Question

Consider a hard disk with a rotational speed of 1500015000 rpm. The time to move the read/write head from a track to its adjacent track is 11 millisecond. Initially, the head is on track 00. The number of sectors per track is 400400. The sector size is 10241024 bytes. It is necessary to transfer data from 1010 randomly located sectors in each of the following tracks in the order: 5,125, 12 and 77.

The total time for the data transfer (in milliseconds) from the hard disk is _________. (rounded off to one decimal place)

Official answer and solution
Official answer
77.3
From
Official GATE 2026 answer key

Solution

Sum the total seek time across tracks 0→5→12→70 \rightarrow 5 \rightarrow 12 \rightarrow 7, the average rotational latency for each of the 3030 random sectors, and the transfer time for 3030 sectors.

  1. Total Seek Time:
    • Track 0→50 \rightarrow 5: ∣5−0∣×1 ms=5 ms|5 - 0| \times 1\text{ ms} = 5\text{ ms}
    • Track 5→125 \rightarrow 12: ∣12−5∣×1 ms=7 ms|12 - 5| \times 1\text{ ms} = 7\text{ ms}
    • Track 12→712 \rightarrow 7: ∣7−12∣×1 ms=5 ms|7 - 12| \times 1\text{ ms} = 5\text{ ms}
    • Total seek time =5+7+5=17 ms= 5 + 7 + 5 = 17\text{ ms}.
  2. Total Rotational Latency:
    • Time for 1 full rotation =60 s15000=0.004 s=4 ms= \frac{60\text{ s}}{15000} = 0.004\text{ s} = 4\text{ ms}.
    • Average rotational latency per randomly located sector =4 ms2=2 ms= \frac{4\text{ ms}}{2} = 2\text{ ms}.
    • Across 33 tracks with 1010 random sectors per track (3030 random sectors total), total rotational latency =30×2 ms=60 ms= 30 \times 2\text{ ms} = 60\text{ ms}.
  3. Total Data Transfer Time:
    • Transfer time per sector =4 ms400 sectors=0.01 ms/sector= \frac{4\text{ ms}}{400\text{ sectors}} = 0.01\text{ ms/sector}.
    • For 3030 sectors: 30×0.01 ms=0.3 ms30 \times 0.01\text{ ms} = 0.3\text{ ms}.
  4. Total Time =17+60+0.3=77.3 ms= 17 + 60 + 0.3 = 77.3\text{ ms}.
Final answer

77.3

Common trap

Counting rotational latency only once per track (3 times) instead of once for each of the 30 randomly located sectors.

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper
GATE 2026 Computer Science and Information Technology - Session 1 (CS1)
Question
59 of 65
Section
Computer Science and Information Technology

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