Let, 1 rotation take x ms to complete.
Now, Total capacity of each track = 32 * 512 B = 16 KB
Therefore, 16 KB data is transmitted in x ms.
Therefore, 1 B data is transferred in (x / 16K) ms.
Therefore, 8 KB data is transferred in (8 K x / 16 K) ms = (x / 2) ms
Transfer Time = (x / 2) ms
Now, we know that
Avg. Read Time = Seek Time + Rotational Latency + Transfer Time
Further, Rotational Latency = Rotation Time / 2 = (x / 2) seconds
Therefore, 20 ms = 10 ms + (x / 2) ms + (x / 2) ms
On solving the above equation, we get
x = 10 ms
This means that in 10 ms, the disk rotates 1 time.
Therefore, in 1 s, the disk rotates 100 times.
Therefore, in 60 s, i.e. 1 min, the disk rotates 6000 times.
Therefore, the rotational speed is 6000 rpm.