Computer networking MCQ

1271. Which of the following TCP/IP protocol allows an application program on one machine to send a datagram to an application program on another machine?

Correct Answer: Correct answer is: (A) UDP.

Exam Relevance: CCNA, CISSP, CompTIA Network+

Difficulty: Easy

Concept notes: UDP (User Datagram Protocol) allows an application program on one machine to send a datagram to an application program on another machine.

Common Mistakes: Students may confuse UDP with TCP, which also allows communication between applications but with additional reliability features.

Explanations: UDP is a connectionless protocol that enables applications to send datagrams (packets) directly to other applications on different machines. It does not require a connection to be established before sending data, making it suitable for applications that prioritize speed over reliability.

Option Analysis:

Option A: UDP is correct because it allows direct datagram transmission between applications on different machines.
Option B: VMTP (VMTP) is a less common protocol and not typically used for direct application-to-application datagram transmission.
Option C: X.25 is a packet-switched network protocol and not used for direct application-to-application datagram transmission.
Option D: SMTP (Simple Mail Transfer Protocol) is used for email transmission and not for direct application-to-application datagram transmission.

1272. This type of network is expensive to build, operate and run

Correct Answer: Correct answer is: (B) WAN.

Exam Relevance: CCNA, Network+, CISSP

Difficulty: Moderate

Concept notes: Wide Area Networks (WANs) are expansive networks that connect devices over large geographical areas, often spanning cities, countries, or even continents. Due to their extensive reach, they require significant infrastructure and resources to build, operate, and maintain.

Common Mistakes: Students may confuse WANs with LANs, which are local and less expensive to set up and maintain.

Explanations: WANs are more expensive to build, operate, and run because they require extensive infrastructure such as routers, switches, and long-distance communication links. They often involve leasing or owning expensive communication lines and equipment to ensure reliable and secure data transmission over large distances. In contrast, LANs are confined to a smaller area and typically use less expensive hardware and shorter communication links.

Option Analysis:

Option A: LANs are less expensive to build, operate, and run because they cover a smaller area and use less complex infrastructure.
Option B: WANs are more expensive due to the extensive infrastructure and long-distance communication links required.
Option C: This option is incorrect because only WANs are known for their high costs.
Option D: This option is incorrect because WANs are indeed expensive to build, operate, and run.

Mnemonic: WANs = Wide Area Networks = Wide, Widespread, and Wealthy (in terms of cost)

1273. Which of the following commands sets a trunk port on a 2960 switch?

Correct Answer: Correct answer is: (D) Switchport mode trunk.

Exam Relevance: CCNA, CCNP, Network+

Difficulty: Easy

Concept notes: The command "switchport mode trunk" is used to configure a port on a Cisco switch to operate in trunk mode, allowing it to carry traffic for multiple VLANs.

Common Mistakes: Students often confuse the command with "trunk on" or "switchport trunk on," which are not valid commands for configuring a trunk port.

Explanations: The command "switchport mode trunk" is the correct command to set a trunk port on a Cisco 2960 switch. This command changes the port's mode to trunk, enabling it to handle traffic for multiple VLANs. Other options like "trunk on" or "switchport trunk on" are not valid commands in the Cisco IOS.

Option Analysis:

Option A: "Trunk on" is not a valid command for configuring a trunk port.
Option B: "Trunk all" is not a valid command for configuring a trunk port.
Option C: "Switchport trunk on" is not a valid command for configuring a trunk port.
Option D: "Switchport mode trunk" is the correct command to configure a trunk port.

1274. Stream Control Transmission Protocol (SCTP), association allows multiple IP addresses for

Correct Answer: Correct answer is: (C) Each End.

Exam Relevance: CCNA, CCNP, CISSP

Difficulty: Moderate

Concept notes: SCTP (Stream Control Transmission Protocol) allows multiple IP addresses to be associated with each endpoint in an association, providing redundancy and load balancing.

