If T is the capacity of a track in bytes, and S is the capacity of each surface in byte, then (T, S) = _____.
(25 K, 50000 K)
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.
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.
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.
Based on our calculations:
Therefore, (T, S) = (25 K, 50000 K).
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.
| 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 |
A hard disk drive stores data magnetically on rotating platters. Understanding its structure is key to calculating storage capacity.
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.
Which of the following can be said about primary storage in comparison with secondary storage?
__________ is the time taken to locate the disk arm to a specified track for data read/write.
For a magnetic disk with concentric circular tracks, the seek latency is not linearly proportional to the seek distance due to
If one track of data can be transferred per revolution, then what is the data transfer rate ?
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: