consider allocation of memory to a new process. Assume that none of the existing holes in the memory will exactly fit the process’s memory requirement. Hence, a new hole of smaller size will be created if allocation is made in any of the existing holes. Which one of the following statements is TRUE?

A
The hole created by worst fit is always larger than the hole created by first fit.
B
The hole created by best fit is never larger than the hole created by first fit.
C
The hole created by first fit is always larger than the hole created by next fit.
D
The hole created by next fit is never larger than the hole created by best fit.
Question 9 Explanation: 
The size of hole created using best fit is never greater than size created by first fit. The best fit chooses the smallest available partition to fit the size of the process. Whereas, first fit and next fit doesn’t consider the size of the holes available.
Question 10

Consider the following statements about process state transitions for a system using preemptive scheduling.

    I. A running process can move to ready state.
    II. A ready process can move to running state.
    III. A blocked process can move to running state.
    IV. A blocked process can move to ready state.

Which of the above statements are TRUE?

A
II and III only
B
I, II and III only
C
I, II, III and IV
D
I, II and IV only
Question 10 Explanation: 
Question 11

Consider the following set of processes, assumed to have arrived at time 0. Consider the CPU scheduling algorithms Shortest Job First (SJF) and Round Robin (RR). For RR, assume that the processes are scheduled in the order P1,P2,P3,P4.

If the time quantum for RR is 4 ms, then the absolute value of the difference between the average turnaround times (in ms) of SJF and RR (round off to 2 decimal places) is _____.

A
5.25
Question 11 Explanation: 
SJF:

Turn Around Time = (21 – 0) + (13 – 0) + (2 – 0) + (6 – 0), Average = 42/4 = 10.50

Turn Around Time (TAT) = (18 – 0) + (21 – 0) + (10 – 0) + (14 – 0), Average = 63/4 = 15.75
Absolute difference = |10.50-15.75| = 5.25
Question 12

Each of a set of n processes executes the following code using two semaphores a and b initialized to 1 and 0, respectively. Assume that count is a shared variable initialized to 0 and not used in CODE SECTION P.

wait (a); count = count+1;
if (count==n) signal (b);
signal (a); wait (b); signal (b);

What does the code achieve?

A
It ensures that all processes execute CODE SECTION P mutually exclusively.
B
It ensures that at most two processes are in CODE SECTION Q at any time.
C
It ensures that no process executes CODE SECTION Q before every process has finished CODE SECTION P.
D
It ensures that at most n-1 processes are in CODE SECTION P at any time.
Question 12 Explanation: 
The wait(a) ensures that the count value is correctly incremented (no race condition) if(count==n) signal (b); // This signal(b) statement is executed by the last (nth) process only. Rest of the n-1 processes are blocked on wait(b). Once the nth process makes signal(b) then the rest of the processes can proceed and enter Code section Q.
Question 13

Consider the following five disk access requests of the form (request id, cylinder number) that are present in the disk scheduler queue at a given time.

         (P, 155), (Q, 85), (R, 110), (S, 30), (T, 115) 

Assume the head is positioned at cylinder 100. The scheduler follows Shortest Seek Time First scheduling to service the requests.

Which one of the following statements is FALSE?

A
The head reverses its direction of movement between servicing of Q and P.
B
T is serviced before P.
C
R is serviced before P.
D
Q is serviced after S, but before T.
Question 13 Explanation: 
(P,155)(Q,85)(R,110)(S,30)(T,115)
Question 14

Consider a paging system that uses a 1-level page table residing in main memory and a TLB for address translation. Each main memory access takes 100 ns and TLB lookup takes 20 ns. Each page transfer to/from the disk takes 5000 ns. Assume that the TLB hit ratio is 95%, page fault rate is 10%. Assume that for 20% of the total page faults, a dirty page has to be written back to disk before the required page is read in from disk. TLB update time is negligible. The average memory access time in ns (round off to 1 decimal places) is ______.

A
154.5 ns
Question 14 Explanation: 
M=100ns
T=20ns
D=5000ns
h=0.95
p=0.1, 1-p=0.9
d=0.2, 1-d=0.8
EMAT = h×(T+M)+(1-h)[(1-p)×2M+p[(1-d)[D+M]+d(2D+M)]+T]
= 0.95×(20+100)+(1-0.95)[(1-0.1)×200+(0.1)[(1-0.2)[5000+100]+(0.2)(10000+100)]+20]
= 154.5 ns
Question 15

The process state transition diagram in below figure is representative of

A
a batch operating system
B
an operating system with a preemptive scheduler
C
an operating system with a non-preemptive scheduler
D
a uni-programmed operating system
Question 15 Explanation: 
Transaction from running → ready present.
So this is preemptive.
Question 16

A critical section is a program segment

A
which should run in a certain specified amount of time
B
which avoids deadlocks
C
where shared resources are accessed
D
which must be enclosed by a pair of semaphore operations, P and V
Question 16 Explanation: 
In CS, share resources are accessed.
Question 17

Which of the following is an example of spooled device?

