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This section includes 228 Mcqs, each offering curated multiple-choice questions to sharpen your Computer Science Engineering (CSE) knowledge and support exam preparation. Choose a topic below to get started.
1. |
A glue between client and server parts of application |
A. | middleware |
B. | firmware |
C. | package |
D. | system software |
Answer» B. firmware | |
2. |
Internet provides for remote login. |
A. | telnet |
B. | http |
C. | ftp |
D. | rpc |
Answer» B. http | |
3. |
provides programmers a familiar programming model by extending the local procedure call to a distributed environment |
A. | distributed environment |
B. | permanent procedure call |
C. | process and file |
D. | remote procedure call |
Answer» E. | |
4. |
is an object acting as a gateway for the client side. |
A. | skeleton |
B. | stub |
C. | remote |
D. | server |
Answer» C. remote | |
5. |
What are the advantages of token(with rings) passing approach?i) One processor as coordinator which handles all requestsii) No starvation if the ring is unidirectionaliii) There are many messages passed per section entered if few users want to get in sectioniv) One processor as coordinator which handles all requestsv) Only one message/entry if everyone wants to get in |
A. | i |
B. | ii and iii |
C. | i, ii and iii |
D. | i, ii and iv |
Answer» E. | |
6. |
consistency is that write operations by the same process are performed in the correct order everywhere. |
A. | weak |
B. | strict |
C. | eventual |
D. | fifo |
Answer» E. | |
7. |
In receivers never acknowledge the successful delivery of multicast message but instead report only when missing the message. |
A. | basic reliable multicasting |
B. | non-hierarchical feedback control |
C. | hierarchical feedback control |
D. | atomic multicast |
Answer» C. hierarchical feedback control | |
8. |
A problem with the protocol is that when the coordinator has crashed, participants may not be able to reach a final decision. |
A. | one-phase commit |
B. | three-phase commit |
C. | two-phase commit |
D. | virtual synchrony |
Answer» D. virtual synchrony | |
9. |
Which algorithm requires "1 to " messages to enter and leave a critical region? |
A. | token ring algorithm |
B. | centralized algorithm |
C. | decentralized algorithm |
D. | distributed algorithm |
Answer» B. centralized algorithm | |
10. |
For each critical section (CS) execution, Ricart-Agrawala algorithm requires messages per CS execution and the Synchronization delay in thealgorithm is . |
A. | 3(n 1), t |
B. | 2(n 1), t |
C. | (n 1), 2t |
D. | (n 1), t |
Answer» C. (n 1), 2t | |
11. |
For each critical section (CS) execution, maekawa s algorithm requires messages per CS execution and the Synchronization delay in thealgorithm is . |
A. | n , t |
B. | 2 n , t |
C. | 3 n , t |
D. | 3 n , 2t |
Answer» E. | |
12. |
In HDFS file System, A serves as the master and there is only one NameNode per cluster |
A. | data node |
B. | namenode |
C. | replication |
D. | data block |
Answer» C. replication | |
13. |
In HDFS file System, NameNode is used when the Primary NameNode goes down. |
A. | rack |
B. | data |
C. | secondary |
D. | primary |
Answer» D. primary | |
14. |
In HDFS, . is the slave/worker node and holds the user data in the form of Data Blocks. |
A. | datanode |
B. | namenode |
C. | data block |
D. | replication |
Answer» B. namenode | |
15. |
Storing file in makes it permanently available |
A. | secondary memory |
B. | ram |
C. | register |
D. | dram |
Answer» B. ram | |
16. |
A client gets a timestamp of 4:12:30.500 from atime server. The elapsed time between the requestand response was 20 msec (0.020 sec).The currenttime on the client is 4:12:30.510. Using Cristian salgorithm, what is the time set to on theclient? |
A. | 1970-01-01 04:12:30 |
B. | 1970-01-01 04:12:30 |
C. | 1970-01-01 04:12:32 |
D. | 1970-01-01 04:12:32 |
Answer» D. 1970-01-01 04:12:32 | |
17. |
is a process that prevents multiple threads or processes from accessing shared resources at the same time. |
A. | critical section |
B. | deadlock |
C. | message passing |
D. | mutual exclusion |
Answer» E. | |
18. |
The file once created can not be changed iscalled |
A. | rigid file |
B. | rex file |
C. | immutable file |
D. | robust file |
Answer» D. robust file | |
19. |
Network file system(NFS) is developedby? |
A. | sun microsystem |
B. | oracle |
C. | apple |
D. | honeywell |
Answer» B. oracle | |
20. |
Session file sharing semantics are suitablefor caching |
A. | complete folder |
B. | complete file |
C. | one byte |
D. | one block |
Answer» C. one byte | |
21. |
Storing file in makes itpermanently available |
A. | secondary memory |
B. | ram |
C. | register |
D. | dram |
Answer» B. ram | |
22. |
If file system is growing without affectingperformance of the system then this feature is called as? |
A. | union |
B. | portable |
C. | robust |
D. | scalability |
Answer» E. | |
23. |
In HDFS, . is the slave/workernode and holds the user data in the form of Data Blocks. |
A. | datanode |
B. | namenode |
C. | data block |
D. | replication |
Answer» B. namenode | |
24. |
In HDFS file System, NameNode is used when the PrimaryNameNode goes down. |
A. | rack |
B. | data |
C. | secondary |
D. | primary |
Answer» D. primary | |
25. |
In HDFS file System, A serves as the master and there is only oneNameNode per cluster |
