Question 4678 – 2012 December UGC NET Paper 1
April 25, 2024Question 10112 – GATE 1996
April 26, 2024Data-Link-Layer
| Question 9 |
Determine the maximum length of the cable (in km) for transmitting data at a rate of 500 Mbps in an Ethernet LAN with frames of size 10,000 bits. Assume the signal speed in the cable to be 2,00,000 km/s.
| 1 | |
| 2 | |
| 2.5 | |
| 5 |
Question 9 Explanation:
500×106 bits —— 1 sec
∴104 bits —–5×108/104 = 104/5×108sec = 1/5×104sec
1 sec——2×105 km
1/5×104—–2×105/5×104 = 4 km
Maximum length of cable = 4/2 = 2 km
∴104 bits —–5×108/104 = 104/5×108sec = 1/5×104sec
1 sec——2×105 km
1/5×104—–2×105/5×104 = 4 km
Maximum length of cable = 4/2 = 2 km
Correct Answer: B
Question 9 Explanation:
500×106 bits —— 1 sec
∴104 bits —–5×108/104 = 104/5×108sec = 1/5×104sec
1 sec——2×105 km
1/5×104—–2×105/5×104 = 4 km
Maximum length of cable = 4/2 = 2 km
∴104 bits —–5×108/104 = 104/5×108sec = 1/5×104sec
1 sec——2×105 km
1/5×104—–2×105/5×104 = 4 km
Maximum length of cable = 4/2 = 2 km
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