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+2 votes
33 views

asked in Operating System by Junior (667 points) | 33 views

2 Answers

+2 votes

2 is correct answer.

..

answered by Loyal (2.6k points)
edited by
0 votes
Virtual address space is in bits And page size and entry size is bytes!
=>Address space : 512MB

No of pages=512MB/4KB=128KB
Page Table size=128KB*4B=512KB !=4KB[Can`t Fit in 1 page]

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$\therefore$ 512KB/4KB=128B < 4KB Second Level Page table which is of size = 128B
answered by Boss (5k points)


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