Question

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

Answer 1

Correct Answer - Option 4 : 1284

Data:

File size is 42797 KB= 42797 ×210 B=85594×29 B.

recording surfaces = 16 (0 to 15)

cylinders = (0-16383)

sectors(0-63
one sector=512 B

Calculation:
The file will be stored in 85594 sectors, that is, it needs to cross 85594 sectors
starting of the file is
number of cylinders to cross = (85594÷16)× 64 = 83 cylinders
remaining sectors to cross = 85594-(83×16× 64) = 602
number of surfaces to cross = 9
so to cross 9 surfaces we need to cross on more cylinder as the file has started at surface 9 and no surface in the cylinder is 16 so
number of the cylinder to cross=83+1=84
so cylinder number = 1200+84=1284