where:
(A) Track
(B) Geometrical sector
(C) Track sector
(D) Cluster
Total number of tracks this program takes = $\frac{64MB}{32MB} = 2$
Program has to be loaded from the disk, it will happen in the following way:
- Head will move at the beginning of the $1^{st}$ track.
To move to that track on the disk it will take 1 Seek Time, and then to reach at the start of that track it will let the disk to rotate and this on an average takes up $\frac{1}{2} \times \text{Rotation Time}$. Then it will read the track and simultaneously transfer data which will take up 1 Rotation Time(in one rotation a head can read a single track)
so, till now $\text{1 Seek Time} + \frac{1}{2} \times \text{Rotation Time} + \text{1 Rotation Time}$ has passed.
- After reading $1^{st}$ track it will move to the second track and to do that same process will be followed which will take exactly the same time as in first case.
Hence,
$\begin{align*} \text{Time taken to load} &= 2 \times \left(\text{1 Seek Time} + \frac{1}{2} \times \text{Rotation Time} + \text{1 Rotation Time} \right )\\ &= 2 \times \left(\text{30ms} + \frac{1}{2} \times \text{20ms} + \text{20ms} \right )\\ &= 120ms \end{align*}$