Access Control MCQ Quiz in বাংলা - Objective Question with Answer for Access Control - বিনামূল্যে ডাউনলোড করুন [PDF]
Last updated on Mar 19, 2025
Latest Access Control MCQ Objective Questions
Top Access Control MCQ Objective Questions
Access Control Question 1:
Which of the following is not a topology in computer field?
Answer (Detailed Solution Below)
Access Control Question 1 Detailed Solution
- Network topology is a schematic description of a network arrangement, connecting various nodes(sender and receiver) through lines of connection.
- Common network topologies used in the computer network is as shown:
Conclusion: Cap is not a topology in the computer field.
Access Control Question 2:
Which of the following is/are false?
I. A firewall is designed to allow "good" packets to pass through and to prevent "bad" packets.
II. In the transport layer, entire connections can be encrypted, end to end, that is, process to process.
Answer (Detailed Solution Below)
Access Control Question 2 Detailed Solution
|
Network layer |
In the network layer, firewalls are typically used to filter and control traffic based on IP addresses, ports, and protocols. Firewalls are designed to allow or block packets based on certain criteria. |
|
Transport Layer |
Transport Layer Security (TLS) and Secure Sockets Layer (SSL) are protocols that provide end-to-end encryption at the transport layer. This ensures that the data transferred between two systems, i.e., process to process, is secure from eavesdropping or tampering. |
Access Control Question 3:
Which of the following operates at network layer?
I. bridge
II. repeater
III. three-layer switch
IV. routerAnswer (Detailed Solution Below)
Access Control Question 3 Detailed Solution
Access Control Question 4:
Which of the following is an example of multiport repeater?
Answer (Detailed Solution Below)
Access Control Question 4 Detailed Solution
An active hub is a multiport repeater. It is normally used to create connections between stations in a physical star topology.
Hubs can also be used to create multiple levels of hierarchy. The hierarchical use of hubs removes the length limitation of 10Base-T (100 m).Access Control Question 5:
A repeater has ____________ capacity.
Answer (Detailed Solution Below)
Access Control Question 5 Detailed Solution
An amplifier cannot discriminate between the intended signal and noise; it amplifies equally everything fed into it. A repeater dots not amplify the signal; it regenerates the signal. When it receives a weakened or corrupted signal, it creates a copy, bit for bit, at the original strength.
A repeater forwards every frame; it has no filtering capacity.Access Control Question 6:
Which of the following entity operates below the physical layer in the internet model?
Answer (Detailed Solution Below)
Access Control Question 6 Detailed Solution
A passive hub is just a connector.
It connects the wires coming from different branches.
In a star-topology Ethernet LAN, a passive hub is just a point where the signals coming from different stations collide; the hub is the collision point.
This type of a hub is part of the media; its location in the Internet model is below the physical layer.
Access Control Question 7:
Which of the following statement is/are false?
I. A mesh topology is less expensive than a star topology.
II. Each device has only one link and one I/O port, in star topology.Answer (Detailed Solution Below)
Access Control Question 7 Detailed Solution
- In star topology, each device needs only one link and one I/O port to connect it to any number of other devices
- In mesh topology, each device is connected to each other directly by a link
- In full duplex mode with ‘n’ nodes are present then number of links needed is (n× (n-1)) ÷ 2)
- Therefore, wiring required in case of mesh topology is more than the star topology, also installation cost is more which makes it expensive
Access Control Question 8:
A pure ALOHA network transmits 400-bit frames on a shared channel of 400 kbps. What is the maximum throughput(in frames) if the system (all stations together) produces 500 frames per second?
Answer (Detailed Solution Below) 91
Access Control Question 8 Detailed Solution
For a pure ALOHA
Efficiency(η) = G × e-2G
If η is maximum then G = 1/2
η = 1/(2e) = 0.1839
Throughput = number of frames × η = 500× 0.1839 = 91.96 frames
This implies only 91 frames will survive.
Note:
Number of frames is always less then 92
Access Control Question 9:
For the CSMA/CA protocol, which of the following statements is/are FALSE?
I. Before sending RTS, source station waits for a period of time called the DIFS.
II. After receiving RTS and waiting a period of time called the SIFS, destination station sends CTS to the source station.
III. Acknowledgment is not needed in CSMA/CA protocol.
Answer (Detailed Solution Below)
Access Control Question 9 Detailed Solution
Before sending a frame, the source station senses the medium.
I)True: After the station is found to be idle, the station waits for a period of time called the distributed interframe space (DIFS); then the station sends a control frame called the request to send (RTS).
II)TRUE: After receiving the RTS and waiting a period of time called the short interframe space (SIFS), the destination station sends a control frame, called the clear to send (CTS), to the source station. This control frame indicates that the destination station is ready to receive data.
III)FALSE: Acknowledgment is needed in CSMA protocol because the station does not have any means to check for the successful arrival of its data at the destination.
Access Control Question 10:
There is a channel with bandwidth 2 MBps, propagation delay 50 ms and packet size 4KB. What will be the channel utilization?
Answer (Detailed Solution Below)
Access Control Question 10 Detailed Solution
Data:
Propagation delay = Tp = 50 ms
Bandwidth = BW = 2 MBps
Packet size = L = 4 KB
Formula:
Transmission delay = Tt = \(\frac{L}{BW}\)
Utilization = \(\frac{T_t}{T_t + 2T_p} \times 100\)
Calculation:
Transmission delay = Tt = \(\frac{4 \;KB}{2\; MB}= 2 \times10^{-3} s\)
∴ Tt = 2 ms
Utilization = \(\frac{2\; ms}{2 \;ms + 100 \;ms} \times 100 = 1.96 \)
Therefore, the channel utilization is 1.96%.