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      Question

      A system has a page size of 2KB. A logical address is 32

      bits. How many bits are used for the page offset?
      A 10 bits Correct Answer Incorrect Answer
      B 11 bits Correct Answer Incorrect Answer
      C 12 bits Correct Answer Incorrect Answer
      D 13 bits Correct Answer Incorrect Answer
      E 16 bits Correct Answer Incorrect Answer

      Solution

      Page size = 2KB = $2 \times 1024$ bytes = $2^{1} \times 2^{10}$ bytes = $2^{11}$ bytes.         The number of bits required for the page offset is $\log_2(\text{page size})$.         So, $\log_2(2^{11}) = 11$ bits.

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