to access some data on disk , disk have to first reach that desired track and then sector.
so to reach desired track the time consumed is called SEEK TIME. so reach desired sector then time consumed is called Rotational latency. and now it only have to transfer data(data transfer time)...
seek time to access 1 sector or n successive sector is same as disk head once have to move desired track then just read , rotational latency would also be same as let suppose you have to access sector 5 and your head is at sector 2 then RL would be to reach 2 to 5 and then to reach 5 to 6 no RL so in that manner to access successive RL=0 , data transfer time is directly proportional to data size...in one rotation , one TRACK data can be transferred ....
so to question no 1 )
RL=5ms ST=5ms DTT=125ms
so total time to access 800 successive sectors is = 5+5+125=135ms
question 2) here random sector accessing so everytime disk head have to seek desired track and then desired sector then DTT so just multiply 800 to ST and RL. but transer time would be same as last as it only depends on data size , which is still same...
8000+125=8125ms...
Seek time =5ms given
Rotational latency =time to complete 1/2 rotation
60sec-------->>>6000rotation
so 1/2 rotation ------>>>5ms
Data transfer time (DTT)= time to transfer data
60sec------->>>>6000 rotation (means 6000 tracks)
so 10ms to transfer data of 1 track, which is 64*103B
10ms--------64*103B
800 sector data size= 800*103B
so to transfer 800 sector data DTT=(10ms/64*103 )*800*103 = 125ms