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  • Computer Networks and Topology MCQs & PYQs Bank | Suraku Academy

Q: 1 Let us assume that there are four (4) network segments in Bus topologies A1—A2, B1—B2—B3, C1—C2 and D1—D2—D3. If A1, B1, C1 and D1 are interconnected in mesh, then how many mesh connections are required?

  1. 45

  2. 6

  3. 10

  4. 12

[ Option B ]

Given 4 network segments represented by points A1, B1, C1, and D1 interconnected in a mesh, we need to find the total number of mesh connections.

  • In a full mesh network, every node is connected to every other node.
  • The formula for the number of connections in a full mesh with n nodes is [n(n-1)/2].

With 4 nodes the number of connections is [4(4-1)/2] = [4*3/2] = 12/2 = 6.

Q: 2 Which of the following statements about twists in twisted pair cables is TRUE?

  1. Increasing the number of twists per unit of length reduces the cable's electrical resistance.

  2. More twists per unit of length in the cable increase signal distortion.

  3. A higher twist rate gives better signal quality.

  4. Fewer twists per unit of length in the cable result in better noise immunity.

[ Option C ]

Twisted pair cables are a type of electrical cable commonly used in networking and telecommunications.
Twisted pair cables consist of two insulated copper conductors twisted together to reduce electromagnetic interference (EMI) and crosstalk between adjacent pairs and improves overall signal quality.

high-twist-rate

Q: 3 What is an infrastructure network as per IEEE 802.11 standard?

  1. A basic service set without an access point.

  2. A basic service set with an access point.

  3. Two or more basic service sets without access point.

  4. Only access points but no basic service set.

[ Option B ]

In wireless networking, IEEE 802.11 is the standard that defines Wi-Fi networks. Under this standard, a Basic Service Set (BSS) is the fundamental building block of a wireless LAN. There are two primary types of network configurations:
Infrastructure Mode : This setup includes a central Access Point (AP) that coordinates communication between wireless devices (stations). Multiple BSS can be interconnected via access points.
Ad hoc Mode (Independent Basic Service Set (IBSS)) :This setup does not have an access point. Wireless devices communicate directly with each other, forming a temporary or decentralized network.

Network TypeAccess Point PresentCommunication TypeExample
Infrastructure NetworkYesThrough Access Point.Home Wi-Fi, Office Wi-Fi.
Ad hoc (IBSS)NoDirect (Peer-to-Peer)Temporary device-to-device sharing.

Q: 4 Which standard is used to define protocol, host computer, port and path on the Internet?

  1. IP address

  2. Uniform Resource Locator (URL)

  3. MAC address

  4. POP address

[ Option B ]

When we access a webpage or any resource on the Internet, we use an address that tells the computer where to find it. This address is called a URL (Uniform Resource Locator). It includes several parts such as the protocol, host computer, port, and path.
URL example, https://www.surakuacademy.com/rpsc-vice-principal-iti-information-technology-paper-solution

PARTDESCRIPTIONEXAMPLE
ProtocolCommunication method such as HTTP, HTTPS, FTP.https
HostDomain name or IP address of the server.www.surakuacademy.com
PortConnection port, optional, tells which service or connection point to use. Common default ports are 80 for HTTP and 443 for HTTPS.443
PathThe specific file or page inside the website. Specifies the exact location or file on the host server./rpsc-vice-principal-iti-information-technology-paper-solution

Q: 5 There are n stations in a slotted LAN. Each station attempts to transmit with a probability p in each time slot. What is the probability that only one station transmits in a given time slot? 

  1. 1 – (1 – p)n-1

  2. (1 – p) n-1

  3. n × p × (1 – p)n-1

  4. p × (1 – p)n-1

[ Option C ]

In a slotted LAN, all n stations get a chance to send data in every time slot. Each station tries to send with a probability p. To find the probability that only one station sends data, we think about two things:

  • One station must send, this happens with probability p.
  • All the other (n – 1) stations must stay silent, each stays silent with probability 1-p, so together it becomes (1−p)n−1.

Because any one of the n stations could be the one that transmits successfully, we multiply by n. So, the probability that only one station transmits is n×p×(1−p)n−1.

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