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GATE 2026 CS1 Q.59: solution and answer

GATE 2026 · CS1Q.59 · 2 marks · NAT

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)

See the question in the paper

Answer and solution

ANSWER

The total data transfer time is 77.3 ms.

Official answer: 77.3Source: Official answer key, GATE 2026 CS1 (IIT Guwahati)

  1. One revolution takes .

  2. The sectors are randomly located, so the average rotational latency before each sector is half a revolution: .

  3. A track holds 400 sectors and one revolution passes all of them, so one sector is transferred in . The sector size (1024 bytes) is not needed.

  4. Time per sector . There are sectors, so the rotational latency and transfer time total .

  5. The head starts on track 0 and visits tracks 5, 12 and 7 in that order, moving tracks. At 1 ms per track, the seek time is .

  6. Total time .

KEY IDEA

Disk access time is seek time plus rotational latency plus transfer time. Random sectors each cost half a revolution of latency, and the seek is the total track distance travelled in the given visiting order.

WATCH OUT FOR

Counting the seeks as 5 + 12 + 7 = 24 tracks, as if the head returned to track 0 each time. The head moves from track 0 to 5, then 5 to 12, then 12 to 7, which is 17 tracks. Another slip is to add a full revolution of latency per sector, or to count the latency only once per track.

From the GATE 2026 CS1 paper. Every question of this paper · Download the full paper · All 6 File systems & disks questions, with notes

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