I am learning computer architecture and organization I am stuck in the following question.
Consider a hard disk with sector size 1024 bytes, 5000 tracks per surface, 64 sectors per track, and 8 surfaces. If the disk platter rotates at 10,000 rpm, and one track of data can be transferred per revolution, the average data transfer rate is ……………. Mbytes/sec.
Correct answer is 10.66 Mbytes/sec.
1 track = 64 x 1024 = 64 Kbytes.
Time for one revolution = 6 msec.
So, in 6 msec the data transferred is 64 Kbytes
Thus, in 1 sec the data transferred is 64 / 6 Mbytes = 10.6 Mbytes/sec
The solution was given along with the question which I think is wrong as the correct answer should be (64 * 1000) / (6 * 1024) = 10.41 Mbytes/sec.