NIC-NIELIT STA 2020
October 4, 2023NIC-NIELIT Scientist-B 2020
October 4, 2023Computer-Networks
|
Question 27
|
Consider a 100 Mbps link between an earth station (sender) and a satellite (receiver) at an altitude of 2100 km. The signal propagates at a speed of 3×10 8 m/s. The time taken (in milliseconds, rounded off to two decimal places) for the receiver to completely receive a packet of 1000 bytes transmitted by the sender is_________
|
7.08 ms
|
Question 27 Explanation:
The time required for the receiver to receive the packet is transmission time + propagation time i.e. Tt +Tp
Tt = L / B => 1000 x 8 bits / 10^8 bps = 0.08 ms
Tp = D / V => 2100 x 1000 m / 3 x10^8 ms = 7 ms
Therefore, Total time = 7.08 ms
Tt = L / B => 1000 x 8 bits / 10^8 bps = 0.08 ms
Tp = D / V => 2100 x 1000 m / 3 x10^8 ms = 7 ms
Therefore, Total time = 7.08 ms
Correct Answer: A
Question 27 Explanation:
The time required for the receiver to receive the packet is transmission time + propagation time i.e. Tt +Tp
Tt = L / B => 1000 x 8 bits / 10^8 bps = 0.08 ms
Tp = D / V => 2100 x 1000 m / 3 x10^8 ms = 7 ms
Therefore, Total time = 7.08 ms
Tt = L / B => 1000 x 8 bits / 10^8 bps = 0.08 ms
Tp = D / V => 2100 x 1000 m / 3 x10^8 ms = 7 ms
Therefore, Total time = 7.08 ms
Subscribe
Login
0 Comments