If the page size in a 32-bit machine is 4K bytes then the size of page table is
(A) 1 M bytes
(B) 2 M bytes

(C) 4 M bytes
(D) 4 K bytes


Answer: (C)

Explanation: Size of physical address = 32 bits
Size of logical address = 32 bits

page size = 4 KBytes
Number of pages = logical address space/ size of each page
= 232/ 212
= 220

page table size = number of pages * size of a page table entry
                   = 220 * 22
                   = 222 

Option (C) is correct.

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