Concern a disk with a sector size of 512 bytes, 2000 tracks per surface, 50…

Step-by-step solution:

• Compute bytes per track: 50 sectors × 512 bytes = 25,600 bytes

• Express in kilobytes (1 KB = 1024 bytes): 25,600 ÷ 1024 = 25 KB

• Interpretation of "one track per revolution": during each revolution the drive can transfer one track (25 KB) from the active surface.

• Use a typical rotational speed to get revolutions per second. A common disk speed is 5,400 rpm which equals 5,400 ÷ 60 = 90 revolutions per second

• Compute transfer rate: 25 KB × 90 rev/s = 2,250 KB/s

Final answer: 2,250 KBytes/second.

Note: The average seek time (10 ms) affects positioning latency between tracks, not the sustained per-revolution transfer rate used above. Also, although there are multiple platter surfaces physically, the calculation assumes one surface (one head) is read at a time; multiplying by all surfaces would be incorrect unless the device can read all surfaces simultaneously.