The Gateway to Computer Science Excellence
First time here? Checkout the FAQ!
x
+13 votes
4.3k views

Consider a disk pack with $16$ surfaces, $128$ tracks per surface and $256$ sectors per track. $512$ bytes of data are stored in a bit serial manner in a sector. The capacity of the disk pack and the number of bits required to specify a particular sector in the disk are respectively:

  1. $256$ Mbyte, $19$ bits
  2. $256$ Mbyte, $28$ bits
  3. $512$ Mbyte, $20$ bits
  4. $64$ Gbyte, $28$ bits
asked in Operating System by Veteran (59.6k points)
edited by | 4.3k views

3 Answers

+29 votes
Best answer

Answer is (A).

$16$ surfaces $= 4$ bits, $128$ tracks $= 7$ bits, $256$ sectors $= 8$ bits, sector size $512$ bytes $= 9$ bits

Capacity of disk $= 2^{4+7+8+9} = 2^{28} = 256 \ MB$

To specify a particular sector we do not need sector size, so bits required $= 4+7+8 = 19$

answered by Loyal (8.2k points)
edited by
+8

address split is-

              4

               7

           8

         9

0
How 2^28 is equal to 256MB.
+1 vote

Disk capacity = total number of surfaces * no. of tracks per surface * amount of data per track  

 16 surfaces (4 bit) * 128 tracks per surface (7 bit) *  256 sectors per track (8 bit) *512 bytes of data

256MB

and to specify a particular sector we do not need sector size, so bits required = 4+7+8 = 19

answered by Active (3.7k points)
–2 votes
answered by Boss (16.2k points)


Quick search syntax
tags tag:apple
author user:martin
title title:apple
content content:apple
exclude -tag:apple
force match +apple
views views:100
score score:10
answers answers:2
is accepted isaccepted:true
is closed isclosed:true

40,871 questions
47,531 answers
146,035 comments
62,297 users