Computer networking MCQ – Change your future

681. The parameter that refers to the set of rules that govern data communications are called

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

Exam Relevance: Networking exams, Computer Science exams, IT certifications

Difficulty: Easy

Concept notes: A protocol is a set of rules that govern data communications.

Common Mistakes: Students may confuse protocol with other terms like standard or forum, which do not specifically refer to the rules governing data communications.

Explanations: A protocol is a formal set of rules that dictate how data is transmitted and received over a network. It defines the format, timing, and sequence of messages exchanged between devices. Standards, forums, and agencies do not specifically refer to these rules governing data communications.

Option Analysis:

Option A: Forum is a place for discussion and exchange of ideas, not a set of rules for data communication.
Option B: Standard is a general term for a set of guidelines or specifications, but it does not specifically refer to the rules governing data communications.
Option C: Agency is an organization or department responsible for a particular function, not a set of rules for data communication.
Option D: Protocol is the correct term for the set of rules that govern data communications.

682. In Low Earth Orbit (LEO) satellites, round-trip time propagation delay is normally less than

Correct Answer: Correct answer is: (C) 20 ms.

Difficulty: Moderate

Concept notes: In Low Earth Orbit (LEO), the round-trip time propagation delay is the time it takes for a signal to travel from the ground station to the satellite and back. This delay is influenced by the distance the signal must travel and the speed of light.

Common Mistakes: Students often underestimate the propagation delay due to the relatively short distance between the ground and the satellite. However, the speed of light is finite, and the delay can be significant enough to be noticeable.

Explanations: The round-trip time propagation delay for LEO satellites is typically less than 20 ms. This is because LEO satellites orbit at an altitude of approximately 160 to 2,000 km. The signal travels this distance twice (to the satellite and back), and the speed of light is approximately 300,000 km/s. Therefore, the delay is calculated as follows: \[ \text{Delay} = \frac{2 \times \text{Distance}}{\text{Speed of Light}} \] For a typical LEO altitude of 1,000 km: \[ \text{Delay} = \frac{2 \times 1,000 \text{ km}}{300,000 \text{ km/s}} = \frac{2,000 \text{ km}}{300,000 \text{ km/s}} = 0.0067 \text{ s} = 6.7 \text{ ms} \] This calculation shows that the delay is well under 20 ms, confirming that the correct answer is 20 ms.

Option Analysis:

Option A: 10 ms is too low for the typical round-trip delay in LEO.
Option B: 15 ms is also too low for the typical round-trip delay in LEO.
Option C: 20 ms is the correct value for the typical round-trip delay in LEO.
Option D: 2

683. Cable modem for data transmission, is similar to

Correct Answer: Correct answer is: (B) Asymmetric DSL modem.

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

Difficulty: Moderate

Concept notes: Cable modems and asymmetric DSL modems both provide high-speed internet access, but they use different technologies. Cable modems use cable television infrastructure, while asymmetric DSL modems use telephone lines.

Common Mistakes: Students may confuse the technologies used by cable modems and DSL modems, leading to incorrect selection of options.

Explanations: Cable modems and asymmetric DSL modems are both used for high-speed data transmission. Both technologies provide asymmetric service, meaning the download speed is typically much higher than the upload speed. This similarity makes the asymmetric DSL modem the closest match to a cable modem in terms of functionality and service type.

Option Analysis:

Option A: High-bit-rate digital subscriber line (HDSL) is a symmetric DSL technology, which means it provides equal upload and download speeds, unlike cable modems and asymmetric DSL modems.
Option B: Asymmetric DSL modem is the correct choice because it provides similar asymmetric data transmission capabilities as a cable modem.
Option C: Very high-bit-rate digital subscriber line (VDSL) is a type of DSL that can provide both symmetric and asymmetric service, but it is not as commonly used for home internet as asymmetric DSL.
Option D: Asymmetric DSL lite is a lower-speed version of asymmetric DSL, but it still provides similar asymmetric service to a cable modem.

Mnemonic: Cable and ADSL both fast, but ADSL uses phone lines.

