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.

What is the capacity of the disk, in bytes ?

This question was previously asked in
UGC NET Computer Science (Paper 2) 2020 Official Paper
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  1. 25,000 K
  2. 500,000 K
  3. 250,000 K
  4. 50,000 K

Answer (Detailed Solution Below)

Option 2 : 500,000 K
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UGC NET Paper 1: Held on 21st August 2024 Shift 1
10.8 K Users
50 Questions 100 Marks 60 Mins

Detailed Solution

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 The correct answer is option 2.

Key Points

capacity of the disk = bytes/sector x sector/track x track/surface x surface/disk=512 x 50 x 2000 x 5 x 2

=500000K Bytes/disk

∴ Hence the correct answer is 500,000 K.

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