The length of cable required for transmitting a data at the rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits and for signal speed 2,00,000 km/s is

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ESE Electrical 2018 Official Paper

Answer 1

Bandwidth = 500 mbps

Frame size = 10,000 bits

Signal speed = 2,00,000 km/sec

Transmission delay = 2 × propagation delay

frame size/bandwidth = 2 × (length of the cable /signal speed)

Length of the cable \(= \frac{{200000 \times 10000}}{{2 \times 500 \times {{10}^6}}} = 2\;km\)