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+2 votes

asked in Operating System by Active (1.2k points) | 75 views

2 Answers

+2 votes

2 is correct answer.


answered by Boss (7.7k 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]

$\therefore$ 512KB/4KB=128B < 4KB Second Level Page table which is of size = 128B
answered by Veteran (11.7k points)

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