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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

1. $\langle 0, 15, 31 \rangle$
2. $\langle 0, 16, 30 \rangle$
3. $\langle 0, 16, 31 \rangle$
4. $\langle 0, 17, 31 \rangle$
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1039th 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 17th 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.

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what is the significance in adding 1 to 1039 ?
couting starts from 0 and 0th sector is counted as 0 itself right ?
So 1039th sectors  is actually numbered 1039 itself right ?

check option C:

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

correct..

point to be noted here that secor 31 means that it is 32th sector since numbering starts from 0.

16*63 =1008 sectors have been crossed and we need to read further 32 sectors to get 1039th sector(sector number 1040)

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 $= 0+ (16 \times 63) + 31 \\ = 1039$

@cse23

edited
@amarVashishth sir,
How could one  interpret that data on disk is stored cross sectionally ? and not latteraly ?

Thus, the 0th sector is addressed as ⟨0,0,0⟩, the 1st sector as ⟨0,0,1⟩, and so on.

If we proceed in a similar fashion we have the address <0,1,0> after all the sectors in a track are addressed. Which means the addresses are numbered in a cross sectional fashion.

+1 vote
track no= (1039/63) =16.492063492063492063492063492063 .... so track no = 16 ..

now i am considering the fractional part ... sector no=0.492063492063492063492063492063* 63 = 31

so the address is <0,16,31> ....

correct me if i am wrong ...