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If the overhead for formatting a disk is $96$ bytes for a $4000$ byte sector,

1. Compute the unformatted capacity of the disk for the following parameters:
• Number of surfaces: $8$
• Outer diameter of the disk: $12$ cm
• Inner diameter of the disk: $4$ cm
• Inner track space: $0.1$ mm
• Number of sectors per track: $20$
2. If the disk in (A) is rotating at $360$ rpm, determine the effective data transfer rate which is defined as the number of bytes transferred per second between disk and memory.

@arjun sir,  I think there should be 401 tracks , as there are 400 gaps and the number of gaps between n+1 tracks is n ( 1 gap requires 2 tracks)
edited

A. Unformatted capacity of the disk:

#of tracks on the surface: $\frac{Storage \ distance}{Inter \ track\ space}$=$\frac{4cm}{0.1mm}$=$\frac{40mm}{0.1mm}=400$

Capacity:  $8$ surface $\times$ $\frac{400\ tracks}{surface}$ $\times$ $\frac{20\ sector}{track}$ $\times$ $\frac{4000\ Bytes}{sector}=8\times400\times20\times 4000$ Bytes $=256MB$

B. Data transfer rate:

$\\ 1R\rightarrow \dfrac{1}{6}sec\\ \\ \dfrac{1}{6}sec\rightarrow 78080B\\ \\ 1sec\rightarrow 6 \times 78080 B\\ \\ means\ 468480 \ B/sec\\ or\ 468.48\ KB/sec$

(Note: 20 sectors $\times$4000 - 20 sectors$\times$96=78080)

While calculating capacity why are we taking M as 10^6, it should be 2^20 right?

For transfer rate 10^6 is fine.

For (A) part :

No of track $=$ Recording width/ inner space between track

Recording width $=$ (Outer Diameter $-$ Inner Diameter )$/ 2= (12-4)/2= 4 \ cm$

Therefore no. of track $= 4 \ \text{cm} / 0.1 \ \text{mm} = 400$ track

Since they have ask capacity of unformatted disk , so no $96$ bytes in $4000$ bytes would be wasted for non data purpose

Whole $4000$ is used

So, total capacity $= 400 \times 8 \times 20 \times 4000 = 256 \times 10^6 \ \text{Bytes} = \mathbf{256 \ MB}$

For (B) part :

Its is given $360$ rotations in $60$ seconds

That is $360$ rotations $= 60$  sec

Therefore, $1$ rotations  will take $(1/6)$ sec

In $(1/6)$ sec - we can read one track $= 20 \times (4000-96) \ B = 20 \times 3904 \ B$

Then, in $1$ sec it will be $= 20 \times 3904 \times 6$ bytes $=$ Data transfer rate $= \mathbf{ 468.480 \ KBps}$  ( when we consider  $1$ Read/Write Head for all surface) .

If we consider 1 Read/Write Heads per surface ( which is default approach ), then number of surfaces = 8

Data transfer rate $= ( 468.480 \times 8 ) \ KBps = 3747.84 \ KBps$

But for our convenience we consider only $1$ surface, it reads from one surface at a time. As data transfer rate is measured wrt a single surface only .

Hence, for part B, the correct answer is $\mathbf{468.480 \ KBps}$.

But for our convenience we consider only 1 surface , it reads from one surface at a time. As data transfer rate is measured wrt a single surface only .

I'm not able to understand why data transfer is considered wrt to a single surface only??

If we consider 1r/w head per surface(by default) then we should have to consider no. of surfaces in data rate calculation also...why not?

i think ans given by spriti1991 is right .... but i want to explain  about formatting of disk
-------- When a disk is formatted, the formatting program creates ID(identification data) areas before and after each sector’s data
that the disk controller uses for sector numbering and for identifying the start and end of each sector.
These areas precede and follow each sector’s data area and consume some of the disk’s total storage
capacity. This accounts for the difference between a disk’s unformatted and formatted capacities.
---disk capacity(unformatted)=no of surfaces* no of tracks on each surface*no of sectors per track*bytes per sector
---disk capacity(formatted)=no of surfaces* no of tracks on each surface*no of sectors per track*(bytes per sector - format overhead)
Now here our main task is to calculate no of tracks on each surface.
{*** we cant also use (outer diameter -inner diameter ) but in whis way we are calculating no of tracks twice so divide by 2 is reauired here}

now tracks on a surface is like (inter/inner track distance)-->(track recording width)-->(inter/inner track distance)-->(track recording width)-->.........(inter/inner track distance)-->(track recording width) till outer track recording width.
note ---  here the term track recording width represents width seperated by inner/inter track distance
now if w=total recording width on surface =4 cm =40 mm
x= inter/inner track distance=0.1 mm
y= track recording width  ;; that is not given in our problem and very small in comparison to (inter/inner track distance) neglect it
now no of tracks =(recording width / ( inter/inner track distance +  track recording width))
=(w/x+y) ≈(w/x)=40/0.1=400 tracks
------for complete solution see ans given above by spriti1991.

:( I am reading it as diameter and thinking it as radius. anyway thanks :)
it happens