Computer networking MCQ – Change your future

21. Most common technique to change an analog signal to digital data is called

Correct Answer: Correct answer is: (D) Pulse code modulation.

Exam Relevance: GATE, ECE, EE, DSP exams

Difficulty: Moderate

Concept notes: Pulse code modulation (PCM) is a method used to convert analog signals into digital form. It involves sampling the analog signal at regular intervals, quantizing the sampled values, and then encoding these quantized values into a digital format.

Common Mistakes: Students often confuse pulse code modulation with other modulation techniques like line code modulation, which is used for encoding digital data rather than converting analog signals.

Explanations: Pulse code modulation (PCM) is the most common technique to change an analog signal to digital data. It involves three main steps: sampling, quantization, and encoding. Sampling captures the amplitude of the analog signal at regular intervals. Quantization converts these sampled values into discrete levels. Encoding then represents these quantized levels as digital codes. This process effectively transforms the continuous analog signal into a discrete digital signal.

Option Analysis:

Option A: Line code modulation is used for encoding digital data, not for converting analog signals to digital form.
Option B: Scrambling is a technique used to randomize data to prevent long sequences of the same bit, but it does not convert analog signals to digital form.
Option C: Block code modulation is a type of error-correcting code used in digital communication, not a method for converting analog signals to digital form.
Option D: Pulse code modulation is the correct technique for converting analog signals to digital form.

Mnemonic: PCM: Pulse, Capture, Modulate

22. In Broadcasting Unknown Server (BUS), destination address is

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

Exam Relevance: Computer Networks, Data Communication, Network Security

Difficulty: Moderate

Concept notes: In Broadcasting Unknown Server (BUS), the destination address is not predetermined or known in advance.

Common Mistakes: Students may mistakenly believe that the destination address is known, as in unicast communication.

Explanations: In BUS, the destination address is unknown because the broadcast is sent to all possible destinations within a network segment. The message is designed to be received by all devices on the network, without specifying a particular recipient. This contrasts with unicast communication, where the destination address is known and specific.

Option Analysis:

Option A: Fixed. This is incorrect because the destination address in BUS is not fixed or predetermined.
Option B: Known. This is incorrect because the destination address in BUS is not known in advance; it is sent to all possible destinations.
Option C: Variable. This is incorrect because while the destination can vary, the term "variable" does not accurately describe the nature of the unknown destination in BUS.
Option D: Unknown. This is correct because the destination address in BUS is not known in advance and is sent to all possible destinations.

23. ..... in the data link layer separates a message from one source to a destination, or from other messages going from other sources to other destinations.

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

Exam Relevance: Computer Networks, Data Communications, Network Fundamentals

Difficulty: Moderate

Concept notes: Framing in the data link layer involves encapsulating data into frames for transmission over a network.

Common Mistakes: Students may confuse framing with digitizing, which is the process of converting analog signals to digital form.

Explanations: Framing is the process in the data link layer that separates a message from one source to a destination from other messages going from other sources to other destinations. This is achieved by encapsulating the data into frames, which are then transmitted over the network. The other options, such as digitizing and controlling, do not specifically address the separation and encapsulation of messages in the data link layer.

Option Analysis:

Option A: Digitizing is the process of converting analog signals to digital form, not related to message separation.
Option B: Controlling refers to the management of data flow, not the separation of messages.
Option C: Framing is the correct process that separates and encapsulates messages for transmission.
Option D: None of the above is incorrect as framing is the correct answer.

Mnemonic: F-R-A-M-E: Frame, Recognize, Arrange, Message, Encapsulate

24. 11i is the same as what?

Correct Answer: Correct answer is: (D) WPA2.

Exam Relevance: CCNA, CISSP, CompTIA Network+

Difficulty: Moderate

Concept notes: WPA2 is the second generation of Wi-Fi Protected Access (WPA) security protocol, designed to secure wireless computer networks.

Common Mistakes: Students may confuse WPA2 with WPA or WEP, which are older security protocols.

