Memory Management Q.8
0. If there are 32 segments, each of size 1 K byte, then the logical address should have
- Option : C
- Explanation : To specify a particular segment, 5 bits are required (since 25 = 32). . Having selected a page, to select a particular byte one needs 10 bits (since 210 = 1 k byte). So, totally 5 + 10 =15 bits are needed.
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