Total Time $=$ Seek $+$ Rotation $+$ Transfer.
Seek Time :
Current Track $1$
Destination Track $8$
Distance Required to travel $= 4-0.5= 3.5 \ Cm$
Time required $= 10 \ m/s == 1\ Cm/ms == \mathbf{3.5 \ ms}$ [ Time= Distance / Speed ]
Rotation Time:
$6000$ RPM in $60$ sec
$100$ RPS in $1$ sec
$1$ Revolution in $10$ ms
$1$ Revolution $=$ Covering entire Track
$1$ Track $= 20$ sector
$1$ sector required $= 10/20 = 0.5 \ ms$
Disk is constantly Rotating so when head moved from inner most track to outer most track total movement of disk $= (3.5/0.5) = 7 $ sectors
Which means that when disk reached outer most track head was at end of $12^{ th}$ sector
Total Rotational Delay $=$ Time required to go from end of $12$ to end of $3 = 11$ sectors
$1$ sector $= 0.5 \ ms$ so $11$ sector $\mathbf{= 5.5 ms}$
Transfer Time
Total Data in Outer most track $= 10 \ MB$
Data in single Sector $= 10 \ MB/20 = 0.5 \ MB$
Data required to read $= 1 \ MB = 2$ sector
Time required to read data $= 2 \times 0.5 = \mathbf{1ms}$
Total Time = Seek + Rotation + Transfer $\mathbf{ = 3.5ms + 5.5ms +1ms = 10\ ms}$
Correct Answer: $B$