GATE 2026 CS (CS1) – Question 59
Consider a hard disk with a rotational speed of 15000 rpm. The time to move the read/write head from a track to its adjacent track is 1 millisecond. Initially, the head is on track 0. The number of sectors per track is 400. The sector size is 1024 bytes. It is necessary to transfer data from 10 randomly located sectors in each of the following tracks in the order: 5, 12 and 7. The total time for the data transfer (in milliseconds) from the hard disk is _________. (rounded off to one decimal place)
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Correct answer: 35
Explanation
1. **Rotation parameters**:
- Rotational speed = $15000\text{ rpm} = \frac{15000}{60} = 250\text{ rev/sec}$.
- One full rotation time: $T_{\text{rot}} = \frac{1}{250}\text{ s} = 4\text{ ms}$.
- Average rotational latency: $T_{\text{lat}} = \frac{T_{\text{rot}}}{2} = 2\text{ ms}$.
2. **Seek time calculation**:
- Track 0 to Track 5: $|5 - 0| \times 1\text{ ms} = 5\text{ ms}$.
- Track 5 to Track 12: $|12 - 5| \times 1\text{ ms} = 7\text{ ms}$.
- Track 12 to Track 7: $|7 - 12| \times 1\text{ ms} = 5\text{ ms}$.
- Total seek time = $5 + 7 + 5 = 17\text{ ms}$.
3. **Track data transfer time**:
On each track, 10 sectors distributed randomly across the track are read. In modern disk scheduling with track buffering, after the initial average rotational latency of $2\text{ ms}$ to reach the first sector, all sectors on that track are read within at most one full revolution ($4\text{ ms}$).
Thus, time spent on each track = $T_{\text{lat}} + T_{\text{rot}} = 2\text{ ms} + 4\text{ ms} = 6\text{ ms}$.
For 3 tracks: $3 \times 6\text{ ms} = 18\text{ ms}$.
4. **Total time**:
$$\text{Total time} = \text{Total seek time} + \text{Total on-track time} = 17\text{ ms} + 18\text{ ms} = 35.0\text{ ms}$$
The correct answer is 35 (or 35.0).