A hard disk has 63 sectors per track, 10 platters each with 2 recording…
2009
A hard disk has 63 sectors per track, 10 platters each with 2 recording surfaces and 1000 cylinders. The address of a sector is given as a triple (c, h, s), where c is the cylinder number, h is the surface number and s is the sector number. Thus, the 0th sector is addressed as (0, 0, 0), the 1st sector as (0, 0, 1), and so on
Consider the data given in previous question. The address of the 1039th sector is
- A.
(0, 15, 31)
- B.
(0, 16, 30)
- C.
(0, 16, 31)
- D.
(0, 17, 31)
Attempted by 87 students.
Show answer & explanation
Correct answer: C
Key method: convert a sequential sector number N into a (cylinder, head, sector) address using the disk geometry.
Given heads H (numbered 0..H-1) and sectors per track S (numbered 1..S), compute sectors per cylinder = H × S.
Compute cylinder = floor((N - 1) / (H × S)).
Compute offset = (N - 1) % (H × S). Then head = floor(offset / S) and sector = (offset % S) + 1.
Application and conclusion:
Using the disk geometry implied by the problem (heads numbered 0–15 and sectors numbered 1–31), the CHS conversion for the 1039th sector yields the address (0, 15, 31).
Why the other choices are incorrect:
Addresses that show head values of 16 or 17 are invalid for a 16-head device because valid head indices are 0 through 15.
Therefore any triple with head = 16 or head = 17 cannot represent a real sector on the specified geometry; the only valid triple among the options is (0, 15, 31).