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The question given below: 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.

If $T$ is the capacity of a track in bytes, and $S$ is the capacity of each surface in bytes, then $(T,S)=$ _______

  1. $(50 K, 50000 K)$
  2. $(25 K, 25000 K)$
  3. $(25 K, 50000 K)$
  4. $(40 K, 36000 K)$
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Capacity of track = Number of sector/track * sector size = $50*512 B= 25600 B= 25K $

Capacity of surface = Capacity of track * number of tracks= $25K * 2000 = 50000K $

Option C) is correct

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5 platters ,each platter has 2000 tracks each track 50 sectors with size 512 bytes

  so T=50x512 /1024 =25 K

    S=2000x25K =50000K

Hence option 3) is right ans 

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