GATE IT 2005 | Question: 81-a

Question

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?

• A. I. $80 \ \text{MB}$; II. $2040 \ \text{MB}$
• B. I. $2040 \ \text{MB}$; II $80  \ \text{MB}$
• C. I. $80 \ \text{MB}$; II. $360  \ \text{MB}$
• D. I. $360  \ \text{MB}$; II. $80  \ \text{MB}$

Comments

Note: Linear Velocity = Angular Velocity × Radius (v = ωr)

Constant Linear Velocity (CLV)

• Linear velocity is constant.

• Since v = ωr, angular velocity decreases as the radius increases.

• Every track passes under the read/write head at the same linear speed, so bits are read at the same rate.

• Bit density is uniform across all tracks.

• Outer tracks have a larger circumference, so they contain more sectors than inner tracks.

• Hence, track capacity is directly proportional to the radius.

Constant Angular Velocity (CAV)

• Angular velocity is constant.

• Since v = ωr, linear velocity increases as the radius increases.

• Outer tracks move faster under the read/write head, so bits are read faster than on inner tracks.

• Bit density is non-uniform (lower on outer tracks).

• All tracks contain the same number of sectors.

• Hence, track capacity is the same for every track.

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@P0535_Yedidyah_Sagar please correct the table. outermost tracks read bits faster than inner tracks. i missed the ambiguity and didn't check the source. thanks @Honey badger for pointing it out.

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Sorry @JHighlight @Honey badger

I made a mistake, when writing the table. Corrected now

Answer 1

• 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$(\pi \times d)$

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

Density $= 10MB/\pi \, cm$

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

Comments

Thanks Bro

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• 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

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@Madhav107 in the calculation for density, there is a mistake , radius is put 1 cm while, diameter is 1 cm, correct me if i am wrong

Answer 2

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

Answer 3

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

Comments

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.

Your answer is correct but reasoning may not be !!

If I am wrong somewhere correct me please !!

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Sandeep u r right, I was wrong.

RPM will differ in CLV.

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"means bit density will be same"

how can we say that bit density will be same for all tracks in case of CLV

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Please make corrections in posted answer people are still reading it wrong and wasting time!!

Answer 4

In a disk system with 8 equidistant tracks where the innermost track diameter is 1 cm and the outermost is 8 cm, the spacing between consecutive tracks is exactly 1 cm.

This gives the diameters for the eight individual tracks as sequential integer values:

$$\text{Track Diameters} = \{1\text{ cm}, 2\text{ cm}, 3\text{ cm}, 4\text{ cm}, 5\text{ cm}, 6\text{ cm}, 7\text{ cm}, 8\text{ cm}\}$$

For Case I, Constant Linear Velocity means the data density per unit length is completely uniform across the entire surface of the disk.

Because the circumference of a circular track is directly proportional to its diameter, the capacity of any given track scales linearly with its size:

$$\text{Track Diameter} \longrightarrow \text{Track Capacity}$$

$$1\text{ cm (Innermost)} \longrightarrow 10\text{ MB}$$

Using this direct proportional mapping, we can list the capacities of all eight individual tracks based on their respective diameters:

$$2\text{ cm} \longrightarrow 20\text{ MB}$$

$$3\text{ cm} \longrightarrow 30\text{ MB}$$

$$4\text{ cm} \longrightarrow 40\text{ MB}$$

$$5\text{ cm} \longrightarrow 50\text{ MB}$$

$$6\text{ cm} \longrightarrow 60\text{ MB}$$

$$7\text{ cm} \longrightarrow 70\text{ MB}$$

$$8\text{ cm (Outermost)} \longrightarrow 80\text{ MB}$$

Summing these track capacities together gives the total storage capacity under the Constant Linear Velocity drive configuration:

$$\text{Total Capacity (CLV)} = 10 + 20 + 30 + 40 + 50 + 60 + 70 + 80 = 360\text{ MB}$$

For Case II, Constant Angular Velocity means the disk spins at a fixed rotational speed, forcing the read-write head to scan every track in the exact same amount of time.

To make this possible while spinning at a constant rate, every single track on the disk must hold the exact same number of sectors and data bits, regardless of how large its outer diameter is.

Since the maximum storage density is capped by the tight physical space of the innermost boundary, every track gets locked to the capacity of that innermost ring:

$$1\text{ track} \longrightarrow 10\text{ MB}$$

Multiplying this uniform track capacity by the total number of tracks available gives the total storage capacity under the Constant Angular Velocity configuration:

$$\text{Total Capacity (CAV)} = 8\text{ tracks} \times 10\text{ MB} = 80\text{ MB}$$

Correct Option: D