Given, No of sectors/track = 8. Size of a sector= 512 B. Rotational rate = 300 rpm. Arm is correctly positioned. Number of rotations required sector 0 under head = 0.5 rotation.
For positioning the arm under desired sector or initial sector ( sector 0) of the track, seek time+ rotational latency is consumed. Since the arm is correctly positioned i.e. it is on the desired track, seek time is 0 and hence it is ignored.
Time for 1 rotation = 1 min/300 = 60/300 sec = 200 ms.
Rotational latency = time to position the arm under desire or initial sector = 0.5*(time for 1 rotation)= 0.5*200 ms = 100 ms.
For reading all the 8 sectors, data transfer time is also considered. Why?
Because for reading data from disk, firstly data is brought from disk to disk controller and then from disk controller to main memory. Secondly, afer data is brought to main memory, data read is performed. Data transfer from disk to disk controller is limited by data transfer rate or bandwidth of disk. And data transfer from disk controller to main memory is limited by bus.
Here only disk to disk controller is considered as nothing is mentioned about other part.
Data transfer rate or bandwidth of disk = no of rotations in 1 sec* size of 1 track= (300/60)*8*512 B/sec = 5*8*512 B/sec
Data transfer time of 8 sectors = (data size transferred)/bandwidth = (8*512)/(5*8*512) sec = 200 ms.
Total time needed to read all 8 sectors = seek time+ rotational latency+ data transfer time= 0 + 100+ 200 ms = 300 ms