In question(b), transfer rate is 120tracks/s

=(120*120sectors)/s=(120*120*512bytes)/s=7200kb/s.

is this answer true?

if it is true, why I do not need consider the cylinder into the transfer rate?

if I consider the cylinder it will become to 1cylinder/s

=(20*120tracks)/s=(20*120*120sectors)/s=(20*120*120*512bytes)=144Mb/s

In question(c), this question ask "access time", it means (seek time)+(average latency)?

OR I/O time= average access time + (amount to transfer / transfer rate) + controller overhead?

this two question describe confuse me, please help.

An IDE hard disk spins at 7200 RPM, has 2 MB internal cache, 5000 cylinders, 20 tracks per cylinder, 120 sectors per track, 512 bytes per sector. (a) Calculate the disk size. (b) Estimate the transfer rate in bytes second. (c) What is the access time in the scale of milliseconds) for reading a file with size 0.36865 MB under the assumption that seek time is 4 milliseconds