Given the drive dimensions 22,164 cylinders per disk, 80 heads per cylinder, and 63 sectors per track, what is the approximate capacity?

Given the drive dimensions 22,164 cylinders per disk, 80 heads per cylinder, and 63 sectors per track, what is the approximate capacity?

• A. 53.26 GB
• B. 11.17 GB
• C. 57.19 GB
• D. 10 GB

Answer: (C)

Disk capacity is found by multiplying the number of cylinders by the number of heads per cylinder, and by the number of sectors per track, then multiplying by the bytes per sector. Using 512 bytes per sector (a common size): - Cylinders × heads per cylinder × sectors per track = 22,164 × 80 × 63 = 111,706,560 sectors - Multiply by 512 bytes per sector: 111,706,560 × 512 = 57,193,758,720 bytes Convert to gigabytes: 57,193,758,720 bytes is about 57.19 × 10^9 bytes, i.e., roughly 57.19 GB (decimal). If you convert to gibibytes (using 1024^3), it’s about 53.26 GiB, but the decimal GB figure aligns with the given option. Therefore, the approximate capacity is 57.19 GB.

Disk capacity is found by multiplying the number of cylinders by the number of heads per cylinder, and by the number of sectors per track, then multiplying by the bytes per sector. Using 512 bytes per sector (a common size):

• Cylinders × heads per cylinder × sectors per track = 22,164 × 80 × 63 = 111,706,560 sectors

• Multiply by 512 bytes per sector: 111,706,560 × 512 = 57,193,758,720 bytes

Convert to gigabytes: 57,193,758,720 bytes is about 57.19 × 10^9 bytes, i.e., roughly 57.19 GB (decimal). If you convert to gibibytes (using 1024^3), it’s about 53.26 GiB, but the decimal GB figure aligns with the given option. Therefore, the approximate capacity is 57.19 GB.