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Consider a disk pack with following specifications -  16 surfaces, 128 tracks/surface, 256 sector/track and 512Bytes/sector. Now, the disk is rotating at 3600RPM, what is the data transfer rate.

Please be as elaborative and simplified in your approach of explaining! Thank you, all the nerds on this site!
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There are 16 surfaces -> The following operation will be carried simultaneously on by 16 read/write heads.

We will find the number of tracks one read write head will cover in one second.

3600 rounds in 60 sec => 60 rounds in 1 sec. 

=> 60 tracks have 60 x 256 x 512 Bytes.

Also 16 surfaces are simultaneously operated => 16 x 60 x 256 x 512 Bytes => 125829120 Bytes => 120 MBps

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Time for one rotation = $\frac{60}{3600}$ sec 
in one rotation it reads one track 
Size of one track = 256*512 bytes

So in $\frac{60}{3600}$ sec it reads -------->  256*512 bytes
           so in 1 sec                        ----------->  ?

Transfer rate =   $\frac{256*512}{{
\frac{60}{3600}} = 7864320 byte/sec
there are 16 surfaces so it will have 16 read & write header so TT = 16 *7864320 bytes/sec
                                                                                                = 125829120 bytes /sec

 

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60/3600 sec ------------256 ×512 byte

1 sec -------------------- 256×512×3600/60=7.5 MBps

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