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Q2. Consider a disk that using FAT table (file allocation table). Block size is 4KB. Disk size is 64 GB. A disk address (pointer, i.e.,

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Q2. Consider a disk that using FAT table (file allocation table). Block size is 4KB. Disk size is 64 GB. A disk address (pointer, i.e., a block number) is 8 bytes long. What is the size of the FAT table in bytes? How may disk blocks will it occupy? Assume FAT entry 0 is for disk block 0, FAT entry 1 for disk block 1 , and so on

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