TEST SERIES OS

Question

THE DISK DRIVE OF 20 GB CONTAIN N PLATTERS AND 640 CYLINDERS .THERE ARE 512 SECTORS PER TRACK AND SECTOR SIZE IS 1024 B. WHAT WILL BE NUMBER OF RECORDING SURFACE??

Answer 1

Size $= 2\times \textbf {number of platters}\times\text{ tracks }\times \textbf{ sectors}\times \textbf{ bytes per sector}$

Let the number of platters $= \text N$ (recording surface)

Given values are:

Size$ = 20\text{GB} = 20 \times 1024 \text{MB} \times 1024 \text{KB} \times 1024 \text {Bytes}$

Number of cylinders $ = 640$

Hence, the Number of tracks$ = 640$ as well, since the number of tracks is the same as the number of cylinders.

The number of sectors per track $= 512$
 
Plucking these values in the given formula, we get:

$\text N = \frac{(20 \times 1024\times 1024\times1024)}{ (2\times 640\times 512\times 1024)}$

On solving it, we get:

$\mathrm N = 32$

Answer: $\therefore$ The number of recording surfaces are $32$