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
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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