Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors…
2007
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:
- A.
256 Mbyte, 19 bits
- B.
256 Mbyte, 28 bits
- C.
512 Mbyte, 20 bits
- D.
64 Gbyte, 28 bit
Attempted by 67 students.
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Correct answer: A
Total sectors: 16 × 128 × 256 = 524,288 sectors (16 = 2^4, 128 = 2^7, 256 = 2^8, so total = 2^(4+7+8) = 2^19).
Total capacity: 524,288 sectors × 512 bytes/sector = 268,435,456 bytes = 256 Mbyte (since 512 = 2^9 and 2^(19+9) = 2^28 bytes = 256 × 2^20).
Bits to specify a particular sector: You need enough bits to address 524,288 sectors, so bits = log2(524,288) = 19 bits.
Final answer: Capacity = 256 Mbyte; number of bits required to specify a sector = 19 bits.
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