UGC NET CSE | October 2020 | Part 2 | Question: 92

Question

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.

What is the capacity of the disk, in bytes?

• A. $25,000 K$
• B. $500,000 K$
• C. $250,000 K$
• D. $50,000 K$

Answer 1

2*no. of platter*no. of tracks on each platter*no. of sector on each track*size of each sector

is capacity of disk

2*5*2000*50*512 bytes

=500000Kbytes

Answer 2

Capacity for 1 platter = Number of track/surface * Number of sector/track * sector size

$= 2000 * 50 * 512 = 51200000 $Bytes

Since it has two surface, it becomes $102400000$ Bytes

For 5 platters, i.e capacity of disk $= 5*102400000 = 512000000$ Bytes  $= 500000$ KB

Option B) is correct