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| Q. |
If there are 32 segments, each of size 1Kb, then the logical address should have |
| A. | 13 bits |
| B. | 14 bits |
| C. | 15 bits |
| D. | 16 bits |
| Answer» C. 15 bits | |
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Operating System (OS)Discussion
Rishu Chaurasia
2 months ago
f there are 32 segments, each of size 1 kilobyte (1KB), then the logical address should have 15 bits. To specify a particular segment, 5 bits are required (since 2^5 = 32). Having selected a page, to select a particular byte, 10 more bits are needed (since 2^10 = 1KB). Therefore, 15 bits in total are needed to address the segments and bytes within them123.
In summary, the logical address for this scenario should consist of 15 bits. 🤓
In summary, the logical address for this scenario should consist of 15 bits. 🤓
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