in Operating System recategorized by
1,175 views
1 vote
1 vote

The question given below concern 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.

If one track of data can be transferred per revolution, then what is the data transfer rate?

  1. $2,850$ KBytes/second
  2. $4,500$ KBytes/second
  3. $5,700$ KBytes/second
  4. $2,250$ KBytes/second
in Operating System recategorized by
1.2k views

2 Answers

1 vote
1 vote
1 rotation takes 60/5400 seconds

in 1 rotation 1 track data can be read

size of 1 track is 50*512B

50*512B takes 60/5400 second

in 1 second  2304000bytes will be transferred

=2250KBytes/second
by
0 votes
0 votes
Given in other part of question, disk makes $5400$ revolution per minute

$60$ sec= $5400$ revolutions

$1$ revolution = $00.11$ sec

Size of track = $50*512$ B = $25600$ B =$25$ KB

If one track of data can be transferred in $1$ revolution that means $25$ KB data is transferred in $0.11$ sec

Data transfer rate(Data transferred/sec) = $25$ KB/$0.11$ = $2272.72$ KB/sec
Answer:

Related questions