Finding the time it would take to read a 1000 byte file

Question

Disk D has one platter(2 surfaces), 200 tracks, 100 sectors/track, and sectors are 1KB. It rotates at 3600 RPM and average seek is 10ms.

1.) In the BEST CASE, how much time would it take to read a 1000 byte file?

I know on AVERAGE CASE I simply need to find the sum of seek time + rotational latency + transfer time. How would I do it with BEST CASE?

2.) If you changed D to 25 sectors per track, each sector is 4 KB: If disk sectors for files are scattered on the disk, would reading a 8000 byte file be faster, slower, or the same.

My answer for this is SLOWER, because it would have more seek time, but apparently it's wrong?

Answer 1

1. Best case is the one sector containing the file is right under the read/write head. So there is no seek time, and no rotational latency time. The disk rotates 60 times/second, so reading an entire track would take 1/60 of a second. But you only want 1/100 of a track, so the answer is 1/(60 * 100) -> .16666 milliseconds.
2. Yes, slower is wrong. If the disk sectors are 4K, an 8K file would be in 2 sectors, which would take 2 seeks and 2 rotational latencies. If the sectors are 1K, the file is in 8 sectors, which is 8 seeks and 8 rotational latencies. Thus 4K sectors are faster if the sectors are scattered.