Given data
Seek time = 4ms (considering its average seek time)
Rotation Speed = 15,000 RPM
Number of Sector = 500
Size of each sector = 512 B
We have to read 2500 Sector which is approx 1.23 MB
We know Disk access time = Avg Seek time + Avg rotation latency + Transfer Time
15,000 Rotation require 60 Sec
1 Rotation we can do in 4 ms
So avg rotation latency = 2 ms
Case 1 : Data are store in continuous way
To read 2500 sector we need to read 5 track and in 1 rotation we are reading 1 track
So to read 5 track we need 5* 4 ms = 20 ms
So transfer time will be 20ms
Since data is store in continuous manner we only have to deal with Avg seek time and Avg rotation latency for first time
So disk access time = 4+2+20 ms = 26 ms
Case 2 : Data store in random way
As we know 1 track or 500 sector can be read in 4 ms
So 2500 sector we can read in 20 ms or 0.02 sec
Since data is store in random way every time we have to deal with seek time = 2500 * 4ms = 10,000 ms or 10 sec
Similarly every time we have to deal with rotation latency = 2500 * 2ms = 5,000 ms or 5 sec
So total disk access time = 0.02 sec + 10 sec + 5 sec = 15.02 sec