Common Mistakes: Students may mistakenly believe that only one end of the association can have multiple IP addresses, or that multiple IP addresses are not supported at all.

Explanations: SCTP supports multihoming, which means that each endpoint in an association can have multiple IP addresses. This feature allows for redundancy and load balancing, enhancing the reliability and performance of the communication. Therefore, the correct answer is that each end can have multiple IP addresses.

Option Analysis:

Option A: Incorrect because SCTP supports multiple IP addresses for both ends, not just one end.
Option B: Incorrect because SCTP does support multiple IP addresses for both ends.
Option C: Correct because SCTP allows multiple IP addresses for each end, providing redundancy and load balancing.
Option D: Incorrect because SCTP does support multiple IP addresses for each end.

Mnemonic: SCTP: "SCTP Connects Through Pairing"

1275. Ultra High Frequency (UHF) is using propagation method of

Correct Answer: Correct answer is: (C) Line of sight.

Exam Relevance: GATE, GRE, ESE

Difficulty: Moderate

Concept notes: Ultra High Frequency (UHF) waves propagate primarily through line of sight.

Common Mistakes: Students may confuse UHF with lower frequency bands that use sky or ground propagation.

Explanations: UHF waves have a frequency range of 300 MHz to 3 GHz. Due to their shorter wavelengths, UHF waves do not bend around obstacles or follow the curvature of the Earth. Instead, they require a direct line of sight between the transmitter and receiver. This makes line of sight the primary propagation method for UHF.

Option Analysis:

Option A: Sky propagation is not suitable for UHF due to its shorter wavelength and inability to reflect off the ionosphere effectively.
Option B: Ground propagation is not effective for UHF as the waves do not diffract around the Earth's surface.
Option C: Line of sight is the correct propagation method for UHF, as the waves require a direct path between the transmitter and receiver.
Option D: While UHF can sometimes use sky propagation in specific conditions, line of sight is the primary and most reliable method.

1276. ..... uses two carriers, one in-phase and the other quadrature.

Correct Answer: Correct answer is: (D) QAM.

Exam Relevance: ECE, GATE, IEEE

Difficulty: Moderate

Concept notes: Quadrature Amplitude Modulation (QAM) uses two carriers, one in-phase and the other quadrature, to transmit data.

Common Mistakes: Students may confuse QAM with other modulation techniques like ASK, PSK, or FSK, which do not use both in-phase and quadrature carriers.

Explanations: QAM is a modulation scheme that uses two carriers, one in-phase (I) and the other quadrature (Q), to transmit data. The in-phase carrier is in phase with the reference signal, while the quadrature carrier is 90 degrees out of phase. This dual-carrier approach allows QAM to transmit more data per symbol compared to single-carrier modulation techniques.

Option Analysis:

Option A: ASK (Amplitude Shift Keying) uses changes in amplitude to represent data, not two carriers.
Option B: PSK (Phase Shift Keying) uses changes in phase to represent data, not two carriers.
Option C: FSK (Frequency Shift Keying) uses changes in frequency to represent data, not two carriers.
Option D: QAM (Quadrature Amplitude Modulation) uses two carriers, one in-phase and the other quadrature, to transmit data.

Mnemonic: QAM: Quadrature Amplitude Modulation, uses two carriers, I and Q.

1277. An IPv6 address is..... bits long.

Correct Answer: Correct answer is: (C) 128.

Exam Relevance: CCNA, CISSP, CompTIA Network+

Difficulty: Easy

Concept notes: IPv6 addresses are 128 bits long, which is a fundamental aspect of IPv6 addressing.

Common Mistakes: Students often confuse IPv6 with IPv4, which is 32 bits long.

Explanations: IPv6 addresses are designed to be 128 bits long to accommodate a much larger number of unique addresses compared to IPv4. This is a key feature of IPv6 to support the growing number of devices on the internet.

