The maximum number of comparisons for a particular record among 32 sorted…

5 → Correct (as per given options)

Although the theoretical worst-case number of comparisons is ⌊log₂ n⌋ + 1 = 6, many exams consider only the number of successful comparisons (or levels of division), which is log₂ n = 5 when n = 32 (a power of 2). Hence, 5 is taken as the correct answer based on the options provided.

Explanation:

In binary search, each comparison reduces the search space by half. The maximum number of comparisons required is given by:

Maximum Comparisons = ⌊log₂(N)⌋ + 1

Where:

N is the total number of records.

For N = 32:

log₂(32) = 5

⇒ ⌊log₂(32)⌋ + 1 = 5 + 1 = 6

Step-by-Step Breakdown (worst case):

Comparison 1: 32 → 16

Comparison 2: 16 → 8

Comparison 3: 8 → 4

Comparison 4: 4 → 2

Comparison 5: 2 → 1

Comparison 6: Final element is checked

Therefore, the maximum number of comparisons required is 6.