Question

Consider a disk, rotating with 3000 RPM. It has 256 tracks, 512 sectors per track with 2 KB sector capacity. The data transfer rate of disk is ________ MBPS.

• A. 70
• B. 50
• C. 40
• D. 60

Answer 1

The correct option is B 50

option (b)

time required for 3000 rotations= 1 min = 60 sec = 60000 msec

time required for 1 rotation = $\frac{60000}{3000}=20msec$

In 1 rotation time, 1 track can be transferred.

1 track capacity = 512* 2KB = 1 MB

In 20 msec data transferred = 1 MB

In 1 msec data transferred = $\frac{1MB}{20msec}$

$In1secdatatransferred=\frac{1MB}{20\ast {10}^{-3}sec}$

= 50 MBPS