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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.

What is the capacity of the disk, in bytes ?

The correct answer is

500,000 K

Understanding Disk Capacity Calculation

The question asks us to determine the total storage capacity of a disk drive given its physical specifications. Disk capacity is calculated by multiplying the number of storage units at each level: sector size, sectors per track, tracks per surface, and the total number of surfaces.

Disk Drive Specifications

Let's list the given specifications of the disk drive:

  • Sector size: 512 bytes
  • Tracks per surface: 2000
  • Sectors per track: 50
  • Number of platters: five double-sided platters
  • Average seek time: 10 milliseconds (This is not used in capacity calculation)

Step-by-Step Disk Capacity Calculation

We will calculate the capacity layer by layer, starting from the smallest unit (sector) up to the entire disk.

Step 1: Calculate Capacity per Track

A track contains multiple sectors. The capacity of one track is the number of sectors in a track multiplied by the size of each sector.

\( \text{Capacity per track} = \text{Sectors per track} \times \text{Sector size} \)

\( \text{Capacity per track} = 50 \text{ sectors/track} \times 512 \text{ bytes/sector} \)

\( \text{Capacity per track} = 25,600 \text{ bytes} \)

Step 2: Calculate Capacity per Surface

Each surface of a platter contains multiple tracks. The capacity of one surface is the number of tracks per surface multiplied by the capacity of each track.

\( \text{Capacity per surface} = \text{Tracks per surface} \times \text{Capacity per track} \)

\( \text{Capacity per surface} = 2000 \text{ tracks/surface} \times 25,600 \text{ bytes/track} \)

\( \text{Capacity per surface} = 51,200,000 \text{ bytes} \)

Step 3: Determine Total Number of Surfaces

The disk has five double-sided platters. Each double-sided platter has two surfaces that can store data.

\( \text{Total surfaces} = \text{Number of platters} \times \text{Surfaces per platter} \)

\( \text{Total surfaces} = 5 \text{ platters} \times 2 \text{ surfaces/platter} \)

\( \text{Total surfaces} = 10 \text{ surfaces} \)

Step 4: Calculate Total Disk Capacity in Bytes

The total capacity of the disk is the capacity per surface multiplied by the total number of surfaces.

\( \text{Total capacity (bytes)} = \text{Capacity per surface} \times \text{Total surfaces} \)

\( \text{Total capacity (bytes)} = 51,200,000 \text{ bytes/surface} \times 10 \text{ surfaces} \)

\( \text{Total capacity (bytes)} = 512,000,000 \text{ bytes} \)

Step 5: Convert Total Capacity to Kilobytes (K)

The options are given in Kilobytes (K). In computing, K typically represents \( 2^{10} = 1024 \) bytes. Let's convert our total capacity using this standard conversion.

\( \text{Total capacity (K)} = \frac{\text{Total capacity (bytes)}}{1024 \text{ bytes/K}} \)

\( \text{Total capacity (K)} = \frac{512,000,000 \text{ bytes}}{1024 \text{ bytes/K}} \)

\( \text{Total capacity (K)} = 500,000 \text{ K} \)

This value matches one of the given options.

Component Calculation Result (bytes)
Sector Given 512
Track 50 sectors \(\times\) 512 bytes/sector 25,600
Surface 2000 tracks \(\times\) 25,600 bytes/track 51,200,000
Disk (Total) 10 surfaces \(\times\) 51,200,000 bytes/surface 512,000,000

Total Capacity Conversion Result (K)
512,000,000 bytes \(\div\) 1024 bytes/K 500,000 K

Therefore, the capacity of the disk is 500,000 K.

Revision Table: Disk Capacity

Term Description Used in Calculation?
Sector Size Smallest unit of data storage on a disk. Yes
Sectors per Track Number of sectors on a single track. Yes
Tracks per Surface Number of tracks on a single data-storing surface of a platter. Yes
Platter A rigid disk inside the drive; usually double-sided. Yes (to find total surfaces)
Surface One side of a platter that stores data. Double-sided platters have two surfaces. Yes
Average Seek Time Average time taken for the read/write head to move to the correct track. No (Relevant to performance, not capacity)
Capacity Total amount of data that can be stored on the disk. The value we calculated.

Additional Information: Hard Disk Drive Structure and Units

A typical hard disk drive (HDD) stores data magnetically on rapidly rotating platters. The platters are organized into tracks (concentric circles) and sectors (divisions of tracks). Multiple platters stacked together form a disk pack. Read/write heads access the data on the surfaces.

Data is read from or written to sectors. Multiple sectors make up a track, and multiple tracks are found on a single surface of a platter. A set of tracks at the same radius across all surfaces is called a cylinder.

Units used for capacity:

  • Byte (B): A standard unit of digital information, typically consisting of 8 bits.
  • Kilobyte (KB or K): Historically and commonly in computing contexts (like RAM or file sizes), 1 KB = \( 1024 \) bytes (\( 2^{10} \) bytes). In storage marketing (like disk size on packaging), 1 KB is sometimes used as \( 1000 \) bytes (\( 10^3 \) bytes). The problem context requires the \( 1024 \) byte conversion.
  • Megabyte (MB or M): Typically \( 1024 \times 1024 \) bytes (\( 2^{20} \) bytes), or sometimes \( 1000 \times 1000 \) bytes (\( 10^6 \) bytes).
  • Gigabyte (GB or G): Typically \( 1024^3 \) bytes (\( 2^{30} \) bytes), or sometimes \( 1000^3 \) bytes (\( 10^9 \) bytes).

It's important to be aware of the potential ambiguity between base-2 (powers of 1024) and base-10 (powers of 1000) prefixes, although in this problem, the options clearly point to the base-2 conversion for K.

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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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