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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 organised cylinder-wise and the addressing format is < cylinder number, surface number, sector number >. 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?
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- 1281
- 1282
- 1283
- 1284
- 1281
Correct Option: D
Starting disk location of the file is <1200, 9,40 >. Since starting number of sectors left on 9th surface is = 24. So, on 9th surface total storage of "12288" B is possible. Now, a part from 9th surface, on cylinder number 1200 only 6 surface is left, Storage possibilities on these 6 surfaces are = 6 × 26 * 29 storage on each sector number of sectors on each surface =196608 B
The total on cylinder no 1200, storage possible = 196608 + 12288 = 208896 B.
Since file size is 42797 KB and out of which 208896 B are stored on cylinder, no 1200. So we are left only with 43615232 B.
Since in 1 cylinder, storage possible is = 24 × 26 × 29 B = 524288
| So, we need about = | = 83.189 more cylinders. | |
| 524288 B |
Hence, we will need 1284th cylinder to completely store the files since after 1283rd cylinder we will be left with data which will 189.
