Aspire Faculty ID #19004 · Topic: UGC NET Computer Science Nov 2020 (Paper II) · Just now
UGC NET Computer Science Nov 2020 (Paper II)

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, then approximately what is the maximum rotational delay?

Solution

Given,

Rotational speed $=5400$ rpm

It means:

$5400$ revolutions in $60$ seconds

So, revolutions per second:

$\frac{5400}{60}=90$

Time taken for one revolution:

$\frac{1}{90}$ seconds

$=0.0111$ seconds

Maximum rotational delay is equal to time for one full revolution.

So, maximum rotational delay is approximately:

$0.011$ seconds

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