Hard disk

Question

If a disk system has an average seek time of 30 ns and rotational rate of 360 rpm. Each track of the disk has 512 sectors each of size 512 bytes.
a)  What is the time required to read 4 successive sectors?
b) What is the data transfer rate?

Answer 1

a) in 1/60 seconds 1 track is read, so in 1/60 seconds 512 sectors are read, so 4 sectors are read in = 4/(60*512) sec.

b) in 1/60 seconds  1 track is read, so in 1/60 seconds 512 sectors are read, each sector 512 kb, so in 1 second --->(2^18)/60 bytes.

Comments

In $\frac{1}{60}sec \rightarrow 2^{18}B$

So, In $1sec \rightarrow 2^{18}*60B = 15MB$

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1/60 seconds or 1/6?Please verify

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it should be 1/6 only