Question

The capacity of a CD-ROM is 800 MB (800 x 20 raised to 20 bytes). It stores information along a spiral track. If each byte uses a space of 9 microns, what is the length of the complete track in m? (Hint: 1 micron= 10 Raised to-6m)

Answer 1

Answer:

The length of the complete track= $7.549\times 10^{23} m$

Step-by-step explanation:

We are given that

Capacity of CD-ROM=800MB=$800\times 20^{20} bytes$

1 byte=9 microns

$1micron=10^{-6}m$

We have to find the  length of the complete track in m.

We change MB into  micron

$800\times 20^{20} bytes=800\times 20^{20}\times 9microns$

=$7200\times 20^{20} microns$

Now, we change micron into m

$7200\times 20^{20} microns=7200\times 10^{-6}\times 20^{20} m$

=$7.549\times 10^{23} m$

Hence, the length of the complete track= $7.549\times 10^{23} m$