The total physical time required for a single random disk access consists of three distinct parts: seek time to move the arm, rotational latency to find the sector, and the data transfer time.
$$\text{Total Time per Access} = \text{Seek Time} + \text{Rotational Latency} + \text{Transfer Time}$$
The problem states that the data transfer time from the disk block once the head is positioned may be completely neglected.
$$\text{Transfer Time} = 0 \text{ ms}$$
The seek time to position the disk head over a random target track is explicitly given as a fixed value:
$$\text{Seek Time} = 10 \text{ ms}$$
To calculate the rotational latency, we first find the total time required for the platter to complete one single 360-degree rotation using the speed of 6000 rpm:
$$\text{Time for 1 rotation} = \frac{60 \text{ seconds}}{6000} = 0.01 \text{ seconds} = 10 \text{ ms}$$
Because the libraries are scattered across random locations, we must use the standard average rotational latency, which corresponds to the time taken for half of a full platter rotation:
$$\text{Average Rotational Latency} = \frac{\text{Time for 1 rotation}}{2} = \frac{10 \text{ ms}}{2} = 5 \text{ ms}$$
Combining these elements gives the average time penalty incurred for every individual random library access:
$$\text{Total Time per Access} = 10 \text{ ms (Seek)} + 5 \text{ ms (Latency)} + 0 \text{ ms (Transfer)} = 15 \text{ ms}$$
The application needs to load a total of 100 independent libraries from completely random disk locations at startup.
We find the total loading time by multiplying the number of libraries by our computed single random access time:
$$\text{Total Time} = 100 \times 15 \text{ ms} = 1500 \text{ ms}$$
Converting this duration from milliseconds into seconds yields the final metric:
$$\text{Total Time} = \frac{1500}{1000} \text{ seconds} = 1.50 \text{ s}$$
Correct Option: B
