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Consider a virtual memory system that uses paging. Virtual and physical addresses are both $32$ bits long, and the page size is $4 \textsf{KB}=2^{12}$ bytes. A process $\text{P}_1$ has the following page table. Frame numbers are given in hexadecimal notation (recall that each hexadecimal digit represents $4$ bits).
$$
\begin{array}{|c|c|}\hline
& \textbf{Frame Number} \\\hline
0 & \textsf{0x0014e} \\
1 & \textsf{0x03b65} \\
2 & \textsf{0x00351} \\
3 & \textsf{0x00875} \\
4 & \textsf{0x06a3f} \\\hline
\end{array}$$

What will be the physical address for the virtual address: $\textsf{0x00003b65}?$

  1. $\textsf{0x00003b65}$
  2. $\textsf{0x00875b65}$
  3. $\textsf{0x00875}$
  4. Translation is not possible as per the given page table
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1 Answer

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page size is correct take log  … 12 is correct dont try to convert into bits. take 12 bits To represent page offset
Answer:

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