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A manufacturer wishes to design a hard disk with a capacity of 30 GB or more (using the
standard definition of 1GB = 2^30 bytes). If the technology used to manufacture the disks allows
1024-byte sectors, 2048 sectors/track, and 4096 tracks/platter, how many platters are required?
(Assume a fixed number of sectors per track.)

1)3
2)5
3)4
4)2
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Let assume we have $x$ number of platters.

Capacity of the hard disk =$30GB$=$30*2^{30} B$

Now the disk has $4096$ tracks/platters.

we have x number of platters so no of tracks we have =$x*4096$

It has $2048$ sector/track.

we have $x*4096$ number of tracks so number of  sector we have =$x*4096*2048$

Each sector has capacity=1024B

so total capacity of the disk =$x*4096*2048*1024 $ Byte

so it is same as $30*2^{30} B$.

By equating them,

$x*4096*2048*1024=30*2^{30}$

$x=\frac{30}{8}=3.75$

so  we need 4 platters for this Disk.

Correct answer is option 3.


If someone has doubt why we need 4 platters why not 3 platters if we have used 3 platters then the capacity of the disk =$3*1024*2048*4096$=24GB but in the question we have given 30 GB as the disk capacity so using 3 platters we cannot get 30GB capacity.

 

 

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