Consider a disk with the following specifications: rotation speed of 6000 RPM, average seek time of 5 milliseconds, 500 sectors/track, 512-byte sectors. A file has content stored in 3000 sectors located $\textit{randomly}$ on the disk. Assuming average rotational latency, the total time ($\textit{in seconds, rounded off to 2 decimal places}$) to read the entire file from the disk is ________

To find the total time to read 3000 randomly located sectors, compute the access time per sector:

1. Rotation Time ($T_{\text{rot}}$):

   $$\text{Rotation speed} = 6000\text{ RPM} = \frac{6000}{60} = 100\text{ rotations/second}$$

   $$T_{\text{rot}} = \frac{1}{100}\text{ s} = 10\text{ ms}$$

2. Average Rotational Latency ($T_{\text{lat}}$):

   $$T_{\text{lat}} = \frac{T_{\text{rot}}}{2} = \frac{10\text{ ms}}{2} = 5\text{ ms}$$

3. Transfer Time per Sector ($T_{\text{trans}}$):

   $$T_{\text{trans}} = \frac{T_{\text{rot}}}{\text{sectors per track}} = \frac{10\text{ ms}}{500} = 0.02\text{ ms}$$

4. Average Time to Read One Random Sector ($T_{\text{sector}}$):

   $$T_{\text{sector}} = \text{Seek Time} + T_{\text{lat}} + T_{\text{trans}} = 5\text{ ms} + 5\text{ ms} + 0.02\text{ ms} = 10.02\text{ ms}$$

5. Total Time for 3000 Random Sectors:

   $$\text{Total Time} = 3000 \times 10.02\text{ ms} = 30060\text{ ms} = 30.06\text{ seconds}$$

Rounded to two decimal places, the total time is 30.06 seconds (official range: 29.50 to 30.50).