Answer:
a. 256MB
b. 11 ms
Explanation:
a. The capacity of the drive is
As we know that
[tex]= Number\ of\ surface \times tracks\ per\ surface \times sectors\ per\ surface \times bytes\ or\ sector[/tex]
[tex]= 4 \times 1024 \times 128 \times 512\ bytes[/tex]
= 256MB
b. Now the access time is
As we know that
Average access time= seek time + average rotational latency
where,
seek time = 5ms
And, average rotational latency is
[tex]= \frac{rotational latency}{2}[/tex]
[tex]= \frac{60}{RPM}[/tex]
[tex]= \frac{60}{5,000}[/tex]
= 12 ms
So, average rotational latency is 6 ms
So, average access time is
= 5 ms + 6 ms
= 11 ms
We simply applied the above formulas
