ISRO Computer Science (CSE) 2015 - Question 31

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If there are 32 segments, each size 1 k bytes, then the logical address should have

  • A. 13 bits
  • B. 14 bits
  • C. 15 bits
  • D. 16 bits

Answer: C. 15 bits

Explanation: 32 segments = 2^5 (5 bits). Size 1k = 2^10 (10 bits offset). Total 15 bits.

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