GATE CSE 2007 | Question: 11, ISRO2009-36, ISRO2016-21

Question

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:

• A. $256$ Mbyte, $19$ bits
• B. $256$ Mbyte, $28$ bits
• C. $512$ Mbyte, $20$ bits
• D. $64$ Gbyte, $28$ bits

Comments

• $16surface\times \dfrac{128tracks}{surface} \times \dfrac{256 sectors}{track}\times\dfrac{512B}{sector}=2^{28}B=256MB$

 

• $\\16surface\times \dfrac{128tracks}{surface} \times \dfrac{256 sectors}{track}=2^{19} sectors \\ \\ \therefore 19\ bits$

Answer 1

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 $= 4+7+8 = 19$

Comments

address split is-

4

7

8

9

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How 2^28 is equal to 256MB.

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(16 surfaces X 128 tracks X 256 sectors X 512 bytes)/(1024*1024) MB= 256MB

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@ilakkiya2000

2^28 B = (2^8 * 2^20) B

=> 256 MB(2^20 = M)

Answer 2

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

Answer 3

Number of surfaces =16
Tracks per surface=128
Sectors per track=256
Data which will store per sector=512 bytes
Capacity of the disk = 16 surfaces X 128 tracks X 256 sectors X 512 bytes = 256 Mbytes.
The number of bits required to access a sector =Total number of sectors.
= 16 surfaces X 128 tracks X 256 sectors
=2^4x2^7x2^8=2^19

Answer 4

correct answer is a) https://gateoverflow.in/1209/gate2007_11