Disk rotation speed= 2400 rpm
Track storage capacity = 62,500
Average Latency = p
Data Transfer rate = q
Data transfer rate is nothing but how much data can be read in one second
WE have 2400 rotation in 1 minute
2400 rotation = 60 *103 ms
therefore 1 rotation = x ms, where,
x= 60 *103 / 2400
therefore we get x=25 ms
Hence 25 mill sec is required to read one complete track (1 rotation )
In 25 m sec, we read track capacity = 62,500 bits
therefore in 1 sec we can read (62,500 bits )*1000/25 = 2.5 Mbits
i.e., Data transfer rate, Q = 2.5 Mbits/ sec
Now come The average Latency
Average latency is nothing but 0.5 * rotation delay
We have 2400 rotations in 1 minute
2400 rotation = 60 *103 ms
therefore 1 rotation = x ms, where,
x= 60 *103 / 2400
therefore we get x = 25 ms
And Rotation delay = (0.5)*25= 12.5 ms
Hence P =12.5 ms
and Q= 2.5 M bits / s