684. Which network topology is connected by a single device?

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

Exam Relevance: CCNA, Network+ Certification

Difficulty: Easy

Concept notes: In a star topology, all devices are connected to a central hub or switch.

Common Mistakes: Students may confuse the star topology with the bus topology, where devices are connected to a single backbone cable.

Explanations: In a star topology, each device is directly connected to a central hub or switch. This central device acts as a single point of connection for all other devices in the network. This makes the star topology distinct from other topologies like bus, ring, and mesh, where devices are connected differently.

Option Analysis:

Option A: Bus topology connects devices to a single backbone cable, not a single device.
Option B: Star topology connects all devices to a central hub or switch, which is a single device.
Option C: Ring topology connects devices in a circular loop, not to a single device.
Option D: Mesh topology connects each device to multiple other devices, not to a single device.

Mnemonic: Star = Central Hub

685. In a..... connection, one TCP connection is made for each request/response.

Correct Answer: Correct answer is: (B) Non-persistent.

Exam Relevance: Computer Networks, Data Communications, Network Protocols

Difficulty: Easy

Concept notes: In a non-persistent connection, a new TCP connection is established for each request/response cycle.

Common Mistakes: Students may confuse non-persistent connections with persistent connections, where a single TCP connection is maintained for multiple request/response cycles.

Explanations: In a non-persistent connection, each request/response cycle requires a new TCP connection. This means that for every request made, a new connection is established, and once the response is received, the connection is closed. This is in contrast to a persistent connection, where a single TCP connection is maintained for multiple request/response cycles.

Option Analysis:

Option A: Persistent connections maintain a single TCP connection for multiple request/response cycles, which does not match the description given in the question.
Option B: Non-persistent connections establish a new TCP connection for each request/response cycle, which matches the description given in the question.
Option C: Both a and b is incorrect because only non-persistent connections fit the description.
Option D: None of the above is incorrect because non-persistent connections do fit the description given in the question.

686. What does the passive command provide to dynamic routing protocols?

Correct Answer: Correct answer is: (B) Stops an interface from sending periodic dynamic updates but not from receiving updates.

Exam Relevance: CCNA, CCNP, JNCIA, JNCIS

Difficulty: Moderate

Concept notes: The passive command in dynamic routing protocols is used to prevent an interface from sending routing updates while still allowing it to receive updates.

Common Mistakes: Students often confuse the passive command with the shutdown command, which completely disables an interface.

Explanations: The passive command is used to stop an interface from sending periodic dynamic updates, but it does not prevent the interface from receiving updates. This is useful in scenarios where an interface should not advertise its routes but should still listen to updates from other routers.

Option Analysis:

Option A: This option is incorrect because the passive command does not stop an interface from receiving updates.
Option B: This option is correct as it accurately describes the function of the passive command.
Option C: This option is incorrect because the passive command does not stop the router from receiving any dynamic updates.
Option D: This option is incorrect because the passive command does not stop the router from sending any dynamic updates, only from sending periodic updates on the specified interface.

687. To avoid collisions on wireless networks, .....was invented.

Correct Answer: Correct answer is: (A) CSMA/CA.

Exam Relevance: Computer Networks, Data Communications, Network Security

Difficulty: Moderate

Concept notes: CSMA/CA (Carrier Sense Multiple Access with Collision Avoidance) is a protocol used in wireless networks to prevent data collisions.

Common Mistakes: Students often confuse CSMA/CA with CSMA/CD (Carrier Sense Multiple Access with Collision Detection), which is used in wired networks.

Explanations: To avoid collisions on wireless networks, CSMA/CA was invented. This protocol ensures that a device listens to the channel before transmitting data. If the channel is busy, the device waits until it is free. Additionally, CSMA/CA uses a random backoff time to further reduce the likelihood of collisions. This approach is different from CSMA/CD, which is used in wired networks and detects collisions after they occur.

Option Analysis:

Option A: Correct. CSMA/CA is the protocol used in wireless networks to avoid collisions.
Option B: Incorrect. CSMA/CD is used in wired networks to detect collisions after they occur.
Option C: Incorrect. Only CSMA/CA is used for wireless networks.
Option D: Incorrect. Both options are not correct for wireless networks.

