Concern a disk with a sector size of 512 bytes, 2000 tracks per surface, 50…
2020
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 the disk platters rotate at 5400 rpm (revolutions per minute), then approximately what is the maximum rotational delay?
Answer: A. 0.011 seconds — Key calculation: convert the rotation speed into time per revolution. Convert rpm to revolutions per second: 5400 rpm ÷ 60 = 90 revolutions per second. Time…
- A.
0.011 seconds
- B.
0.11 seconds
- C.
0.0011 seconds
- D.
1.1 seconds
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Correct answer: A
Key calculation: convert the rotation speed into time per revolution.
Convert rpm to revolutions per second: 5400 rpm ÷ 60 = 90 revolutions per second.
Time per revolution = 1 ÷ 90 ≈ 0.011111... seconds (≈11.11 ms).
Maximum rotational delay is one full revolution, so ≈0.01111 s. (Average rotational latency is half that, ≈0.00556 s or 5.56 ms.)
Answer: 0.011 seconds (about 11.11 ms).
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