1. Average rotational latency in hard drive is
(1 / (RPM / 60)) * 0.5 * 1000 ms
Or, (30000 / RPM) ms = 30000 / 6000 = 5 ms , this makes statement I correct .
2. The burst data rate is average track capacity * rotations per second = 128 * 0.001 * 100 MB/sec = 12.5 MB per second ( as 1 KB = 0.001 MB ) , this makes statement II correct.
3. The average capacity of each track is average number of sectors per track * the size per sector = (128 * 1) KB =128 KB, which makes statement III false .
So, only statement I and II are correct , which is option C .