Comprehension

Comprehension: Questions 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 one track of data can be transferred per revolution, then what is the data transfer rate ? 

This question was previously asked in
UGC NET Computer Science (Paper 2) 2020 Official Paper
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  1. 2,850 KBytes/second
  2. 4,500 KBytes/second
  3. 5,700 KBytes/second
  4. 2,250 KBytes/second

Answer (Detailed Solution Below)

Option 4 : 2,250 KBytes/second
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UGC NET Paper 1: Held on 21st August 2024 Shift 1
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Detailed Solution

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The correct answer is option 4.

Key Points

The capacity of a track is=bytes / sector x sector / track= 512x50=25k

Since one track of data can be transferred per revolution,

Data transfer rate = tracksize÷rotational delay

if the disk platters rotate at 5400rpm, the time required for one complete rotation will be, 1/5400 x60= 0.011 seconds

Rotational delay=0.011

Data transfer rate=2250K÷ 0.011= 2,250 KBytes/second
∴ Hence the correct answer is 2,250 KBytes/second.

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