ISRO | ISRO CS 2015 | Question 31

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)
Explanation: Number of bits required for 32 segments, each of 1 k bytes
= 2⁵ * 2¹⁰
= 2¹⁵
So, 15 bits are required.
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