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Question

Comprehension: Questions concern a disk with a sector size of 512 bytes, 2000 tracks per surface, 50 sectors per track, five double-sided platters, and average seek time of 10 milliseconds.

If T is the capacity of a track in bytes, and S is the capacity of each surface in byte, then (T, S) = _____.

The correct answer is

(25 K, 50000 K)

Calculating Disk Track and Surface Capacity

This problem involves calculating the storage capacity of a single track and a single surface of a hard disk drive based on the given specifications. We are provided with the sector size, the number of sectors per track, and the number of tracks per surface.

Given Disk Specifications

  • Sector size: 512 bytes
  • Sectors per track: 50
  • Tracks per surface: 2000
  • Number of double-sided platters: 5 (This implies 10 surfaces, but is not needed for calculating single track or surface capacity)
  • Average seek time: 10 milliseconds (Not needed for capacity calculation)

Calculate Track Capacity (T)

The capacity of a track is the total size of all sectors it contains. We can calculate this by multiplying the number of sectors per track by the size of each sector.

Formula:

\[ T = \text{Number of sectors per track} \times \text{Sector size} \]

Substituting the given values:

\[ T = 50 \text{ sectors/track} \times 512 \text{ bytes/sector} \] \[ T = 25600 \text{ bytes} \]

The question asks for the capacity in Kilobytes (K). We know that 1 K = 1024 bytes. So, we convert the capacity from bytes to K:

\[ T = \frac{25600 \text{ bytes}}{1024 \text{ bytes/K}} \] \[ T = 25 \text{ K} \]

So, the capacity of a track is 25 K.

Calculate Surface Capacity (S)

The capacity of a surface is the total size of all tracks it contains. We can calculate this by multiplying the number of tracks per surface by the capacity of each track.

Formula:

\[ S = \text{Number of tracks per surface} \times \text{Track capacity (T)} \]

Substituting the given values:

\[ S = 2000 \text{ tracks/surface} \times 25600 \text{ bytes/track} \] \[ S = 51200000 \text{ bytes} \]

Again, the question asks for the capacity in Kilobytes (K). We convert the capacity from bytes to K:

\[ S = \frac{51200000 \text{ bytes}}{1024 \text{ bytes/K}} \] \[ S = 50000 \text{ K} \]

So, the capacity of a surface is 50000 K.

Summary of Capacities

Based on our calculations:

  • Track Capacity (T) = 25 K
  • Surface Capacity (S) = 50000 K

Therefore, (T, S) = (25 K, 50000 K).

Comparison with Options

Let's compare our calculated values with the provided options:

Option (T, S) Matches Calculation?
1 (50 K, 50000 K) No (T is incorrect)
2 (25 K, 25000 K) No (S is incorrect)
3 (25 K, 50000 K) Yes
4 (40 K, 36000 K) No (Both T and S are incorrect)

The calculated capacities (T, S) = (25 K, 50000 K) match Option 3.

Revision Table: Disk Capacity Calculations

Component Formula Calculation Capacity
Sector Given 512 bytes 512 bytes
Track (T) Sectors per track $\times$ Sector size \(50 \times 512\) bytes \( = 25600\) bytes
\(\frac{25600}{1024}\) K \( = 25\) K
25 K
Surface (S) Tracks per surface $\times$ Track capacity (T) \(2000 \times 25600\) bytes \( = 51200000\) bytes
\(\frac{51200000}{1024}\) K \( = 50000\) K
50000 K

Additional Information: Hard Disk Structure and Capacity

A hard disk drive stores data magnetically on rotating platters. Understanding its structure is key to calculating storage capacity.

  • Platter: A circular disk, typically made of glass or ceramic, coated with magnetic material. Data is stored on both the top and bottom surfaces.
  • Surface: Each side of a platter is a surface. A drive with 5 double-sided platters has \(5 \times 2 = 10\) surfaces.
  • Track: Data is recorded in concentric circles on each surface, similar to grooves on a vinyl record. Each circle is called a track. All tracks at the same radial distance across all surfaces form a cylinder.
  • Sector: Each track is divided into smaller units called sectors. This is the smallest addressable unit of storage on the disk. Data is read from or written to sectors.
  • Cluster: Operating systems group sectors into clusters (or allocation units) for managing storage space. A cluster is the minimum amount of disk space used to hold a file.
  • Total Capacity: The total storage capacity of a hard disk can be calculated as: \[ \text{Total Capacity} = \text{Number of cylinders (tracks per surface)} \times \text{Number of surfaces} \times \text{Sectors per track} \times \text{Sector size} \] In this case, Total Capacity \( = 2000 \times 10 \times 50 \times 512\) bytes \( = 512,000,000\) bytes. In K: \(\frac{512,000,000}{1024}\) K \( = 500,000\) K. In MB (1 MB = 1024 K): \(\frac{500,000}{1024}\) MB \(\approx 488.28\) MB. In GB (1 GB = 1024 MB): \(\frac{488.28}{1024}\) GB \(\approx 0.477\) GB. Note that manufacturers often use base 10 (1 KB = 1000 bytes) for capacity which would give higher numbers. Here, we used base 2 (1 K = 1024 bytes).

Seek time is the time it takes for the read/write head to move to the correct track, which is related to disk performance, not capacity.

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Important Questions from Secondary Memory

  1. Which of the following can be said about primary storage in comparison with secondary storage?

  2. __________ is the time taken to locate the disk arm to a specified track for data read/write.

  3. For a magnetic disk with concentric circular tracks, the seek latency is not linearly proportional to the seek distance due to

  4. If one track of data can be transferred per revolution, then what is the data transfer rate ?

  5. Given below are two statements:

    Statement I: The disk has a total number of 2000 cylinders.

    Statement II: 51200 bytes is not a valid block size for the disk.

    In the light of the above statements, choose the correct answer from the options given below:

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