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GATE | GATE CS 2013 | Question 65

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  • Last Updated : 10 Sep, 2018

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 . 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?
(A) 1281
(B) 1282
(C) 1283
(D) 1284

Answer: (D)

Explanation: File size is 42797KB= 4279*2^10B=85594*2^9B.
Now one sector=512B
so file will be stored in 85594 sectors i.e we need 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 surface we need to cross on more cylinder as file has started at surface 9 and no of surface in cylinder is 16 so
number of cylinder to cross=83+1=84
so cylinder no. 1200+84=1284

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