A
A line printer used to print the output of a number of jobs.
B
A terminal used to enter input data to a running program.
C
A secondary storage device in a virtual memory system.
D
A graphic display device.
Question 17 Explanation: 
Example of spooled device is a line printer used to print the output of a number of jobs.
Question 18

The correct matching for the following pairs is

         A. Activation record	   1. Linking loader
         B. Location counter	   2. Garbage collection
         C. Reference counts	   3. Subroutine call
         D. Address relocation	   4. Assembler 
A
A – 3 B – 4 C – 1 D – 2
B
A – 4 B – 3 C – 1 D – 2
C
A – 4 B – 3 C – 2 D – 1
D
A – 3 B – 4 C – 2 D – 1
Question 18 Explanation: 
Each time a subroutine is called its activation record is created.
An assembler uses location counter value to give address to each instruction which is needed for relative addressing as well as for jump labels.
Reference count is used by garbage collector to clear the memory whose reference count be comes 0.
Linker loader is a loader which can load several compiled codes and link them together into a single executable. Thus it needs to do relocation of the object codes.
Question 19

A 1000 Kbyte memory is managed using variable partitions but to compaction. It currently has two partitions of sizes 200 Kbytes and 260 Kbytes respectively. The smallest allocation request in Kbytes that could be denied is for

A
151
B
181
C
231
D
541
Question 19 Explanation: 
200 and 260 are already hold by some other processes. Now we have to model the partition in such a way so that smallest allocation request could be denied. So, we can do the division as,

So, smallest allocation request which can be denied is 181 KB.
Question 20

A solution to the Dining Philosophers Problem which avoids deadlock is

A
ensure that all philosophers pick up the left fork before the right fork
B
ensure that all philosophers pick up the right fork before the left fork
C
ensure that one particular philosopher picks up the left fork before the right fork, and that all other philosophers pick up the right fork before the left fork
D
None of the above
Question 20 Explanation: 
In the Dining philosopher problem, each philosopher needs exactly two chopsticks to eat food but the problem is: each philosopher is going to take one chopstick at a time, which is placed at its right-hand side or at its left-hand side, but remember all should choose in the same manner like if first one chooses in a clockwise manner then each one should choose in clockwise, this type of picking cause, a circular waiting loop because each one is depending on other. This is also called as circular waiting and it leads to deadlock.
To avoid this, atleast one philosopher should choose its first chopstick in different way so that circular loop is not formed.
Question 21

Four jobs to be executed on a single processor system arrive at time 0+ in the order A, B, C, D. their burst CPU time requirements are 4, 1, 8, 1 time units respectively. The completion time of A under round robin scheduling with time slice of one time unit is

A
10
B
4
C
8
D
9
Question 21 Explanation: 
Time quantum is 1 unit.

∴ Completion time of A is 9 unit.
Question 22

A demand paged virtual memory system uses 16 bit virtual address, page size of 256 bytes, and has 1 Kbyte of main memory. LRU page replacement is implemented using a list whose current status (page number in decimal) is

For each hexadecimal address in the address sequence given below

 00FF, 010D, 10FF, 11B0

indicate
(i) the new status of the list
(ii) page faults, if any, and
(iii) page replacements, if any

A
Theory Explanation.
Question 23

The concurrent programming constructs fork and join are as below:
fork

   N = 2
   M = 2
   fork L3
   fork L4
   S1
L1:join N
   S3
L2:join M
   S5
L3:S2
   goto L1
L4:S4
   goto L2
next: 
A
Theory Explanation.
Question 24

A computer system uses the Banker’s Algorithm to deal with deadlocks. Its current state is shown in the table below, where P0, P1, P2 are processes, and R0, R1, R2 are resources types.

(a) Show that the system can be in this state.
(b) What will the system do on a request by process P0 for one unit of resource type R1?