Explanations: 11i is the IEEE 802.11i standard, which defines the security enhancements for Wi-Fi networks. This standard is implemented as WPA2, making it the correct answer.

Option Analysis:

Option A: WPA is the first generation of Wi-Fi Protected Access, not the same as 11i.
Option B: WEP is the Wired Equivalent Privacy protocol, an older and less secure standard than 11i.
Option C: 802.11g is a wireless networking standard that specifies a throughput of up to 54 Mbit/s, unrelated to security protocols.
Option D: WPA2 is the correct implementation of the 802.11i standard, providing enhanced security for Wi-Fi networks.

Mnemonic: 11i is WPA2, think "i" for "improved" security.

25. IEEE 802.11 Frequency-Hopping Spread Spectrum (FHSS) uses Band rate of

Correct Answer: Correct answer is: (A) 2.4 GHz.

Exam Relevance: IEEE Certification Exams, CompTIA Network+, CCNA

Difficulty: Easy

Concept notes: IEEE 802.11 Frequency-Hopping Spread Spectrum (FHSS) operates in the 2.4 GHz band.

Common Mistakes: Students may confuse the frequency band with other wireless standards or technologies that operate in different frequency ranges.

Explanations: IEEE 802.11 FHSS is designed to operate in the 2.4 GHz band, which is a globally available unlicensed frequency band. This band is commonly used for various wireless communication technologies, including Wi-Fi. The 2.4 GHz band is chosen for its ability to provide adequate coverage and penetration through walls and other obstacles.

Option Analysis:

Option A: Correct. IEEE 802.11 FHSS operates in the 2.4 GHz band.
Option B: Incorrect. 5.75 GHz is not the correct band for IEEE 802.11 FHSS.
Option C: Incorrect. 5.725 GHz is not the correct band for IEEE 802.11 FHSS.
Option D: Incorrect. 2.9 GHz is not the correct band for IEEE 802.11 FHSS.

Mnemonic: 2.4 GHz, Global Wi-Fi

26. The model that shows how the network function of a computer ought to be organised?

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

Exam Relevance: CCNA, CISSP, CompTIA Network+

Difficulty: Moderate

Concept notes: The OSI (Open Systems Interconnection) model is a conceptual framework used to describe the functions of a networking system.

Common Mistakes: Students may confuse the OSI model with other standards organizations like ITU-T, ISO, or ANSI, which do not specifically define a network function model.

Explanations: The OSI model provides a theoretical framework for understanding how data is transmitted over a network. It divides network functions into seven layers, each with specific responsibilities. This model is used to standardize and simplify the design and implementation of network protocols. The other options (ITU-T, ISO, ANSI) are standards organizations that do not define a network function model.

Option Analysis:

Option A: ITU-T is a standards organization that develops telecommunication standards, not a network function model.
Option B: OSI is the correct answer as it is the model that defines the network function organization.
Option C: ISO is a standards organization that publishes the OSI model, but it is not the model itself.
Option D: ANSI is a standards organization that does not define a network function model.

Mnemonic: OSI: Open Systems Interconnection

27. Find bandwidth for AM if we need to modulate a 5-KHz voice

Correct Answer: Correct answer is: (A) 10 KHz.

Exam Relevance: GATE, ECE, EE

Difficulty: Easy

Concept notes: In amplitude modulation (AM), the bandwidth required is twice the highest frequency of the modulating signal.

Common Mistakes: Students may mistakenly think the bandwidth is equal to the modulating signal frequency, not twice the frequency.

Explanations: The bandwidth for AM is given by 2 times the highest frequency of the modulating signal. Here, the modulating signal is a 5-KHz voice signal. Therefore, the bandwidth is 2 * 5 KHz = 10 KHz.

Option Analysis:

Option A: Correct. The bandwidth is 10 KHz.
Option B: Incorrect. This would be the case if the bandwidth was four times the modulating signal frequency.
Option C: Incorrect. This would be the case if the bandwidth was six times the modulating signal frequency.
Option D: Incorrect. This would be the case if the bandwidth was eight times the modulating signal frequency.

