NIELIT 2016 MAR Scientist C - Section C: 8

Question

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

• A. $13 \text{ bits}$
• B. $14 \text{ bits}$
• C. $15 \text{ bits}$
• D. $16 \text{ bits}$

Answer 1

No of segments = $2^5$, Size of each segment = $1k = 2^{10}$

So size of memory = $2^5 * 2^{10} = 2^{15}$

So we need 15 bits to address this logical space.

Hence C is correct.

Answer 2

yes nos of segment 32 = 2^5 correspond to 5 bits

again Segnent size is equal to 1 KB =1024 B= 2^10 BYTES  which correspond to 10 bits for addressing.

hENCE NOS OF TOTAL BITS ARE 5+10=15 Bits  as per option (C).is correct.