5.7k views

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
edited | 5.7k views

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

Capacity of disk $= 2^{4+7+8+9} = 2^{28} = 256 \ MB$

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

answered by Loyal (8.1k points)
edited
+9

 4 7 8 9
0
How 2^28 is equal to 256MB.
0
(16 surfaces X 128 tracks X 256 sectors X 512 bytes)/(1024*1024) MB= 256MB
0

@ilakkiya2000

2^28 B = (2^8 * 2^20) B

=> 256 MB(2^20 = M)

+1 vote

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 Active (4k points)
answered by Boss (15.9k points)

1
2