Final answer:
To calculate the number of bits for the logical address with 32 segments of 1Kb each, add 5 bits for segment number and 10 bits for the offset, giving a total of 15 bits. The correct option is c) 15 bits.
Step-by-step explanation:
The student has asked how many bits are needed for the logical address if there are 32 segments, each of size 1Kb. Kb stands for kilobytes, which implies that each segment is 1024 bytes in size. To determine this, we need to calculate how many bits are required to address a segment and how many bits to address within a segment.
Since there are 32 segments, we need 5 bits to uniquely identify each segment (as 2^5 = 32). For the offset within each 1Kb segment, we need an additional 10 bits because 1Kb is 1024 bytes, and 2^10 = 1024. By adding these together (5 bits for segment number + 10 bits for offset within the segment), we get a total of 15 bits needed for the logical address.
Therefore, the correct answer is c) 15 bits.