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Consider a $4-TB$ disk that uses $4-KB$ blocks and the free-list method. How many block addresses can be stored in one block?
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The size of disk$=4\;\text{TB}=4\times 2^{40}$

Size of a block$=4\;\text{KB} = 4 \times 2^{10}$

So, the number of blocks on the disk $=\frac{\text{Size of disk}}{\text{Size of block}} \\= \frac{2^{40}}{2^{10}} \\= 2^{30}$

So, Total number of blocks$=2^{30} \;\text{Blocks}$

by Boss (19.2k points)
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