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Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data is 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 respectively 
are:
  • a)
    256 Mbytes, 19 bits
  • b)
    256 Mbytes, 28 bits
  • c)
    512 Mbytes, 20 bits
  • d)
    64 Gbytes, 28 bits
Correct answer is option 'A'. Can you explain this answer?
Most Upvoted Answer
Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 ...
Capacity of the disk pack:
The capacity of the disk pack can be calculated by multiplying the number of surfaces, tracks per surface, sectors per track, and the number of bytes per sector.

Given:
- Number of surfaces = 16
- Tracks per surface = 128
- Sectors per track = 256
- Bytes per sector = 512

Capacity = Number of surfaces × Tracks per surface × Sectors per track × Bytes per sector
= 16 × 128 × 256 × 512 bytes

Converting bytes to megabytes:
1 megabyte = 1,048,576 bytes

Capacity = (16 × 128 × 256 × 512) / 1,048,576 megabytes
= 256 megabytes

Therefore, the capacity of the disk pack is 256 megabytes.

Number of bits required to specify a particular sector in the disk:
To specify a particular sector, we need to consider the number of surfaces, tracks per surface, and sectors per track.

Given:
- Number of surfaces = 16
- Tracks per surface = 128
- Sectors per track = 256

To specify a sector, we need to specify the surface, track, and sector number.

Number of bits required to specify a surface = log2(number of surfaces) = log2(16) = 4 bits
Number of bits required to specify a track = log2(tracks per surface) = log2(128) = 7 bits
Number of bits required to specify a sector = log2(sectors per track) = log2(256) = 8 bits

Therefore, the number of bits required to specify a particular sector in the disk is 4 + 7 + 8 = 19 bits.

Hence, the correct answer is option A) 256 Mbytes, 19 bits.
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Community Answer
Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 ...
Number of surfaces = 16
Tracks = 16 × 128 × 256 = 24 × 27 × 28
Number of bits required by a sector = 219
So, 19 lines are required to address all the sectors.
Bytes = 219 × 512 B = 219 × 29 = 228
Bytes = 256 MB
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Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data is 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 respectivelyare:a)256 Mbytes, 19 bitsb)256 Mbytes, 28 bitsc)512 Mbytes, 20 bitsd)64 Gbytes, 28 bitsCorrect answer is option 'A'. Can you explain this answer?
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