A 2.5 inch diameter disk has 8 platters with each platter having two data recording surfaces, each platter on disk has 4084 tracks, each track has 400 sectors and one sector can store 1 MB of data. Calculate the storage capacity of this disk in Bytes. If this disk has a seek time of 2 milli-seconds and rotates at the speed of 6000 rpm, find the Access time for the disk. Make suitable assumptions, if any.

Let's break the problem into parts for clarity.

### Part 1: **Storage Capacity Calculation**
Given:
- Diameter of the disk = 2.5 inches (not directly needed for capacity calculation)
- 8 platters, and each platter has two data recording surfaces.
- Each platter has 4084 tracks.
- Each track has 400 sectors.
- Each sector can store 1 MB of data.

We can calculate the storage capacity step by step:

1. **Number of surfaces** = \(8 \text{ platters} \times 2 \text{ surfaces per platter} = 16 \text{ surfaces}\)
2. **Number of tracks per surface** = 4084 tracks
3. **Number of sectors per track** = 400 sectors
4. **Size of each sector** = 1 MB = \(1 \times 10^6\) bytes

Now, the total storage capacity \(C\) in bytes is calculated as:

\[
C = (\text{Number of surfaces}) \times (\text{Number of tracks per surface}) \times (\text{Number of sectors per track}) \times (\text{Size of each sector})
\]

Substituting the values:

\[
C = 16 \times 4084 \times 400 \times 10^6 \, \text{bytes}
\]

Let's calculate this.

\[
C = 16 \times 4084 \times 400 \times 10^6 = 26,937,600 \times 10^6 \, \text{bytes}
\]

So, the total storage capacity is:

\[
C = 26,937,600,000,000 \, \text{bytes} \quad \text{or} \quad 26.94 \, \text{TB}
\]

### Part 2: **Access Time Calculation**

The access time \(T_{\text{access}}\) consists of two components:
1. **Seek Time**: The time to move the read/write head to the appropriate track.
   - Given seek time \(T_{\text{seek}} = 2 \, \text{milliseconds} = 2 \times 10^{-3} \, \text{seconds}\)
   
2. **Rotational Latency**: The average time it takes for the desired sector to rotate under the read/write head.
   - The disk rotates at 6000 revolutions per minute (rpm).
   - To find the time per revolution:
     \[
     \text{Time per revolution} = \frac{60 \, \text{seconds}}{6000 \, \text{rpm}} = 0.01 \, \text{seconds per revolution}
     \]
   - The **average rotational latency** is half of the time per revolution:
     \[
     T_{\text{rotation}} = \frac{0.01}{2} = 0.005 \, \text{seconds} = 5 \, \text{milliseconds}
     \]

Now, the total access time \(T_{\text{access}}\) is the sum of the seek time and the average rotational latency:

\[
T_{\text{access}} = T_{\text{seek}} + T_{\text{rotation}} = 2 \, \text{ms} + 5 \, \text{ms} = 7 \, \text{milliseconds}
\]

### Final Answers:
1. **Storage Capacity** = 26.94 TB
2. **Access Time** = 7 milliseconds

Would you like more details or have any questions?

#### Related Questions:
1. How does the seek time affect the overall disk performance?
2. What is the role of rotational speed (RPM) in access time calculation?
3. Can rotational latency be reduced, and how?
4. What are other factors influencing disk performance besides seek time and rotational latency?
5. How do modern solid-state drives (SSDs) compare to this type of disk in terms of access time?

#### Tip:
Increasing the RPM of a disk reduces rotational latency, significantly improving the access time for random reads.

Learn how to calculate the storage capacity and access time of an 8-platter disk with a 2.5 inch diameter. Understand key concepts like seek time, rotational latency, and access time calculation, with detailed steps and formulas provided. Suitable for college-level computer science and engineering students.

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