Computer networking MCQ – Change your future

601. In multiplexing, channels are separated by unused strips of bandwidth guard bands-to prevent

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

Exam Relevance: GATE, ECE, EE

Difficulty: Moderate

Concept notes: In multiplexing, guard bands are used to prevent overlapping of adjacent channels.

Common Mistakes: Students may confuse guard bands with synchronization mechanisms, which is incorrect.

Explanations: Guard bands are strips of unused bandwidth placed between active channels in a multiplexed system. Their primary purpose is to prevent the signals from adjacent channels from overlapping, which can cause interference and degrade signal quality. Synchronization and random motion of electrons are not directly related to the function of guard bands in multiplexing.

Option Analysis:

Option A: Synchronization is not the purpose of guard bands in multiplexing.
Option B: Guard bands prevent overlapping of adjacent channels, which is the correct function.
Option C: Random motion of electrons is unrelated to the function of guard bands in multiplexing.
Option D: Since overlapping is the correct function, combining it with random motion of electrons is incorrect.

Mnemonic: Guard bands = Gap to avoid overlap

602. The original IEEE 802.11, has a data rate of.....Mbps.

Correct Answer: Correct answer is: (A) 1.

Exam Relevance: Computer Networks, Wireless Communications, IEEE Standards

Difficulty: Very Easy

Concept notes: The original IEEE 802.11 standard specified a data rate of 1 Mbps.

Common Mistakes: Students may confuse the original 802.11 standard with later versions, such as 802.11b, which introduced higher data rates.

Explanations: The original IEEE 802.11 standard, released in 1997, specified a data rate of 1 Mbps. This was the initial standard for wireless local area networks (WLANs) and laid the foundation for subsequent enhancements and higher data rates.

Option Analysis:

Option A: Correct. The original IEEE 802.11 standard specified a data rate of 1 Mbps.
Option B: Incorrect. 6 Mbps is the data rate specified by the 802.11b standard, not the original 802.11.
Option C: Incorrect. 11 Mbps is the data rate specified by the 802.11b standard, not the original 802.11.
Option D: Incorrect. 22 Mbps is not a standard data rate for any IEEE 802.11 standard.

603. In internetwork, network layer at destination is responsible for

Correct Answer: Correct answer is: (D) Address Verification.

Exam Relevance: CCNA, CISSP, CompTIA Network+

Difficulty: Moderate

Concept notes: In internetworking, the network layer at the destination is responsible for verifying the destination address to ensure the packet is delivered to the correct node.

Common Mistakes: Students may confuse address verification with address authentication, which is a different process typically handled at the security layer.

Explanations: The network layer at the destination is responsible for verifying the destination address to ensure the packet is delivered to the correct node. This process involves checking the destination address against the routing table to confirm the packet's intended recipient. Address verification is crucial for ensuring that data packets are correctly routed and delivered to the appropriate destination within the network.

Option Analysis:

Option A: Address Authentication is not the responsibility of the network layer; it is more related to security protocols.
Option B: Address Node is not a standard term in networking and does not describe a specific function of the network layer.
Option C: Address format is typically defined by the protocol and not a function performed by the network layer at the destination.
Option D: Address Verification is the correct function performed by the network layer to ensure packets are delivered to the correct destination.

604. Walsh table is a two-dimensional table in which number of rows and columns are

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

Exam Relevance: Computer Science Exams, Data Structures Exams, Discrete Mathematics Exams

Difficulty: Easy

Concept notes: A Walsh table is a two-dimensional table where the number of rows and columns are equal.

Common Mistakes: Students may confuse the structure of a Walsh table with other types of tables where rows and columns can differ.

Explanations: A Walsh table is specifically designed to have an equal number of rows and columns. This symmetry is a defining characteristic of the table, ensuring that the structure is balanced and consistent.

Option Analysis:

Option A: Correct. The number of rows and columns in a Walsh table are the same.
Option B: Incorrect. The number of rows and columns in a Walsh table are not different.
Option C: Incorrect. The number of rows and columns are not specified as 'N' without context.
Option D: Incorrect. The number of rows and columns cannot be zero as it would not form a table.

605. Which of the following protocol is connection-oriented?

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

Exam Relevance: CCNA, CISSP, CompTIA Network+

Difficulty: Easy

Concept notes: TCP (Transmission Control Protocol) is a connection-oriented protocol, meaning it establishes a connection before data transmission and ensures reliable delivery of data.

Common Mistakes: Students often confuse TCP with UDP, which is connectionless and does not establish a connection before data transmission.

Explanations: TCP is connection-oriented because it establishes a connection before data transmission. This connection is maintained throughout the data transfer process, ensuring reliable delivery of data packets. The protocol uses a three-way handshake to establish a connection and employs mechanisms like sequence numbers, acknowledgments, and flow control to ensure data integrity and reliability.

Option Analysis:

Option A: IPX (Internetwork Packet Exchange) is a connectionless protocol, similar to UDP, and does not establish a connection before data transmission.
Option B: UDP (User Datagram Protocol) is a connectionless protocol, meaning it does not establish a connection before data transmission and does not guarantee reliable delivery of data.
Option C: NetBEUI (Network Basic Input/Output System Extended User Interface) is a connectionless protocol designed for small local area networks and does not establish a connection before data transmission.
Option D: TCP (Transmission Control Protocol) is a connection-oriented protocol, establishing a connection before data transmission and ensuring reliable delivery of data.

