GATE 2015 CS (CS2) – Question 42
A disk rotates at 15000 RPM and transfers $50\times10^6$ bytes/s. Average seek time is twice average rotational delay. Controller transfer time is ten times disk transfer time. Find average time in milliseconds to read or write a 512-byte sector.
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Correct answer: 6.1 to 6.2
Explanation
One rotation takes 60/15000 seconds=4 ms, giving average rotational delay two ms and average seek four ms. Disk sector transfer takes 512/(50×10⁶) seconds=0.01024 ms. Controller time is 0.1024 ms. Total is 4+2+0.01024+0.1024=6.11264 ms, within the published 6.1–6.2 range.