Bit length for 0 to 255 – UGC NET Question

To represent 0 to 255, we need 256 distinct binary codes. The number of patterns provided by n bits is 2^n. Since 2^7 is only 128 and 2^8 is 256, 8 bits are necessary. Thus, 8 bits form the standard byte that exactly represents this range.

Option A:
Option A suggests 5 bits, which give only 2^5 = 32 patterns. These patterns can represent decimal values only from 0 to 31. This is far less than the required upper limit of 255, so this option is not correct.

Option B:
Option B suggests 6 bits, which give 2^6 = 64 patterns. With 64 combinations, the range is only 0 to 63. Because 255 is much larger than 63, 6 bits cannot represent all the required integers.

Option C:
Option C suggests 7 bits, which give 2^7 = 128 patterns. That represents values from 0 to 127. Since 255 is beyond 127, 7 bits still fall short of the needed representation range.

Option D:
Option D uses 8 bits, producing 2^8 = 256 unique patterns. These map naturally to the integers 0 through 255. This makes 8 bits the minimal and correct choice for representing all values in this interval.