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 cyclinder-wise and the addressing format is <cylinder no., surface no., sector no.>. A file of size 42797KB is stored in the disk and the starting disk location of the file is <1200,9,40>. What is cylinder number of the last sector of the file, if it is stored in a contiguous manner?
Sectors per cylinder = 16 surfaces × 64 sectors = 1024.
File size = 42797 KB = \(\frac{42797 \times 1024}{512}\) = 85594 sectors.
The start location ⟨1200, 9, 40⟩ is sector \(9 \times 64 + 40 = 616\) within cylinder 1200 (counting from 0).
The last sector is at offset \(616 + 85594 - 1 = 86209\) sectors from the start of cylinder 1200.
\(86209 \div 1024 = 84\) remainder 193.
Last cylinder = 1200 + 84 = 1284.
Answer: (A) 1284
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