A. | data node |
B. | namenode |
C. | replication |
D. | data block |
Answer» C. replication | |
26. |
In which file system mapreduce function isused? |
A. | afs |
B. | nfs |
C. | hdfs |
D. | none of the above |
Answer» D. none of the above | |
27. |
datanodes and namenode are two elementsof which file system? |
A. | afs |
B. | hdfs |
C. | nfs |
D. | none of the above |
Answer» C. nfs | |
28. |
which command is used to create adirectory |
A. | rmdir |
B. | symlink |
C. | mkdir |
D. | open |
Answer» D. open | |
29. |
in NFS which funtion is used for creatingnew file? |
A. | open() |
B. | create() |
C. | develope() |
D. | null() |
Answer» C. develope() | |
30. |
The dynamic replication algorithmtakes into account |
A. | to reduce load on server |
B. | files on server can bemigrated anywhere |
C. | schedule processmigration |
D. | resource sharing |
Answer» B. files on server can bemigrated anywhere | |
31. |
Staleness deviations relate to thea replica was updated. |
A. | first time |
B. | most recent time |
C. | last time |
D. | previous time |
Answer» D. previous time | |
32. |
Optimistic logging protocols need to keeptrack of |
A. | independencies |
B. | checking points |
C. | dependencies |
D. | erroneous state |
Answer» D. erroneous state | |
33. |
A problem with the protocol is that when the coordinator has crashed, participants may not be able to reach afinal decision. |
A. | one-phase commit |
B. | three-phase commit |
C. | two-phase commit |
D. | virtual synchrony |
Answer» D. virtual synchrony | |
34. |
In receivers never acknowledge the successful delivery of multicast message but instead report onlywhen missing the message. |
A. | basic reliable multicasting |
B. | non-hierarchical feedback control |
C. | hierarchical feedback control |
D. | atomic multicast |
Answer» C. hierarchical feedback control | |
35. |
replicas are used only toimprove access time on data |
A. | client initiated |
B. | server initiated |
C. | permanent |
D. | dynamic |
Answer» B. server initiated | |
36. |
consistency is that write operations by the same process are performed in the correct ordereverywhere. |
A. | weak |
B. | strict |
C. | eventual |
D. | fifo |
Answer» E. | |
37. |
In which of the following consistency model all writes become perceptible to allprocesses |
A. | strict |
B. | weak |
C. | casual |
D. | sequential |
Answer» B. weak | |
38. |
In Casual consistency model all processesshared accesses in |
A. | random order |
B. | same order |
C. | sequential order |
D. | specific order |
Answer» C. sequential order | |
39. |
Absolute time ordering of all sharedaccesses matters in |
A. | sequential consistency |
B. | casual consistency |
C. | strict consistency |
D. | weak consistency |
Answer» D. weak consistency | |
40. |
When the process issues an I/O request |
A. | it is placed in an i/o queue |
B. | it is placed in a waitingqueue |
C. | it is placed in the readyqueue |
D. | it is placed in the job queue |
Answer» B. it is placed in a waitingqueue | |
41. |
How is access to resources of variousmachines is done? |
A. | remote logging using ssh or telnet |
B. | zone are configured forautomatic access |
C. | ftp is not used |
D. | ftp is used |
Answer» B. zone are configured forautomatic access | |
42. |
Which algorithm requires "1 to ∞" messages to enter and leave a criticalregion? |
A. | token ring algorithm |
B. | centralized algorithm |
C. | decentralized algorithm |
D. | distributed algorithm |
Answer» B. centralized algorithm | |
43. |
Suzuki-Kasami's Broadcast Algorithm isan |
A. | non- token based algorithm. |
B. | token based algorithm. |
C. | centralized basedalgorithm |
D. | physical clocksynchronization algorithm. |
Answer» C. centralized basedalgorithm | |
44. |
RAYMOND'S TREE BASEDALGORITHM is an |
A. | non- token based algorithm. |
B. | token based algorithm. |
C. | centralized basedalgorithm |
D. | physical clocksynchronization algorithm. |
Answer» C. centralized basedalgorithm | |
45. |
For each critical section (CS) execution, maekawa’s algorithm requires messages per CS execution and the Synchronization delay in thealgorithm is . |
A. | √n , t |
B. | 2√n , t |
C. | 3√n , t |
D. | 3√n , 2t |
Answer» E. | |
46. |
For each critical section (CS) execution, Ricart-Agrawala algorithm requires messages per CS execution and the Synchronization delay in thealgorithm is . |
A. | 3(n − 1), t |
B. | 2(n − 1), t |
C. | (n − 1), 2t |
D. | (n − 1), t |
Answer» C. (n − 1), 2t | |
47. |
Which of the following is an example ofelection algorithm. |
A. | berkley algorithm: |
B. | bully algorithm. |
C. | cristian’s algorithm |
D. | lamport’s |
Answer» C. cristian’s algorithm | |
48. |
NTP is layer protocol. |
A. | application |
B. | session |
C. | transport |
D. | physical |
Answer» B. session | |
49. |
A client gets a timestamp of 4:12:30.500 from atime server. The elapsed time between the requestand response was 20 msec (0.020 sec).The currenttime on the client is 4:12:30.510. Using Cristian’salgorithm, what is the time set to on theclient? |
A. | 1970-01-01 04:12:30 |
B. | 1970-01-01 04:12:30 |
C. | 1970-01-01 04:12:32 |
D. | 1970-01-01 04:12:32 |
Answer» D. 1970-01-01 04:12:32 | |
50. |
Which event is concurrent with the vectorclock (2, 8, 4)? |
A. | (3,9,5) |
B. | (3,8,4) |
C. | (1,7,3) |
D. | (4,8,2) |
Answer» E. | |