01-08-2012, 02:21 AM
Q.1 SOLUTION:
The floppy drive has a 16-bit data bus; it can transfer two bytes at one time.
Its transfer rate is 48/2 = 24K per second.
This corresponds to an overhead of 10 instructions
or 24K x 10 = 24 x 210 x 10 = 245760 instructions per second.
Q.2 SOLUTION:
512 bytes x 512
Sectors=0.2MB/track
0.2MB x 512 tracks
=0.1GB/platter
so the hard disk has the total capacity of 10 x 0.1=1GB
Q.3 SOLUTION:
The capacity of one platter
= 2048 x 2048 x 2048
= 8GB
For an 80GB hard disk, we need 80/8
= 10 such platters.
The floppy drive has a 16-bit data bus; it can transfer two bytes at one time.
Its transfer rate is 48/2 = 24K per second.
This corresponds to an overhead of 10 instructions
or 24K x 10 = 24 x 210 x 10 = 245760 instructions per second.
Q.2 SOLUTION:
512 bytes x 512
Sectors=0.2MB/track
0.2MB x 512 tracks
=0.1GB/platter
so the hard disk has the total capacity of 10 x 0.1=1GB
Q.3 SOLUTION:
The capacity of one platter
= 2048 x 2048 x 2048
= 8GB
For an 80GB hard disk, we need 80/8
= 10 such platters.