Mnemonic: TCP: "Tightly Connected Protocol"

606. In a..... name space, a name is assigned to an address. A name in this space is a sequence of characters without structure.

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

Exam Relevance: Computer Networks, Data Communication, Network Security

Difficulty: Easy

Concept notes: In a flat name space, names are assigned to addresses without any hierarchical structure.

Common Mistakes: Students may confuse a flat name space with a hierarchical name space, where names have a structured hierarchy.

Explanations: A flat name space assigns names to addresses without any structured hierarchy. This means that a name in this space is simply a sequence of characters without any inherent structure or levels. Therefore, the correct answer is a flat name space.

Option Analysis:

Option A: Correct. A flat name space assigns names to addresses without any structured hierarchy.
Option B: Incorrect. A hierarchical name space has a structured hierarchy, which is not the case here.
Option C: Incorrect. An organized name space implies some form of structure, which is not present in a flat name space.
Option D: Incorrect. The correct answer is explicitly provided as a flat name space.

Mnemonic: Flat = No Structure

607. In IPv4, fragmented datagram's fragment size should have first byte number divisible by

Correct Answer: Correct answer is: (C) 8.

Exam Relevance: CCNA, CISSP, CompTIA Network+

Difficulty: Moderate

Concept notes: In IPv4, the fragment offset field in the IP header specifies the position of the fragment in the original datagram. This offset is measured in units of 8 bytes (64 bits).

Common Mistakes: Students often confuse the fragment offset unit with other byte multiples, such as 4 or 16 bytes.

Explanations: The fragment offset field in the IPv4 header is 13 bits long and specifies the position of the fragment in the original datagram. This position is measured in units of 8 bytes (64 bits). Therefore, the first byte number of each fragment must be divisible by 8 to align with the fragment offset field's unit.

Option Analysis:

Option A: Incorrect, as the fragment offset is not measured in units of 2 bytes.
Option B: Incorrect, as the fragment offset is not measured in units of 4 bytes.
Option C: Correct, as the fragment offset is measured in units of 8 bytes.
Option D: Incorrect, as the fragment offset is not measured in units of 16 bytes.

608. The participating computers in a network are referred to as:

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

Exam Relevance: CCNA, CompTIA Network+, A+ Certification

Difficulty: Easy

Concept notes: In a network, a node is any device or endpoint that is connected to the network and can send, receive, or forward information.

Common Mistakes: Students may confuse nodes with clients or servers, which are specific types of nodes with particular roles in a network.

Explanations: A node is a general term for any device connected to a network, including computers, printers, routers, and switches. Clients and servers are specific types of nodes, but not all nodes are clients or servers. CPUs are components within a node, not the node itself.

Option Analysis:

Option A: Clients are specific types of nodes that request services from servers. They are not the general term for all connected devices.
Option B: Servers are specific types of nodes that provide services to clients. They are not the general term for all connected devices.
Option C: Nodes are the general term for any device connected to a network, including clients, servers, and other devices.
Option D: CPUs are components within a node, not the node itself.

Mnemonic: N for Node, Network's Nucleus.

609. Telephone companies implement multiplexing

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

Exam Relevance: Networking exams, Telecommunications exams, Computer Science exams

Difficulty: Easy

Concept notes: Time Division Multiplexing (TDM) is a method used by telephone companies to transmit multiple signals over a single communication channel by dividing the channel into time slots.

Common Mistakes: Students may confuse TDM with FDM, WDM, or DWDM, which are other multiplexing techniques but not the primary method used by telephone companies for voice communication.

Explanations: Telephone companies use Time Division Multiplexing (TDM) to efficiently transmit multiple voice signals over a single communication line. In TDM, each signal is allocated a specific time slot, allowing multiple signals to be transmitted sequentially without interference. This method is well-suited for voice communication because it ensures that each signal is transmitted in a timely and orderly manner.

Option Analysis:

Option A: TDM is the correct answer as it is the primary multiplexing technique used by telephone companies for voice communication.
Option B: FDM (Frequency Division Multiplexing) is not the primary method used by telephone companies for voice communication.
Option C: WDM (Wavelength Division Multiplexing) is used in optical communication systems, not typically in traditional telephone networks.
Option D: DWDM (Dense Wavelength Division Multiplexing) is used in high-capacity optical networks, not in traditional telephone voice communication.

Mnemonic: TDM: Time Division Multiplexing, Think "Time Slots"

610. What is the error (if any) in the following representation-11100010.23.14.67?

Correct Answer: Correct answer is: (D) None of the mentioned.

Exam Relevance: CCNA, CISSP, CompTIA Network+

Difficulty: Moderate

Concept notes: An IPv4 address consists of four bytes, each ranging from 0 to 255, separated by periods.

Common Mistakes: Students often misunderstand the format of an IPv4 address, particularly the range of values each byte can take.

Explanations: The representation 11100010.23.14.67 is an IPv4 address. Each byte in an IPv4 address can range from 0 to 255. The first byte, 11100010, is a binary representation of the number 226, which is within the valid range. The other bytes (23, 14, and 67) are also within the valid range. Therefore, there is no error in the given representation.

Option Analysis:

Option A: Leading zeros are allowed in IPv4 addresses, so this is not an error.
Option B: An IPv4 address can have exactly four bytes, which is correct here.
Option C: Each byte in the address is within the valid range of 0 to 255.
Option D: The representation is correct, so none of the mentioned options are errors.

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Computer networking MCQ - Set 61

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