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Consider a hard disk with 16 recording surfaces (0-15) having 16384 cylinders (0-16383) and each cylinder contains 64 sectors (0-63). Data storage capacity in each sector is 512 bytes. Data are organized cylinder-wise and the addressing format is < cylinder no., surface no., sector no. >. A file of size 42797 KB is stored in the disk and the starting disk location of the file is <1200, 9, 40>. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?

1. 1281
2. 1282
3. 1283
4. 1284

Correct answer: 1284

Solution

Sectors per cylinder = 16*64 = 1024. File = 42797 KB / 0.5 KB = 85594 sectors. Start linear address = 1200*1024 + 9*64 + 40 = 1229416; last = 1229416 + 85594 - 1 = 1315009; cylinder = 1315009 / 1024 = 1284.

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