UGC NET Questions (Paper – 1)

The decimal number 4095 is equal to 2¹² − 1, which is the largest number that can be represented using 12 bits. Any n-bit unsigned binary representation can represent numbers from 0 to 2ⁿ − 1. Since 4095 matches 2¹² − 1 exactly, 12 bits are both necessary and sufficient. Therefore, the minimum number of bits required is 12.

Option A:
If only 10 bits were used, the maximum representable number would be 2¹⁰ − 1 = 1023. This is much smaller than 4095, so 10 bits are insufficient.

Option B:
With 11 bits, the maximum value is 2¹¹ − 1 = 2047. This still falls short of 4095, so 11 bits also cannot represent the number.

Option C:
Using 12 bits allows values up to 2¹² − 1 = 4095, which matches the given decimal exactly. This shows that 12 is the smallest bit length that can store this number in unsigned binary form.

Option D:
13 bits would increase the range to 2¹³ − 1 = 8191, which is more than enough for 4095. However, the question asks for the minimum number of bits, so 13 is not the best answer even though it is technically sufficient.