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Size of the Disk $=4\;\text{TB}=4\times 2^{40} = 2^{42} \ Bytes$

Size of a Block $=4\;\text{KB}=4\times 2^{10} = 2^{12} \ Bytes$

Number of Blocks present in the Disk $= Size \ of \ Disk / Size \ of \ Block = 2^{42} / 2^{12} = 2^{30}$

Number of Bits needed to represent address of one Block $= \lg 2^{30} \ bits = 30 \ bits$  $\approx$ $4 \ Byte$

Number of block addresses that can be stored in one block $= 2^{12} / 4 = 2^{10} = 1024$

 

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Answer:

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$=2^{30} \;\text{Blocks}$

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Number of blocks=2^42/2^12 =2^30

So each disk block  address size =30 bits=4 bytes

number of  disk block address = 2^30/4 =2^28

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