Question 16141 – GATE 2022
May 30, 2024NTA-UGC-NET 2021 Dec & 2022 June Paper-2
May 30, 2024Question 13946 – Computer-Organization
Comprehension:
Question 91-95 concerns a disk with a sector size of 512 bytes, 2000 tracks per surface. 50 sectors per track, five double-sided platters, and average seek time of 10 milliseconds.
Q91: If one track of data can be transferred per revolution, then what is the data transfer rate?
Correct Answer: D
Question 659 Explanation:
It is given that the disk platters rotate at 5400 RPM.
In 60s – 5400 rotations
Time for one rotation = 60s/5400 = (1/90) seconds
In (1/90)seconds one track size can be transferred. Each track is of size 25KB.
In 1/90 seconds 25KB can be transferred.
In 1 seconds the amount of data that can be transferred = 25KB*90 = 2250KB
So the data transfer rate = 2250 KB/s
In 60s – 5400 rotations
Time for one rotation = 60s/5400 = (1/90) seconds
In (1/90)seconds one track size can be transferred. Each track is of size 25KB.
In 1/90 seconds 25KB can be transferred.
In 1 seconds the amount of data that can be transferred = 25KB*90 = 2250KB
So the data transfer rate = 2250 KB/s
2,850 KBytes/second
4,500 KBytes/second
5,700 KBytes/second
2,250 KBytes/second