What is the size of the physical address space in a paging system which has a page table containing 64 entries of 11 bit each (including valid and invalid bit) and a page size of 512 bytes?
- A. $2^{11}$
- B. $2^{15}$
- C. $2^{19}$
- D. $2^{20}$
Answer: C. $2^{19}$
Explanation: Page table entries = 64 (Virtual pages).\nEntry size = 11 bits. (Includes frame number + control).\nValid/Invalid = 1 bit. Frame bits = 10 bits.\nPhysical Frames = $2^{10}$.\nPage Size (Frame Size) = 512 bytes = $2^9$.\nPhysical Address Space = Frames $\times$ Size = $2^{10} \times 2^9 = 2^{19}$ bytes.