Consider a disk with an average seek time of 4 ms, rotational delay of 2 ms, a rotation speed of 15000 r.p.m and 512-byte sectors with 500 sectors per track. A file occupies all of the sectors on 5 adjacent tracks. After reading the first track, if remaining tracks can be read with no seek time, then the time required in the sequential organization to transfer the file will be nearly

Concept:Average total TIME is defined as the SUM of:Average seek + Average Rotational Delay + Transfer timeLet us first read the first TRACK:Average seek time = 4 msAverage Rotational Delay = 2 msReading 500 sectors per track is also givenThe rotational speed is 15000 rpm, i.e.The time taken in 15000 revolution = 1 MIN = 60 secThe time taken in 1 revolution will be:\(\frac{1}{{15000}} \times 60\;sec\) Now, Reading 500 sectors = Time for 1 Revolution = 4 msecThe above is the complete track reading time.Hence the time required to read the first track will be:4 ms(average seek) + 2 ms (average rotational delay) + 4 ms (Reading 500 sector) = 10 msecNow, it is given that sequential organization is followed and no seek time is required for further reading of the remaining 4 tracks.But the remaining 4 tracks will have a rotational delay and reading time of tracks, i.e.= 4 × (2 + 4) msec4 = No. of remaining tracks2 ms = rotational delay4 ms = reading time of 500 sectors or 1 track= 4 × 6 = 24 sec∴ The total time will be:= 10 + 24 = 34 msec = 0.034 sec