In a system with a page size of 4 KB and a physical address space of 64 GB, what is the number of bits required for the page offset?

Asked In Exam: BPSC TRE 3.0

Options

  • A. 10 bits
  • B. 12 bits
  • C. 14 bits
  • D. More than one of the above
  • E. None of the above

Correct Answer (Detailed Explanation is Below)

B. 12 bits

Detailed Explanation

4 KB equals 4096 bytes, which is 2^12 bytes. Therefore, 12 bits are required for the page offset.

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BPSC TRE 4 operating system practice set 1

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