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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 MB; II. 2040 MB
2. I. 2040 MB; II 80 MB
3. I. 80 MB; II. 360 MB
4. I. 360 MB; II. 80 MB
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What is track length?
What is meant by degree measure of a track?

• 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*8 = 80 so answer should be d

Edit:- for CLV disk capacity

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

As described that density is uniform.

So all tracks has equal storage density.

Track capacity=storage density* circumference(2*pi*r)

For 1st track. 10 MB = density* pi*1

Density = 10/pi.  MB/cm

For 2nd track capacity = density* circumference

= (10/pi) *(pi*2) MB = 20MB

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

edited by
getting the answr

but ny one more clearify why to add 10+20+..+80?
Great explanation :-)
Thanks Bro
@sachin bro,

Not able to understand the solution given in the pdf.
• 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. )

But as per the below document :

http://www.cse.chalmers.se/edu/year/2010/course/EDA092/SLIDES/12-disk_storage.pdf

Constant angular velocity (CAV). The rotational speed of the disk is constant. To use the platter in an efficient way, the outer tracks have more sectors than the inner tracks. Used in hard disks.

In that case, don't you think that the answer should change?

Also, since the diameter of outer tracks is more than that of inner tracks, outer tracks can be made to hold more data as compared to inner tracks.

So, to efficiently use the space, most hard disks divide the tracks into zones with 0th zone is starts from the outer edge of the platter and the last zone is the innermost tracks of the platter.

So, the data rate changes as the read write head changes moves from outer edge to inner edge.

More Reading on this : http://www.pcguide.com/ref/hdd/geom/tracksZBR-c.html

But disk with constant linear velocity, density of bits per track is constant and to get higher data rate the speed of the disk is increased.

So, in case of Constant Linear Velocity, I think all 8 equidistant tracks, all can store only 10MB of data.

So, in case of CLV data that can be stored should be 80MB.

And in case of CAV, it should be 360MB.

Please let me know if anyone agrees on this

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
How the RPM will be same for Constant Linear Velocity ??

Because if it is, Outer track will traverse a longer length per rotation as compared to the inner track. So they get to have different Linear lengths traversed for the same time period making Linear Velocity Not constant any longer.

If I am wrong somewhere correct me please !!
Sandeep u r right, I was wrong.

RPM will differ in CLV.
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