Question 58

1. Time for one complete rotation:

   $$T_{\text{rot}} = \frac{60 \times 1000\text{ ms}}{10000\text{ RPM}} = 6\text{ ms}.$$

2. Average rotational latency:

   $$T_{\text{latency}} = \frac{T_{\text{rot}}}{2} = \frac{6\text{ ms}}{2} = 3\text{ ms}.$$

3. Transfer time per sector:
   Since each track has 600 sectors, the time to pass over one sector is:

   $$T_{\text{transfer}} = \frac{T_{\text{rot}}}{600} = \frac{6\text{ ms}}{600} = 0.01\text{ ms}.$$

4. Total access time per sector:
   Given seek time $T_{\text{seek}} = 4\text{ ms}$:

   $$T_{\text{sector}} = T_{\text{seek}} + T_{\text{latency}} + T_{\text{transfer}} = 4 + 3 + 0.01 = 7.01\text{ ms}.$$

5. Total time for 2000 sectors:

   $$T_{\text{total}} = 2000 \times 7.01\text{ ms} = 14020\text{ ms}.$$

The total time needed is $14020$ milliseconds.