A 2 inch disk has 8 recording surfaces (4 plates), this disk has 512 tracks with each track having 512 sectors of 512 bytes each. Calculate the capacity and recording density for the disk. This disk rotates at a speed of 6000 rpm and has an average seek time of 10ms; what will be the average access time of the disk?
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Suppose a HDD having n plates,
total no: of recording surfaces (m)=8,
tracks per surface(t)=512,
sector per track(p)=512,
bytes per sector (s)=512,
π=3.14,
d=2,
storage capacity of the disk
=(m*t*p*s)bytes
=(8*512*512*512)bytes
=(8*512*512*512)/(1024*1024*1024)
=1 GB//
density =(s*p)/(π*d)bytes/inch
=(512*512)/(3.14*2)=41742.68 bytes per inch//
total no: of recording surfaces (m)=8,
tracks per surface(t)=512,
sector per track(p)=512,
bytes per sector (s)=512,
π=3.14,
d=2,
storage capacity of the disk
=(m*t*p*s)bytes
=(8*512*512*512)bytes
=(8*512*512*512)/(1024*1024*1024)
=1 GB//
density =(s*p)/(π*d)bytes/inch
=(512*512)/(3.14*2)=41742.68 bytes per inch//
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