GATE CSE 2013 | Question: 29

Question

Consider a hard disk with $16$ recording surfaces $(0-15)$ having $16384$ cylinders $(0-16383)$ and each cylinder contains $64$ sectors $(0-63)$. Data storage capacity in each sector is $512$ bytes. Data are organized cylinder-wise and the addressing format is $\langle \text{cylinder no.}, \text{surface no.}, \text{sector no.} \rangle$ . A file of size $42797$ KB is stored in the disk and the starting disk location of the file is $\langle 1200, 9, 40\rangle$. What is the cylinder number of the last sector of the file, if it is stored in a contiguous manner?

• A. $1281$
• B. $1282$
• C. $1283$
• D. $1284$

Comments

@priyanshu.jha 

 I hope you know difference between a cylinder and a track..

I like your confidence here 
cause ...gatecse is the creator of this site

and your link was helpful for me thanks!!

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Each cylinder contains 64 sectors ⟹ at each surface within the cylinder there are 64 sectors

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 Important point that should also be noted here

 

Answer 1

Data are organized cylinder-wise

It means sectors of a full cylinder will be recorded before going to the next cylinder.

Now,sectors of a full cylinder means set of tracks(from different surface) which is situated in same R/W head positioned.16 recording surface means there are 16 R/W head which is recording simultaneously from 16 surfaces.(see the basic diagram of a disk).

So, a cylinder size = (set of tracks)*sector/track = (16 * 64)

suppose R/w head is starting from inner-most level cylinder.then after recording (16*64) sectors,R/w head will be moved into 2nd innermost level cylinder,again after (16*64) sectors R/w head will be moved towards 3rd inner-most cylinder & so on.

So, 42797 KB = 85,594 sectors.

No. of cylinders we have to cross for the file = $\frac{85,594}{16*64}$ = 83.58

clearly the last sector of the file is situated in (1200 + 84) = 1284 cylinder.

Answer 2

As per the defination of cylinder the given statement is clearly incorrect with respect to answers, but since options are there, may be this question belongs to reverse engineering more than disk platter 🥲

Ans is (d) as per best answer