Answer:
The answer is "2.95 GB"
Explanation:
Given value:
Surfaces = 6
The tracks per surface =16,383
The sectors per track = 63
bytes/sector = 512
Track-to-track seek time = 8.5ms
The Rotational speed = 7,200rpm
[tex]\text{Drive capacity= surface} \times \text{tracks per surface} \times \text{sectors per track} \times \frac{bytes}{sector}[/tex]
[tex]= 6 \times 16383 \times 63 \times 512 \\\\= 3170700288 \ B\\\\ = 3023.81543 \ MB \\\\= 2.95294 \ GB\\[/tex]
The drive capacity is 2.95 GB