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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

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.

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.

Hii @Kabir5454

Platter can store data on both surfaces right??

In practice Platters store data on both surfaces but in theoretically there may be instances where data was stored only one surface then that can be implementation dependent as well.

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