A
Theory Explanation.
Question 25

A file system with a one-level directory structure is implemented on a disk with disk block size of 4K bytes. The disk is used as follows:

 Disk-block 0: File Allocation Table, consisting of one 8-bit entry per date block, 
               representing the data block address of the next date block in the file
 Disk block 1: Directory, with one 32 bit entry per file:
 Disk block 2: Data block 1;
 Disk block 3: Data block 2; etc. 

(a) What is the maximum possible number of files?
(b) What is the maximum possible file size in blocks?

A
Theory Explanation.
Question 26

Locality of reference implies that the page reference being made by a process

A
will always be to the page used in the previous page reference
B
is likely to be to one of the pages used in the last few page references
C
will always be to one of the pages existing in memory
D
will always lead to a page fault
Question 26 Explanation: 
Locality of reference is also called as principle of locality. There are three different principle of locality:
(1) Temporal locality
(2) Spatial locality
(3) Sequential locality
→ In programs related data are stored in consecutive locations and loops in same locations are used repeatedly.
Question 27

The correct matching for the following pairs is

(A) Disk Scheduling        (1) Round robin
(B) Batch Processing       (2) SCAN
(C) Time sharing           (3) LIFO
(D) Interrupt processing   (4) FIFO  
A
A – 3 B – 4 C – 2 D – 1
B
A – 4 B – 3 C – 2 D – 1
C
A – 2 B – 4 C – 1 D – 3
D
A – 3 B – 4 C – 3 D – 2
Question 27 Explanation: 
Disk scheduling - SCAN
Batch processing - FIFO
Time sharing - Round Robin
Interrupt processing - LIFO
Question 28

I/O redirection

A
implies changing the name of a file
B
can be employed to use an existing file as input file for a program
C
implies connection 2 programs through a pipe
D
None of the above
Question 28 Explanation: 
Redirection is known as capturing output from a file, command, program (or) even code block within a script and sending it as input to another file, command, program (or) script.
Question 29

Thrashing

A
reduces page I/O
B
decreases the degree of multiprogramming
C
implies excessive page I/O
D
improve the system performance
Question 29 Explanation: 
Thrashing implies excessive page I/O.
Question 30

Dirty bit for a page in a page table

A
helps avoid unnecessary writes on a paging device
B
helps maintain LRU information
C
allows only read on a page
D
None of the above
Question 30 Explanation: 
The dirty bit allows for a performance optimization i.e., Dirty bit for a page in a page table helps to avoid unnecessary writes on a paging device.
Question 31

An operating system contains 3 user processes each requiring 2 units of resource R. The minimum number of units of r such that no deadlocks will ever arise is

A
3
B
5
C
4
D
6
Question 31 Explanation: 
For the system to cause deadlock give each process 1 resource less than they require. Since in this case they require 2 resource each, so just give them 1 resource each. So if at max if 3 resource will be available then there can be deadlock. So by adding one more resource deadlock will never occur. So in total minimum 4 resources are required so that deadlock will never occur.
Question 32

Each Process Pi, i = 1.......9 is coded as follows

    repeat 
           P(mutex)
           {Critical section}
           V(mutex)
    forever   

The code for P10 is identical except it uses V(mutex) in place of P(mutex). What is the largest number of processes that can be inside the critical section at any moment?

A
1
B
2
C
3
D
None of the above
Question 32 Explanation: 
Since the both code (i.e., P1 to P9 and P10) can be executed any number of times and code for P10 is
repeat
{
V(mutex)
C.S.
V(mutex)
}
forever
Now, let me say P1 is in CS then P1 is in CS
then P10 comes and executes CS (upon mutex).
Now, P2 comes (down on mutex).
Now P10 moves out of CS (again binary semaphore will be 1).
Now P3 comes (down on mutex).
Now P10 comes (upon mutex).
Now P4 comes (down mutex).
⁞
So, if we take P10 'in' and 'out' of CS recursively, all 10 processes can be in CS at same time using binary semaphore only.
Question 33

Threads of a process share

A
global variables but not heap.
B
heap but not global variables.
C
neither global variables nor heap.
D
both heap and global variables.
Question 33 Explanation: 
Thread share all other resources of process except local data like – register, stack.
Question 34

Consider the following CPU processes with arrival times (in milliseconds) and length of CPU bursts (in milliseconds) as given below:

If the pre-emptive shortest remaining time first scheduling algorithm is used to schedule the processes, then the average waiting time across all processes is __________ milliseconds.