688. A root server's zone is entire

Correct Answer: Correct answer is: (B) DNS Tree.

Exam Relevance: CCNA, CISSP, CompTIA Network+

Difficulty: Moderate

Concept notes: A root server's zone encompasses the entire DNS Tree, which is the hierarchical structure of the Domain Name System.

Common Mistakes: Students may confuse the DNS Tree with other types of trees, such as the STP Tree used in network protocols.

Explanations: The DNS Tree is the hierarchical structure that organizes domain names. Root servers are responsible for the top level of this hierarchy, which includes all domain names. Therefore, a root server's zone is the entire DNS Tree.

Option Analysis:

Option A: STP Tree is not related to DNS and is used in Spanning Tree Protocol for network loop prevention.
Option B: DNS Tree is the correct hierarchical structure managed by root servers.
Option C: HostTree is not a standard term in DNS or networking.
Option D: Sub-Tree would only represent a portion of the DNS Tree, not the entire structure.

Mnemonic: DNS Root: Top of the Tree

689. A modulation technique that improves channel bandwidth utilization.

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

Exam Relevance: GATE, ECE, EE, Telecommunications Engineering

Difficulty: Moderate

Concept notes: Multiplexing is a technique that allows multiple signals to be transmitted over a single channel, thereby improving channel bandwidth utilization.

Common Mistakes: Students may confuse modulation with multiplexing, as both techniques are used in communication systems to improve efficiency.

Explanations: Multiplexing improves channel bandwidth utilization by allowing multiple signals to share the same channel. This is achieved by dividing the channel into multiple sub-channels or time slots, each carrying a different signal. This technique maximizes the use of available bandwidth, making it an effective method for improving channel utilization.

Option Analysis:

Option A: Direct modulation does not inherently improve channel bandwidth utilization.
Option B: Modulation is the process of encoding information onto a carrier signal, but it does not directly improve channel bandwidth utilization.
Option C: Demodulation is the reverse process of modulation, extracting the original information from the carrier signal, and does not improve channel bandwidth utilization.
Option D: Multiplexing allows multiple signals to share a single channel, thereby improving channel bandwidth utilization.

Mnemonic: Mnemonic: "M for Multiplexing, M for Maximizing Bandwidth."

690. In..... there is always a transition at the middle of the bit, but the bit values are determined at the beginning of the bit. If the next bit is 0, there is a transition; if the next bit is 1, there is none.

Correct Answer: Correct answer is: (B) Differential Manchester.

Difficulty: Moderate

Concept notes: Differential Manchester encoding is a line code in which the presence or absence of a transition at the middle of a bit interval is used to encode the data. The bit value is determined by the presence or absence of a transition at the beginning of the bit interval.

Common Mistakes: Students often confuse Manchester encoding with Differential Manchester encoding. In Manchester encoding, the transition occurs at the middle of the bit interval to represent the bit value, whereas in Differential Manchester encoding, the transition at the middle of the bit interval is always present, and the bit value is determined by the presence or absence of a transition at the beginning of the bit interval.

Explanations: In Differential Manchester encoding, there is always a transition at the middle of the bit interval. The bit value is determined by the presence or absence of a transition at the beginning of the bit interval. If the next bit is 0, there is a transition at the beginning; if the next bit is 1, there is no transition at the beginning. This makes Differential Manchester encoding distinct from Manchester encoding, where the transition at the middle of the bit interval represents the bit value.

Option Analysis:

Option A: Manchester encoding does not always have a transition at the middle of the bit interval to determine the bit value. Instead, the transition at the middle of the bit interval represents the bit value.
Option B: Differential Manchester encoding always has a transition at the middle of the bit interval, and the bit value is determined by the presence or absence of a transition at the beginning of the bit interval.
Option C: Both Manchester and Differential Manchester encoding have different characteristics for determining bit values, so this option is incorrect.
Option D: Neither Manchester nor Differential Manchester encoding fits the description given in the question, so this option is incorrect.

Mnemonic: Differential Manchester: "Transition at the middle, bit value at the start."

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Computer networking MCQ - Set 69

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