Aspire Faculty ID #12655 · Topic: JECA MCA 2024 · Just now
JECA MCA 2024

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

Solution

Solution: 32 segments ⇒ segment bits = $\log_2 32 = 5$. Size = 1 KB ⇒ offset bits = $\log_2 1024 = 10$.
Total address bits $= 5 + 10 = \boxed{15}$.
Answer: (D) 15 bits
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