Bit length for 0 to 1023 – UGC NET Question

To represent integers from 0 to 1023, we need 1024 distinct codes. A binary pattern of n bits can represent 2^n distinct values. The smallest power of two greater than or equal to 1024 is 2^10. Therefore, 10 bits are required to cover the entire range from 0 to 1023.

Option A:
Option A uses 8 bits, which give 2^8 = 256 patterns. These patterns represent only the range 0 to 255. Since 1023 is much larger than 255, 8 bits are insufficient for this range.

Option B:
Option B uses 9 bits, which give 2^9 = 512 patterns. This allows representation only from 0 to 511. Because 1023 is greater than 511, 9 bits still do not cover the full required range.

Option C:
Option C uses 10 bits, which give 2^10 = 1024 patterns. These patterns can represent all integers from 0 to 1023 without any gap. Hence 10 bits are exactly sufficient and not excessive.

Option D:
Option D uses 11 bits, which give 2^11 = 2048 patterns. This covers a much larger range than necessary for 0 to 1023. It is technically sufficient but not the minimum, so it is not the best answer.