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Consider a UNIX-like file system implemented with i-nodes that resides on a disk of size $512\text GB$. Each i-node has a total of $15$block addresses consisting of direct and indirect lock addresses.

Suppose we configure the file system to use a block size of $32$KB. How many bytes are needed to store all $15$ block addresses in an I-node?

a). $15$ Bytes                                                b). $29$ Bytes

c). $45$ Bytes                                                d). $75$ Bytes

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Ans is C

Total Number of Blocks in Disk is = Size of Disk / Size of each block

Total Number of Blocks in Disk is =239/215 =224

So, to address Each Block we need 24bits  ==>3 bytes

So space required by 15 address to block is 15*3=  45 Bytes

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