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