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current track = 4 and sector is 25 thus to move to 36 track it is given that 1msec is required

thus now we reached 36 track 25 sector now moving forward to 153 sector

for 1 sector time will be 20ms/256 thus for moving 25 to 153 there are 128 sectors thus 128* 20/256=10ms

now for each sector it contains 0.5k data thus it will take 128 sectors for 64k data thus time will be  128*20/256 =10ms

thus total time will be 32ms+10ms+10ms = 52ms
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WE HAVE….

1)AVG ROTATIONAL LATENCY=1/2*(TIME TAKEN FOR ONE COMPLETE ROTATION)

3000 RPM IMPLIES THAT 3000 ROTATIONS IN 60 SEC(COZ 1 MIN=60 SEC)

SO, 1 COMPLETE ROTATION IS DONE IN = 60/3000 i.e 1/50 SEC

SO,AVG  ROTATIONAL LATENCY= ½ X 1/50 = 1/100 = 10 msec

NOW,

2)IN ONE COMPLETE ROTATION ONE ENTIRE TRACK IN READ i.e ALL THE SECTORS PRESENT IN THE TRACK IS READ...SO ACC TO THE QUESTION 256 SECTORS IN READ IN (1/50 ) SECONDS

SO 1 SECTOR IS READ IN = (1/50)/256  SEC  -------------------------------( 1 )

SEEK TIME= TIME TAKEN BY THE R/W HEADER TO MOVE TO THE DESIRED TRACK….

SO MOVING FROM TRACK NUMBER 4 TO TRACK NUMBER 36, TIME REQUIRED IS =32 X 1 = 32 msec

SO, SEEK TIME= 32 msec

 

3)NOW REPOSITIONING  THE HEADER FROM 25th SECTOR TO 153th SECTOR...NUMBER OF SECTORS WE HAVE TO TRAVERSE IS= 128

TIME TAKEN TO TRAVERSE 128 SECTORS(TO REACH 153th SECTOR) = (1/50)/256 X 128 = 1/100= 10msec……...USING  1

 

4)NOW WE HAVE TO FETCH 64 KB OF DATA...WHICH IS NOTHING BUT (2^6 X 2^10) BYTES OF DATA…..( 2 )

AND, WE HAVE SIZE OF EACH SECTOR = 512 BYTES = 2^9 BYTES OF DATA……...( 3 )

SO, NUMBER OF SECTORS WE HAVE TO READ(TRAVERSE) TO FETCH 64 KB OF DATA IS =  ( 2 ) / ( 3 )

WHICH RESULTS INTO = 128 SECTORS

AND TIME TAKEN TO TRAVERSE 128 SECTORS = 10 msec.

 

SO TOTAL TIME TAKEN = (32 + 10 + 10 + 10) msec = 62 msec

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