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If the disk is rotating at 3600rpm, determine the effective data transfer rate which is defined as the number of bytes transferred per second between
disk and memory. (Given the size of track = 512 bytes)?
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Given rpm= 3600 rpm

so in 1 minute it rotates 3600 times,

So, time for 1 rotation = 60/3600 = 1/60 sec

In one rotation, it traverses one track

So, in 1 rotation ,it transfers 512 bytes,

So, 1 second it transfers 60 * 512 bytes= 30720 bytes

Now convert it into Kb = 30720/2^10

=30 KBps

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