Question

Question 7. A hard disk has the following parameters 8 surfaces 40,000 tracks per surface • 512 sectors per track 1024 bytes per sector Rotational speed 10,000 rpm a) What is the capacity of the drive? b) What is the rotational latency? c) Once a sector has been accessed, what is the rate at which data is read?

Answer 1

a)

Given,

Given,

Number of surfaces = 8

Number of tracks per surface = 40000

Number of sectors per track = 512

Number of bytes per sector = 1024 Bytes

Rotational speed = 10000 rpm

a)

Capacity of the drive = Number of surfaces * Number of tracks per surface * Number of sectors per track * Sector Size

= 8 * 40000 * 512 * 1024 Bytes

= 167772160000 Bytes

Hence, the capacity of the drive is 167772160000 bytes.

b)

Rotational Latency = ( ( 60 / Rotational speed ) * 1000 ) ms

= ( ( 60 / 10000 ) * 1000 ) ms

= 6 ms

Hence, the rotational latency is 6 ms.

c)

The rotational latency obtained is 6 ms.

Size of a track = 512 sectors

This means in 6 ms, one complete track is read.

In other words,

1 track takes 6 ms to be read.

512 sectors take 6 ms to be read.

1 sector takes ( 6 ms / 512 ) time to be read which equals to 0.011718 ms

Hence, the time taken to read data when 1 sector is accessed is 0.011718 ms.