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45 votes
45 votes

A disk has $8$ equidistant tracks. The diameters of the innermost and outermost tracks are $1$ cm and $8$ cm respectively. The innermost track has a storage capacity of $10$ MB.

What is the total amount of data that can be stored on the disk if it is used with a drive that rotates it with

  1. Constant Linear Velocity
  2.  Constant Angular Velocity?
  1. I. $80 \ \text{MB}$; II. $2040 \ \text{MB}$
  2. I. $2040 \ \text{MB}$; II $80  \ \text{MB}$
  3. I. $80 \ \text{MB}$; II. $360  \ \text{MB}$
  4. I. $360  \ \text{MB}$; II. $80  \ \text{MB}$
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3 Answers

Best answer
73 votes
73 votes
  • With Constant Linear Velocity, CLV, the density of bits is uniform from cylinder to cylinder. Because there are more sectors in outer cylinders, the disk spins slower when reading those cylinders, causing the rate of bits passing under the read-write head to remain constant. This is the approach used by modern CDs and DVDs.
  • With Constant Angular Velocity, CAV, the disk rotates at a constant angular speed, with the bit density decreasing on outer cylinders. ( These disks would have a constant number of sectors per track on all cylinders. )
  • CLV$=10+20+30+40+..80=360$
  • CAV$=10\times8 = 80$ so answer should be (D) 

Edit:- for CLV disk capacity

let track diameters like $1$cm, $2$cm... $8$cm.

As described that density is uniform.

So all tracks has equal storage density.

Track capacity$=$storage density $\times$ circumference$(2 \times pi \times r)$

For $1$st track. $10 \text{MB} = \text{density} \times 2 \times pi \times 1$

Density $= 10/pi.$  MB/cm

For $2$nd track capacity = density $\times$ circumference

$= (10/pi) \times(pi \times 2) \text{MB} = 20 \text{MB}$

Now each track capacity can be calculated and added for disk capacity

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18 votes
18 votes
Constant linear velocity :

Diameter of inner track = d = 1cm
Circumference of inner track :
= 2 * 3.14 * (d/2)
= 3.14 cm

Storage capacity = 10 MB (given)
Circumference of all equidistant tracks :
= 2 * 3.14 *(0.5 + 1 + 1.5 + 2 + 2.5 + 3+ 3.5 + 4)
= 113.14cm

Here, 3.14 cm holds 10 MB.
Therefore, 1 cm holds 3.18 MB.
113.14 cm holds 113.14 * 3.18 = 360 MB.
Total amount of data that can be stored on the disk = 360 MB

 
Constant angular velocity :

In case of CAV, the disk rotates at a constant angular speed.
Same rotation time is taken by all the tracks.
Total amount of data that can be stored on the disk = 8 * 10 = 80 MB
12 votes
12 votes
Constant linear velocity - RPM will be same, means bit density will be same So outer track has more data than inner

10+20+30....+80 = 360 MB

Constant Angular velocity - RPM will not be same, Density will not be same, Same amount of data per track

8*10 = 80 MB
Answer:

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