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This section includes 301 Mcqs, each offering curated multiple-choice questions to sharpen your Internet of things (IoT) knowledge and support exam preparation. Choose a topic below to get started.
1. |
In NAT, the range of addresses for private networks from 10.0.0.0 to 10.255.255.255 makes a total of |
A. | 2^24 |
B. | 2^25 |
C. | 2^26 |
D. | 2^28 |
Answer» B. 2^25 | |
2. |
In IPv4 addressing, a block of addresses can be defined as |
A. | f.g.h.i.tln |
B. | c.d.e.f.tln |
C. | a.b.c.tln |
D. | x.y.z.tln |
Answer» E. | |
3. |
In classful addressing, a large part of the available addresses are |
A. | Dispersed |
B. | Blocked |
C. | Wasted |
D. | Reserved |
Answer» D. Reserved | |
4. |
In IPv6, each packet is composed of a mandatory base header followed by the |
A. | Data Header |
B. | Payload |
C. | Off load |
D. | Type |
Answer» C. Off load | |
5. |
The dotted-decimal notation is used to make an IPv4 address more compact and easier to |
A. | Communication |
B. | Understand |
C. | Read |
D. | none of the given |
Answer» D. none of the given | |
6. |
In multicast address of IPv6, the value of the prefix is |
A. | 11111111 |
B. | 11110000 |
C. | 10101010 |
D. | 1111 |
Answer» B. 11110000 | |
7. |
A sender can choose a route so that its datagram does not travel through |
A. | Competitor's Network |
B. | Stations |
C. | Backbone Link |
D. | Competitor's Router |
Answer» B. Stations | |
8. |
Datagram network uses the universal addresses defined in the network layer to route packets from the source to the |
A. | Same source |
B. | Layers |
C. | Destination |
D. | Application |
Answer» D. Application | |
9. |
The Configuration management can be divided into two subsystems which are |
A. | Reconfiguration and documentation |
B. | Management and configuration |
C. | Documentation and dialing up |
D. | both a and c |
Answer» B. Management and configuration | |
10. |
In the IPv4 layer, the datagrams are the |
A. | Frames |
B. | Addresses |
C. | Protocol |
D. | Packets |
Answer» E. | |
11. |
An IPv4 address uniquely and universally defines the connection of a device to the |
A. | Media |
B. | Internet |
C. | Monitoring devices |
D. | User |
Answer» C. Monitoring devices | |
12. |
The value of the checksum must be recalculated regardless of |
A. | De-fragmentation |
B. | Fragmentation |
C. | Transfer |
D. | Size |
Answer» C. Transfer | |
13. |
A system with 8-bit addresses has address space of |
A. | 32 |
B. | 256 |
C. | 720 |
D. | 65535 |
Answer» C. 720 | |
14. |
ICANN stands for |
A. | Internet Corporation for Assigned Names and Addresses |
B. | International Corporation for Assigned Names and Addresses |
C. | Internet Cells for Assigning Names and Addresses |
D. | Internet Corporation for Assigning Nodes and Addresses |
Answer» B. International Corporation for Assigned Names and Addresses | |
15. |
In IPv6 Addresses, the field of the prefixes for provider-based unicast address that defines the identity of the node connected to a subnet is called |
A. | Subnet identifier |
B. | Provider identifier |
C. | Subscriber Identifier |
D. | Node identifier |
Answer» E. | |
16. |
An IPv6 address of the Internet consists of |
A. | 16 bytes |
B. | 12 bytes |
C. | 32 bytes |
D. | 48 bytes |
Answer» B. 12 bytes | |
17. |
A connectionless communication of the Internet is established at |
A. | Data Link Layer |
B. | Network Layer |
C. | Physical Layer |
D. | Transport Layer |
Answer» C. Physical Layer | |
18. |
The header of the IPv4 packet changes with each |
A. | Visited Data |
B. | Visited Address |
C. | Visited Router |
D. | Visited Header |
Answer» D. Visited Header | |
19. |
In classfull addressing, the smaller networks are called |
A. | Supernet |
B. | linear net |
C. | Subnet |
D. | Multinet |
Answer» D. Multinet | |
20. |
While transmission of packets over Internet, when all the packets have been delivered, the connection is |
A. | Static |
B. | Established |
C. | Terminated |
D. | On-Pause |
Answer» C. Terminated | |
21. |
The initial value of checksum is |
A. | -1 |
B. | 2 |
C. | 1 |
D. | 0 |
Answer» E. | |
22. |
In classless addressing the size of the block, varies based on the nature and size of the |
A. | Routers |
B. | Mask |
C. | Entity |
D. | Data |
Answer» D. Data | |
23. |
In IPv4, formula for Length of data = |
A. | Length + header |
B. | Total length - header length |
C. | total length - footer length |
D. | length + footer |
Answer» C. total length - footer length | |
24. |
The first fragment has an offset field value of |
A. | 10 |
B. | 5 |
C. | 2 |
D. | 0 |
Answer» E. | |
25. |
