Concern a disk with a sector size of 512 bytes, 2000 tracks per surface, 50…
2020
Concern a disk with a sector size of 512 bytes, 2000 tracks per surface, 50 sectors per track, five double-sided platters, and average seek time of 10 milliseconds.
What is the capacity of the disk, in bytes?
Answer: B. 500,000𝐾 — Answer: 500,000K (which equals 512,000,000 bytes) Step 1: Compute number of surfaces: 5 platters × 2 sides = 10 surfaces. Step 2: Total sectors = 50…
- A.
25,000𝐾
- B.
500,000𝐾
- C.
250,000𝐾
- D.
50,000𝐾
Attempted by 200 students.
Show answer & explanation
Correct answer: B
Answer: 500,000K (which equals 512,000,000 bytes)
Step 1: Compute number of surfaces: 5 platters × 2 sides = 10 surfaces.
Step 2: Total sectors = 50 sectors/track × 2000 tracks/surface × 10 surfaces = 1,000,000 sectors.
Step 3: Total capacity in bytes = 1,000,000 sectors × 512 bytes/sector = 512,000,000 bytes.
Step 4: Convert to kilobytes using K = 1024 bytes: 512,000,000 ÷ 1024 = 500,000 K.
Therefore the disk capacity shown as 500,000K is correct. Note: 'K' here is being used as 1024 bytes (KiB).