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Can someone give a detailed solution of this problem

asked in Computer Networks by (487 points) | 70 views

2 Answers

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This question cant be of CN, i think it is of Comp.Org
answered by (245 points)
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No of Addresses that can fit in a single block = Block Size / Disk Address size = 4KB/4B(32 bit) = 1024

In total there are 4 Blocks

therefore it can support -> 4 * 1024 Files = 4096 Files
answered by Active (1.5k points)

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