Multiple choice

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 organised cylinder–wise and the addressing format is <cylinder 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
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D Correct answer
Explanation

42797 KB $\equiv \dfrac{72797 \times 1025}{512}$= 85594 sectors Starting of 1200, 9, 40 contains total 24 + (6 × 64) = 408 sectors Next, 1201, --------, 1283 cylinders contains total 1024 × 83 = 84992 sectors ($\because$ each cylinder contains 16641024 sectors) $\therefore$Total = 408 + 84992 = 85400 sectors $\therefore$ The required cylinder number is (1284) which will contain the last sector of the file.

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