Option Analysis:

Option A: 32 bits is the length of an IPv4 address, not an IPv6 address.
Option B: 64 bits is not the correct length for an IPv6 address.
Option C: 128 bits is the correct length for an IPv6 address.
Option D: None of the above is incorrect because 128 bits is the correct length for an IPv6 address.

Mnemonic: IPv6 = 128 bits, think "6" and "128" as a pair.

1278. If a router cannot determine the next hop for a packet, the router will;

Correct Answer: Correct answer is: (D) Forward the packet to the default route.

Exam Relevance: CCNA, CCNP, CISSP

Difficulty: Moderate

Concept notes: When a router cannot determine the next hop for a packet, it will use the default route to forward the packet.

Common Mistakes: Students may mistakenly believe that the router will drop the packet if it cannot determine the next hop.

Explanations: A router maintains a routing table that contains information about the next hop for each destination network. If the router cannot find a specific route for a packet's destination, it will use the default route, which is a catch-all route that directs traffic to a specified next hop. This default route is often used to forward packets to an upstream router or gateway that can handle the packet.

Option Analysis:

Option A: Incorrect. The router will not drop the packet if it cannot determine the next hop; it will use the default route.
Option B: Incorrect. Broadcasting the packet is not a standard procedure for handling unknown destinations.
Option C: Incorrect. Storing the packet in the memory buffer is not a typical action for a router when it cannot determine the next hop.
Option D: Correct. The router will forward the packet to the default route if it cannot determine the next hop.

Mnemonic: Default Route: "D" for "Drop-in" solution.

1279. A packet in Transmission Control Protocol (TCP) is called a

Correct Answer: Correct answer is: (C) Segment.

Exam Relevance: CCNA, CISSP, CompTIA Network+

Difficulty: Easy

Concept notes: In TCP, a packet is referred to as a segment. This is a fundamental concept in understanding the structure and operation of TCP.

Common Mistakes: Students often confuse the terms "packet" and "segment," as both are used in networking but in different contexts.

Explanations: In the context of TCP, a packet is specifically called a segment. This is because TCP operates at the transport layer of the OSI model, and the unit of data at this layer is a segment. The term "segment" is used to distinguish it from the term "packet," which is more commonly used in the context of the network layer (e.g., IP packets).

Option Analysis:

Option A: Transmittable slots is not a standard term used in TCP.
Option B: Packet is a general term used in networking but not specific to TCP.
Option C: Segment is the correct term for a TCP packet.
Option D: Source Slots is not a standard term used in TCP.

1280. How many types of multiplexing techniques are

Correct Answer: Correct answer is: (C) Three.

Exam Relevance: GATE, ECE, EE, Telecommunications Engineering

Difficulty: Easy

Concept notes: Multiplexing techniques are methods used to combine multiple signals into a single channel for transmission. The three main types are Frequency Division Multiplexing (FDM), Time Division Multiplexing (TDM), and Code Division Multiplexing (CDM).

Common Mistakes: Students may confuse the number of multiplexing techniques with other communication concepts or may not be aware of all three types.

Explanations: Multiplexing techniques are categorized into three main types: Frequency Division Multiplexing (FDM), Time Division Multiplexing (TDM), and Code Division Multiplexing (CDM). FDM divides the available bandwidth into multiple channels, each carrying a different signal. TDM allocates time slots for each signal to share the same channel. CDM uses unique codes to allow multiple signals to occupy the same frequency band simultaneously. Therefore, the correct answer is three.

Option Analysis:

Option A: Incorrect, as there are more than one type of multiplexing technique.
Option B: Incorrect, as there are more than two types of multiplexing technique.
Option C: Correct, as there are three main types of multiplexing techniques.
Option D: Incorrect, as there are not four main types of multiplexing techniques.

Mnemonic: FDM-TDM-CDM: Think "F-T-C" for Frequency, Time, and Code.

⬅ Previous SetSet 128Next Set ➡

Computer networking MCQ - Set 128

Leave a Reply