A
3
B
4
C
5
D
6
Question 34 Explanation: 
Question 35

A multithreaded program P executes with x number of threads and used y number of locks for ensuring mutual exclusion while operating on shared memory locations. All locks in the program are non-reentrant, i.e., if a thread holds a lock l, then it cannot re-acquire lock l without releasing it. If a thread is unable to acquire a lock, it blocks until the lock becomes available. The minimum value of x and the minimum value of y together for which execution of P can result in a deadlock are:

A
x = 1, y = 2
B
x = 2, y = 1
C
x = 2, y = 2
D
x = 1, y = 1
Question 35 Explanation: 
First you have to know multithreading, mutual exclusion and reentrant mutex. The reentrant mutex (recursive mutex, recursive lock) is a particular type of mutual exclusion (mutex) device that may be locked multiple times by the same process/thread, without causing a deadlock.
Here a non-reentrant process can’t own the same lock multiple times, so if the process tries to acquire an already owned lock, it will get blocked, and deadlock will happen.
From the above options x=1 (a single thread) and y=1 (a single lock) deadlock is possible when we consider the given situations in question.
Question 36
Recall that Belady’s anomaly is that the page-fault rate may increase as the number of allocated frames increases. Now, consider the following statements:
S1: Random page replacement algorithm (where a page chosen at random is replaced) suffers from Belady’s anomaly
S2: LRU page replacement algorithm suffers from Belady’s anomaly
Which of the following is CORRECT?
A
S1 is true, S2 is true
B
S1 is true, S2 is false
C
S1 is false, S2 is true
D
S1 is false, S2 is false
Question 36 Explanation: 
FIFO may suffer from Belady's anomaly not always FIFO suffer from Belady's anomaly.
Page replacement algorithm suffers from Belady's anomaly when it is not a stack algorithm.
S1: Random page replacement algorithm is not a stack algorithm. So, S1 is true.
S2: LRU is a stack algorithm. Therefore, it doesn't suffer from Belady's anomaly. S2 is false.
Question 37

A linker reads four modules whose lengths are 200, 800, 600 and 500 words, respectively. If they are loaded in that order, what are the relocation constants?

A
0, 200, 500, 600
B
0, 200, 1000, 1600
C
200, 500, 600, 800
D
200, 700, 1300, 2100
Question 37 Explanation: 
First module loaded starting at address 0. Size is 200. Hence it will occupy first 200 address, last address being 199.
Now 2nd will start loading at 200, since size is 800, so last address is 999.
Now 3rd module will start loading at 1000, since size is 600. So last address is 1599.
Now 4th module will start loading at 1600 and go till 2099.
Question 38

Which of the following is an example of a spooled device?

A
The terminal used to enter the input data for the C program being executed
B
An output device used to print the output of a number of jobs
C
The secondary memory device in a virtual storage system
D
The swapping area on a disk used by the swapper
Question 38 Explanation: 
Spooling is a technique in which an intermediate device such as disk is interposed between process and low speed i/o device.
For example in printer if a process attempt to print a document but printer is busy printing another document, the process, instead of waiting for printer to become available, write its output to disk. When the printer becomes available the data on disk is printed.
Spooling allows process to request operation from peripheral device without requiring that the device is ready to service the request.
Question 39

When  the  result  of  a  computation  depends  on  the  speed  of  the  processes involved there is said to be

A
cycle stealing
B
rare condition
C
a time lock
D
a deadlock
Question 39 Explanation: 
When first result depends on ordering of processes it is called race condition.
Speed of processes corresponds to ordering of processes.
Question 40

A counting semaphore was initialized to 10. Then 6P (wait) operations and 4V (signal) operations were completed on this semaphore. The resulting value of the semaphore is

A
0
B
8
C
10
D
12
Question 40 Explanation: 
Let the semaphore be S which is initially 10.
S = 10
Now 6P operations and uv operations were completed on this semaphore. So final value of S will be
S = 10 - 6 + 4 = 8
Question 41

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

A
6
B
5
C
4
D
3
Question 41 Explanation: 
Each process needs 2 drives. So for deadlock just give each process one drive. So total 6 process can be given 1 drive each and can cause deadlock. So to break deadlock just reduce 1 process.
So maximum no. of process for the system to be deadlock free is 5.
Question 42

The overlay tree for a program is as shown below:

What will be the size of the partition (in physical memory) required to load (and run) this program?

