UGC NET Questions (Paper – 1)

To represent an unsigned decimal number N in binary, the minimum number of bits n must satisfy 2^n > N−1. For N = 100, 2^6 = 64 which is less than 100, while 2^7 = 128 which is greater than or equal to 100. Therefore 7 bits can represent all numbers from 0 to 127, including 100, and 6 bits are insufficient. Hence, 7 bits is the smallest bit-length that can encode the decimal number 100 in binary.

Option A:
Option A suggests 6 bits, but 6 bits can represent numbers only from 0 to 63, which does not include 100. Since 100 exceeds 63, 6 bits are not enough to encode it in binary. Therefore, 6 bits underestimate the required capacity.

Option B:
Option B gives 7 bits, and 7 bits can represent values from 0 to 127, easily covering 100. It is also the smallest n such that 2^n ≥ 101, so it is the minimal correct choice. Thus, 7 bits is the correct minimum bit-length for representing 100 in binary.

Option C:
Option C suggests 8 bits, which can represent values from 0 to 255 and therefore can represent 100, but it is not the minimum number of bits required. While 8 bits are sufficient, the question asks specifically for the minimum, making this option excessive. Hence, 8 bits is correct in range but not minimal.

Option D:
Option D suggests 9 bits, which can represent an even larger range from 0 to 511, including 100. However, it is even more excessive than 8 bits and clearly not minimal. Therefore, although 9 bits would work, it does not satisfy the requirement of minimum bit-length.

For a positive integer N, the number of bits required is tied to the highest power of 2 not exceeding N. If 2^(k−1) ≤ N ≤ 2^k − 1, the number uses k bits. Taking base-2 logarithms gives k − 1 ≤ log₂N < k, so k = ⌊log₂N⌋ + 1. This matches the theoretical formula for bit-length. Option A: Option A expresses this relationship exactly, using the floor function to convert the real logarithm to an integer count of bits. It ensures that every value in the closed interval of integers with the same magnitude uses the same bit-length. Option B: Option B, ⌈log₂N⌉, overestimates the bit-length when N is a power of 2 because log₂N is then an integer and the ceiling does not add 1. It thus fails to capture the inclusive nature of the range. Option C: Option C, log₂(N−1), omits rounding and the +1 term, and can be undefined for N = 1. It does not yield a discrete measure of bit-length. Option D: Option D, 2×log₂N, doubles the logarithmic value and has no standard interpretation as a digit count. It grossly misestimates the number of bits required.

We require n such that 2ⁿ − 1 ≥ 1000 for unsigned representation. Checking nearby powers, 2⁹ = 512 and 2¹⁰ = 1024. Since 512 ≤ 1000 ≤ 1023, 9 bits are insufficient, but 10 bits cover up to 1023. Thus, 10 bits are the minimum needed to represent 1000 in binary.

Option A:
Option A correctly applies the inequality 2ⁿ−1 ≥ N with N = 1000. It identifies that 2⁹ − 1 = 511 is too small, while 2¹⁰ − 1 = 1023 is adequate. Therefore, 10 is the least n satisfying the condition, making this option correct.

Option B:
Option B, 9 bits, corresponds to a maximum value of 511, which does not reach 1000. Using only 9 bits would fail to encode the number 1000, so it cannot be minimal or sufficient.

Option C:
Option C, 11 bits, is technically sufficient but not minimal, since 10 bits already handle values up to 1023. The question specifically asks for the minimum, so including an extra bit is unnecessary.

Option D:
Option D, 8 bits, yields a maximum of 255, which is far below 1000. It clearly cannot represent 1000 and hence is not even sufficient, let alone minimal.