In hexadecimal colon notation, the address consists of 32 hexadecimal digits with every four digits separated by a |
A. | dot |
B. | semicolon |
C. | coma |
D. | colon |
Answer» E. | |
26. |
Each number in dotted-decimal notation is a value ranging from |
A. | 0 to 32 |
B. | 0 to 255 |
C. | 0 to 525 |
D. | 0 to 680 |
Answer» C. 0 to 525 | |
27. |
Based on the predefined policy of Network management, controlling access to the network is the task of |
A. | Fault Management |
B. | Performance Management |
C. | Active Management |
D. | Security Management |
Answer» E. | |
28. |
The packet is fragmented in the network layer, if the packet is |
A. | Too Small |
B. | Too Large |
C. | Too busy |
D. | Null |
Answer» C. Too busy | |
29. |
In IPv6, the header checksum is eliminated because the checksum is provided by |
A. | Lower Layer |
B. | Session layer |
C. | Upper Layer |
D. | All of the above |
Answer» D. All of the above | |
30. |
Hardware documentation normally involves two sets of |
A. | Reconciliation |
B. | Documentation |
C. | Restoration |
D. | Reconfiguration |
Answer» C. Restoration | |
31. |
A sequence of packets with same source and destination addresses can be sent one after another, shows the |
A. | Relationship Between Packets |
B. | Independency of packets |
C. | Origin of the Packets |
D. | All of the above |
Answer» B. Independency of packets | |
32. |
The flag field that does fragmentation of IPv4 segment is the |
A. | 1 bit field |
B. | 2 bit field |
C. | 3 bit field |
D. | 4 bit field |
Answer» D. 4 bit field | |
33. |
The size of the IPv4 datagram may increase, as the underlying technologies allows |
A. | Greater bandwidth |
B. | Fixed bandwidth |
C. | Reserved bandwidth |
D. | Greater Header |
Answer» B. Fixed bandwidth | |
34. |
Network addresses are the very important concepts of |
A. | Routing |
B. | Mask |
C. | IP Addressing |
D. | Classless Addressing |
Answer» D. Classless Addressing | |
35. |
A reactive fault management system is responsible for |
A. | Detecting and Isolating |
B. | Correcting, and Recording |
C. | Security and Recording |
D. | Both a and b |
Answer» E. | |
36. |
In IPv6, the base header can be followed by, up to |
A. | Six Extension Layers |
B. | Six Extension Headers |
C. | Eight Extension headers |
D. | Eight Extension layers |
Answer» C. Eight Extension headers | |
37. |
The first address in a block is used as the network address that represents the |
A. | Class Network |
B. | Entity |
C. | Organization |
D. | DataCodes |
Answer» D. DataCodes | |
38. |
In using One IP address for the Network Address Translation, then the translation table has |
A. | one column |
B. | two columns |
C. | three columns |
D. | four columns |
Answer» C. three columns | |
39. |
In an IPv6 datagram, the M bit is 0, the value of HLEN is 5, the value of total length is 200 and the offset value is |
A. | 400 |
B. | 350 |
C. | 300 |
D. | 200 |
Answer» E. | |
40. |
In its simplest form, a flow label can be used to speed up the processing of a packet by a |
A. | Packet Switch |
B. | Router |
C. | Data Switch |
D. | Protocol |
Answer» C. Data Switch | |
41. |
Port number for (Simple Network Management Protocol) SNMP? |
A. | 161 |
B. | 149 |
C. | 144 |
D. | 150 |
Answer» B. 149 | |
42. |
In IPv4, two devices on the Internet can never have the |
A. | Different address |
B. | Same address |
C. | Unknown address |
D. | Fixed address |
Answer» C. Unknown address | |
43. |
The Simple Network Management Protocol (SNMP), uses the concept of manager and |
A. | Host |
B. | Server |
C. | Agent |
D. | Query |
Answer» D. Query | |
44. |
In an IPv4 packet, the value of HLEN is |
A. | 7 |
B. | 6 |
C. | 5 |
D. | 4 |
Answer» D. 4 | |
45. |
In IPv4, the checksum covers only the |
A. | Header |
B. | Layers |
C. | Datagrams |
D. | Fragmentation |
Answer» B. Layers | |
46. |
The network layer was designed to solve the problem of delivery through |
A. | Single Link |
B. | Multilevel Link |
C. | Several Link |
D. | Unicast Link |
Answer» D. Unicast Link | |
47. |
To represent the IPv4 address, there are |
A. | two notations |
B. | three notations |
C. | four notations |
D. | five notations |
Answer» B. three notations | |
48. |
In classless addressing, there are no classes but the addresses are still granted in |
A. | Sections |
B. | Blocks |
C. | Codes |
D. | All of the Above |
Answer» C. Codes | |
49. |
In IPv4 protocol, each datagram is handled |
A. | Independently |
B. | dependently |
C. | Priority basis |
D. | Systematically |
Answer» B. dependently | |
50. |
A switch can allow fast handling of the |
A. | Tokens |
B. | Frames |
C. | Packets |
D. | Slots |
Answer» D. Slots | |