UGC NET CSE | October 2020 | Part 2 | Question: 91

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.

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

• A. $(50 K, 50000 K)$
• B. $(25 K, 25000 K)$
• C. $(25 K, 50000 K)$
• D. $(40 K, 36000 K)$

Answer 1

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

Answer 2

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 

Answer 3

tried to add complete questions of UGC NTA NET..

But only admins can do that..!!!!

 

So finally, being big fan of Sanjay Sharma Sir, putting the link for complete question paper and official answer key here: https://ugcntanetcomputerscience2020paper.blogspot.com/