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+15 votes

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 $\langle c, h, s \rangle$, where $c$ is the cylinder number, $h$ is the surface number and $s$ is the sector number. Thus, the 0$^{th}$ sector is addresses as $\langle 0, 0, 0 \rangle$, the 1$^{st}$ sector as $\langle 0, 0, 1 \rangle$, and so on

The address of the 1039$^{th}$ sector is

- $\langle 0, 15, 31 \rangle$
- $\langle 0, 16, 30 \rangle$
- $\langle 0, 16, 31 \rangle$
- $\langle 0, 17, 31 \rangle$

+21 votes

Best answer

1039^{th} sector will be stored in track number (1039 + 1)/63 = 16.5 (as counting starts from 0 as given in question) and each track has 63 sectors. So, we need to go to 17^{th} track which will be numbered 16 and each cylinder has 20 tracks (10 platters * 2 recording surface each) . Number of extra sectors needed = 1040-16*63 = 32 and hence the sector number will be 31. So, option C.

couting starts from 0 and 0th sector is counted as 0 itself right ?

So 1039th sectors is actually numbered 1039 itself right ?

+4 votes

a sector address is given like this $\langle c, h, s \rangle$ so it means that, we traverse in this way:

this all is happening in a single cylinder:

means first move sector wise(covering a track on a surface),

when all sector are over change the surface (also means a single track is over)

then move in all surfaces(a cylinder has $2\times 10 = 20$ total surfaces) as done previously,

when all surface are over then at last move into next cylinder

for a single track, it has 63 sectors

for a single surface, it has 1 track = 63 sectors

this means that for a single cylinder, it has 20 surfaces = 20 tracks = 20 $\times$ 63 sectors

on checking **option C**:

$\langle0,16,31\rangle$

Number of sectors to be crossed $= 0+ (16 \times 63) + 31 \\ = 1039$

**option C** is the answer

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