In a token ring network the transmission speed is $10^7$ bps and the propagation speed is $200\;\text{meters}/\mu \text{s}.$ The $1$-bit delay in this network is equivalent to:
$500$ meters of cable.
$200$ meters of cable.
$20$ meters of cable.
$50$ meters of cable.
As, token ring network
So, transmission delay = length / bandwidth
transmission delay = 1/10^7 = 0.1micro sec.
as, propagation speed =200m/micro sec.
In , 1micro sec. it covers 200m
than in 0.1micro sec. it is 20metres.(C) answer
@smartmeet & @Regina Phalange
It works on CSMA/CD and note that there we deal only with one sender and one receiver at distant location (so wait 2Tp for both collision S-R & R-S direction)
but in ring each station(sender/receiver) is connected with another adjacent station. So we don't require that much time to wait for.
L/B = d/v
So it will work but remove 2 from your from your TRICK
for another example working
https://www.avatto.com/computer-science/test/mcqs/networking/questions/81/10.html
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THIS IS MY OBSERVATION ANY KIND OF CORRECTION IS WELCOMED
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