Mnemonic: Double the voice, not the choice.

28. In Integrated Services, when a source makes a reservation, it needs to define a

Correct Answer: Correct answer is: (D) Flow Specification.

Exam Relevance: CCNA, CCNP, Network+

Difficulty: Moderate

Concept notes: In Integrated Services (IntServ), a source making a reservation must define a flow specification to describe the traffic characteristics and the required service level.

Common Mistakes: Students often confuse flow specification with flow control, which is a different concept related to managing data flow rates.

Explanations: In Integrated Services, a flow specification is crucial because it defines the traffic characteristics and the required service level for the data flow. This specification is necessary for the network to allocate resources appropriately and ensure the desired quality of service. Flow control, timers, and error solutions are not directly related to defining the reservation parameters in IntServ.

Option Analysis:

Option A: Flow Control is a mechanism to manage the rate of data transmission, not a reservation parameter in IntServ.
Option B: Timer is a component used in various protocols for timing events, but it is not a reservation parameter in IntServ.
Option C: Error Solution refers to methods for handling errors, which is not a reservation parameter in IntServ.
Option D: Flow Specification is the correct term for defining traffic characteristics and service requirements in IntServ reservations.

Mnemonic: Flow Spec for IntServ, not Flow Control.

29. Why are standards important in networking?

Correct Answer: Correct answer is: (A) They allow hardware and software to interact across different manufacturers.

Exam Relevance: CCNA, CISSP, CompTIA Network+

Difficulty: Moderate

Concept notes: Standards in networking ensure interoperability between different systems and devices from various manufacturers.

Common Mistakes: Students may confuse standards with proprietary protocols, which are designed to work only within a specific manufacturer's ecosystem.

Explanations: Standards in networking are crucial because they establish common protocols and specifications that enable hardware and software from different manufacturers to communicate and work together seamlessly. This interoperability is essential for the functioning of the Internet and other networked systems, ensuring that devices from various vendors can exchange data and services without compatibility issues.

Option Analysis:

Option A: Correct. Standards ensure that hardware and software from different manufacturers can interact effectively.
Option B: Incorrect. Standards do not aim to keep systems private; they promote openness and interoperability.
Option C: Incorrect. Standards do not provide government control over the Internet; they are developed by consensus among industry stakeholders.
Option D: Incorrect. Standards are not created for bureaucratic purposes but to ensure effective communication and interoperability.

30. In Internet, domain name tree is divided into three

Correct Answer: Correct answer is: (D) Different Sections.

Exam Relevance: AP Computer Science Principles, SAT Subject Test in Computer Science

Difficulty: Easy

Concept notes: The domain name system (DNS) in the Internet is organized into a hierarchical structure known as the domain name tree. This tree is divided into different sections, which include top-level domains (TLDs), second-level domains, and subdomains.

Common Mistakes: Students may confuse the hierarchical structure of the domain name tree with layers or stages, which are not accurate terms in this context.

Explanations: The domain name tree is divided into different sections, which are organized hierarchically. These sections include top-level domains (TLDs) such as .com, .org, and .edu, second-level domains like example.com, and subdomains like blog.example.com. The term "sections" accurately describes these divisions within the domain name tree.

Option Analysis:

Option A: "Different Stages" is incorrect because the domain name tree is not organized by stages but by a hierarchical structure.
Option B: "Different Layers" is incorrect because the term "layers" is not used to describe the divisions within the domain name tree.
Option C: "Different Parts" is incorrect because while the domain name tree is divided into parts, the term "sections" is more precise and commonly used.
Option D: "Different Sections" is correct because the domain name tree is divided into sections such as TLDs, second-level domains, and subdomains.

Mnemonic: DNS Tree: TLD, SLD, Sub

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Computer networking MCQ - Set 3

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