A
12 KB
B
14 KB
C
10 KB
D
8 KB
Question 42 Explanation: 
To enable a process which is larger than the amount of memory allocated to it, we can use overlays. The idea of overlays is to heap only those instructions and data that are needed at any given time. When some instructions are needed, they are loaded into space occupied previously by instructions that are no longer needed.
For the above program, maximum memory will be required when running code portion present at leaves.
Maximum requirement
= MAX(12, 14, 10, 14)
= 14
Question 43

Consider n processes sharing the CPU in a round-robin fashion. Assuming that each process switch takes s seconds, what must be the quantum size q such that the overhead resulting from process switching is minimized but at the same time each process is guaranteed to get its turn at the CPU at least every t seconds?

A
q ≤ t-ns/n-1
B
q ≥ t-ns/n-1
C
q ≤ t-ns/n+1
D
q ≥ t-ns/n+1
Question 43 Explanation: 
Question 44

If an instruction takes i microseconds and a page fault takes an additional j microseconds, the effective instruction time if on the average a page fault occurs every k instruction is:

A
i + (j/k)
B
i + (j*k)
C
(i+j)/ k
D
(i+j)*k
Question 44 Explanation: 
Effective memory access time
= service time + page fault rate * page fault service time
= i + 1/k * j
= i + j/k

or
The effective instruction time, if on the average a page fault occurs every k instructions, is:
i + (j / k)
This formula accounts for the normal instruction time (i) and the additional time for a page fault (j), considering the average frequency of page faults (k instructions between page faults).
Question 45

(a) Free disk space can be kept track of using a free list or a bit map. Disk addresses require d bits. For a disk with 13 blocks, F of which is free, state the condition under which the free list uses less space than the bit map.

(b) Consider a disk with C cylinders, t tracks per cylinder, s sectors per track and a sector length sl. A logical file dl with fixed record length rl is stored continuously on this disk starting at location (cL,tL,sL), where cL, tL and SL are the cylinder, track and sector numbers, respectively. Derive the formula to calculate the disk address (i.e. cylinder, track and sector) of a logical record n assuming that rl=sl.

A
Theory Explanation.
Question 46

Listed below are some operating system abstractions (in the left column) and the hardware components or mechanism (in the right column) that they are abstractions of. Which of the following matching of pairs is correct?

    A. Thread                      1. Interrupt
    B. Virtual address space       2. Memory
    C. File system                 3. CPU
    D. Signal                      4. Disk 
A
(A) – 2 (B) – 4 (C) – 3 (D) - 1
B
(A) – 1 (B) – 2 (C) – 3 (D) – 4
C
(A) – 3 (B) – 2 (C) – 4 (D) - 1
D
(A) – 4 (B) – 1 (C) – 2 (D) – 3
Question 46 Explanation: 

⇒ Threads are handled by CPU.
⇒ Virtual address is a memory type.
⇒ File system is used to manage the disk.
⇒ Interrupt is a signal.
Question 47

Which of the following disk scheduling strategies is likely to give the best through put?

A
Farthest cylinder next
B
Nearest cylinder next
C
First come first served
D
Elevator algorithm
Question 47 Explanation: 
Farthest cylinder next - Longest job first
Nearest cylinder next - SSTF
First come first served - FCFS
Elevator - SCAN
SSTF always serves the request of nearest cylinder first. Due to which the necessary movement gets reduced.
Question 48

System calls are usually invoked by using

A
a software interrupt
B
polling
C
an indirect jump
D
a privileged instruction
Question 48 Explanation: 
Software interrupts are implementing device drivers (or) transitions between protected mode of operations, such as system calls.
Question 49

A  multi-user,  multi-processing  operating  system  cannot  be  implemented  on hardware that does not support

A
Address translation
B
DMA for disk transfer
C
At least two modes of CPU execution (privileged and non-privileged)
D
Demand paging
E
Both A and C
Question 49 Explanation: 
Address translation and atleast two modes of CPU execution (Privileged and non-privileged) are needed to implement multiuser and multiprocessing operating system, because address translation provides memory protection which ensures that a given process does not interfere with another, and we need privileged and non-privileged instruction, so that user and OS interconnects properly.
Question 50

Which of the following is/are advantage of virtual memory?

A
Faster access to memory on an average.
B
Processes can be given protected address spaces.
C
Linker can assign addresses independent of where the program will be loaded in physical memory.
D
Programs larger than the physical memory size can be run.
E
Both B and D
Question 50 Explanation: 
A) False. Because in virtual memory concept address translation is required due to which access is slow.
B) True. Because in virtual memory concept of address translation provides protected address space so that one process do not interfere the other process.
C) False.
D) True.
There are 50 questions to complete.

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