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Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data are stored in a bit serial manner in a sector. The capacity of the disk pack and the number of bits required to specify a particular sector in the disk are respectively:

  1. 256 Mbyte, 19 bits
  2. 256 Mbyte, 28 bits
  3. 512 Mbyte, 20 bits
  4. 64 Gbyte, 28 bits

 

asked in Operating System by Veteran (69k points)
retagged by | 3.2k views

3 Answers

+26 votes
Best answer
ans is A.

16 surfaces= 4 bits, 128 tracks= 7 bits, 256 sectors= 8 bits, sector size 512 bytes = 9 bits

capacity of disk = 2^(4+7+8+9) = 2^28 = 256MB

to specify a particular sector we do not need sector size, so bits required = 4+7+8 = 19
answered by Boss (8.5k points)
selected by

address split is-

              4

               7

           8

         9

 

0 votes

Disk capacity = total number of surfaces * no. of tracks per surface * amount of data per track  

 16 surfaces (4 bit) * 128 tracks per surface (7 bit) *  256 sectors per track (8 bit) *512 bytes of data

256MB

and to specify a particular sector we do not need sector size, so bits required = 4+7+8 = 19

answered by Loyal (4.8k points)
–2 votes
answered by Veteran (17.7k points)


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