option (D)
Explanation
Avg. time to transfer $=$ Avg. seek time $+$ Avg. rotational delay $+$ Data transfer time
Avg Seek Time
given that : time to move between successive tracks is $1 \ ms$
time to move from track $1$ to track $1 : 0ms$
time to move from track $1$ to track $2 : 1ms$
time to move from track $1$ to track $3 : 2ms$
..
..
time to move from track $1$ to track $500 : 499 ms$
Avg Seek time $= \frac{\sum 0+1+2+3+...+499}{500}$
$\mathbf{= 249.5 \ ms}$
Avg Rotational Delay
RMP $: 600$
$600$ rotations in $60$ sec
one Rotation takes $60/600$ sec $= 0.1$ sec
Avg Rotational Delay $= \frac{0.1}{2}$ { usually $\frac{\text{Rotation time}}{2}$ is taken as Avg Roational Delay }
$= .05 sec$
$\mathbf{= 50 \ ms}$
Data Transfer Time
One $1$ Roatation we can read data on one complete track .
$= 100 \times 500 = 50,000$ B data is read in one complete rotation
one complete rotation takes $0.1$ s ( we seen above )
$ 0.1 \rightarrow 50,000$ bytes.
$ 250$ bytes $ \rightarrow 0.1 \times 250 / 50,000 = \mathbf{ 0.5 \ ms}$
Avg. time to transfer $=$ Avg. seek time $+$ Avg. rotational delay $+$ Data transfer time
$\mathbf{= 249.5+50+0.5}$
$\mathbf{= 300 \ ms}$