Register

GATE IT 2005 | Question: 81-b

• edited by
25,291 views
99 99 votes

A disk has $8$ equidistant tracks. The diameters of the innermost and outermost tracks are $1$ cm and $8$ cm respectively. The innermost track has a storage capacity of $10$ MB.

If the disk has $20$ sectors per track and is currently at the end of the $5^{th}$ sector of the inner-most track and the head can move at a speed of $10$ meters/sec and it is rotating at constant angular velocity of $6000$ RPM, how much time will it take to read $1$ MB contiguous data starting from the sector $4$ of the outer-most track?

  1. $13.5 \ ms$
  2. $10 \ ms$
  3. $9.5 \ ms$
  4. $20 \ ms$

7 Answers

Best answer
150 150 votes

Total Time $=$ Seek $+$ Rotation $+$ Transfer.

Seek Time :

Current Track $1$

Destination Track $8$

Distance Required to travel $= 4-0.5= 3.5 \ Cm$

Time required $= 10 \ m/s == 1\ Cm/ms == \mathbf{3.5 \ ms}$ [ Time= Distance / Speed ]

Rotation Time:

$6000$ RPM in $60$ sec

$100$ RPS in $1$ sec

$1$ Revolution in $10$ ms

$1$ Revolution $=$ Covering entire Track

$1$ Track $= 20$ sector

$1$ sector required $= 10/20 = 0.5 \ ms$

Disk is constantly Rotating so when head moved from inner most track to outer most track total movement of disk $= (3.5/0.5) = 7 $ sectors

Which means that when disk reached outer most track head was at end of $12^{ th}$ sector

Total Rotational Delay $=$ Time required to go from end of $12$ to end of $3 = 11$ sectors

$1$ sector $= 0.5 \ ms$ so $11$ sector $\mathbf{= 5.5 ms}$

Transfer Time

Total Data in Outer most track $= 10 \ MB$

Data in single Sector $= 10 \ MB/20 = 0.5 \ MB$

Data required to read $= 1 \ MB = 2$ sector

Time required to read data $= 2 \times 0.5 = \mathbf{1ms}$

Total Time = Seek + Rotation + Transfer $\mathbf{ = 3.5ms + 5.5ms +1ms = 10\  ms}$

Correct Answer: $B$

• edited by
60 60 votes

May be it will help :

17 17 votes

Radius of inner track is 0.5 cm (where the head is standing ) and the radius of outermost track is 4 cm.

So the header has to seek (4-0.5)=3.5 cm. It will take (3.5x1000/1000)=3.5 ms.

now angular velocity is constant and header is now end of 5th sector . To start from front of 4th sector it must rotate upto 18 sector.   

6000 rotation in 60000ms

1 rotation in 10 ms.( same time to traverse 20 sector)

So to traverse sector 18 it takes 9 ms.

in 10 ms 10 MB data is read.

1 MB can be read in 1 ms.

So total time is =(1+9+3.5) ms =13.5ms  

7 7 votes

$Seek\ time=\frac{distance\ from\ innermost\ track\ to\ outermost\ track}{10m/sec}$

$=\frac{3.5cm\times sec}{10\times 100cm}=$ $3.5msec$

$Rotation\ time:$

$60sec\leftarrow6000\ R$

$?\leftarrow 1R$

$?=10msec$

$1R=20\ sectors$

$10msec\leftarrow20\ sectors$

$?\leftarrow 1\ sector$

$?=0.5msec$

When you are coming from the innermost track to the outermost track, your sectors are also changing in parallel.

Means

$1^{st}\ track-5^{th}\ sector$

$2^{nd}\ track-6^{th}\ sector$

$3^{rd}\ track-7^{th}\ sector$

.

.

$8^{th}\ track-12^{th}\ sector$

Now you need to go to the 4th sector

$12-13-14-15-16-17-18-19-20-1-2-3$

$Time- taken=11 sectors \times 0.5msec=$ $5.5msec$

$Transfer\ time:$

$Track\ consist\ of\ 10MB\ and\ 20\ sectors$

$1\ sector=0.5MB$

$We\ need\ to\ read\ 1MB\ data\ means\ 2\ sectors$

$Time-taken=2\times 0.5msec=$ $1msec$

$Ans: 10msec$

• edited by
1 1 vote

According to the key of all new gate answers, to move from one track to another, we count the seek of intermediate tracks, 
Just watch this 2007 question: https://gateoverflow.in/3479/gate-it-2007-question-44-isro2015-34?show=55144#a55144
To traverse all 500 tracks, in worst case we consider the seek of 499 intermediate tracks
So here it would be better to calculate seek delay of 7 intermediate tracks
So out of all possible answers, only 9.5ms survives

Anyway it is very unfair to ask question which differ by 0.5 ms and no concrete proof to challenge an answer as correct, these are better as NAT’s with a range of answers, as in any case there is no logic to get a pin point accuracy in 0.5 ms of disk access irl, as there are many mechanical delays and other delays
 

Page:
Voting & Accepting an answer helps improve the community
Answer:
← Previous · GATE IT 2005 | Quest...Next → · GATE IT 2005 | Quest...
← Previous in category · GATE IT 2005 | Quest...Next in category → · CPU Scheduling
Position:
Show:

Related questions

54 54 votes
4 answers 4 answers
17.5k
17.5k views
Ishrat Jahan asked Nov 3, 2014
17,522 views
GATE IT 2005 | Question: 81-a
A disk has $8$ equidistant tracks. The diameters of the innermost and outermost tracks are $1$ cm and $8$ cm respectively. The innermost track has a storage capacity of $...
49 49 votes
3 answers 3 answers
11.1k
11.1k views
Ishrat Jahan asked Nov 3, 2014
11,051 views
GATE IT 2005 | Question: 63
In a computer system, four files of size $11050$ bytes, $4990$ bytes, $5170$ bytes and $12640$ bytes need to be stored. For storing these files on disk, we can use either...
55 55 votes
7 answers 7 answers
15.2k
15.2k views
Ishrat Jahan asked Nov 3, 2014
15,178 views
GATE IT 2005 | Question: 62
Two shared resources $R_1$ and $R_2$ are used by processes $P_1$ and $P_2$. Each process has a certain priority for accessing each resource. Let $T_{ij}$ denote the prior...
31 31 votes
3 answers 3 answers
9.1k
9.1k views
Ishrat Jahan asked Nov 3, 2014
9,137 views
GATE IT 2005 | Question: 60
We wish to schedule three processes $P1$, $P2$ and $P3$ on a uniprocessor system. The priorities, CPU time requirements and arrival times of the processes are as shown be...