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69 votes
69 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. $20 \ ms$
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5 Answers

Best answer
124 votes
124 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: $B$

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42 votes
42 votes

May be it will help :

17 votes
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  

5 votes
5 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

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

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


$Transfer\ time:$

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

$1\ sector=0.5MB$

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

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

$Ans: 10msec$

Answer:

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