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ISRO | ISRO CS 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)

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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Last Updated : 04 Apr, 2018
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