GATE CSE
First time here? Checkout the FAQ!
x
+12 votes
901 views

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. $\langle 0, 17, 31 \rangle$
asked in Operating System by Veteran (87.2k points)  
retagged by | 901 views

3 Answers

+18 votes
Best answer

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. 

answered by Veteran (19.5k points)  
selected by
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 $= 0+ (16 \times 63) + 31 \\ = 1039$

option C is the answer

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)
+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

@cse23

answered by Veteran (28.7k points)  
edited by
@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 ...
answered by Loyal (2.8k points)  


Top Users Sep 2017
  1. Habibkhan

    6362 Points

  2. Warrior

    2234 Points

  3. Arjun

    2212 Points

  4. nikunj

    1980 Points

  5. manu00x

    1726 Points

  6. SiddharthMahapatra

    1718 Points

  7. Bikram

    1716 Points

  8. makhdoom ghaya

    1660 Points

  9. A_i_$_h

    1552 Points

  10. rishu_darkshadow

    1512 Points


25,991 questions
33,561 answers
79,413 comments
31,029 users