Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors…
20092018
Consider a disk pack with 16 surfaces, 128 tracks per surface and 256 sectors per track. 512 bytes of data are stores in a bit serial manner in a sector. The capacity of the disk pack and the number of bits required to specify a particular sector in the disk are respectively
- A.
256 Mbyte, 19 bits
- B.
256 Mbyte, 28 bit
- C.
512 Mbyte, 20 bits
- D.
64 Gbyte, 28 bits
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Correct answer: A
First, calculate the total capacity. Multiply surfaces (16), tracks per surface (128), sectors per track (256), and bytes per sector (512). In powers of 2: 16=2^4, 128=2^7, 256=2^8, 512=2^9. Total bytes = 2^(4+7+8+9) = 2^28 bytes. Since 1 MB = 2^20 bytes, capacity is 2^8 * 1 MB = 256 MB. Next, determine bits for addressing. Surfaces need log2(16)=4 bits. Tracks need log2(128)=7 bits. Sectors need log2(256)=8 bits. Total addressing bits = 4 + 7 + 8 = 19. Thus, the capacity is 256 Mbytes and addressing bits are 19.
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