UGC NET Questions (Paper – 1)

To convert 3F₁₆ to octal, we can first convert it to decimal. 3F₁₆ equals 3×16 + 15 = 48 + 15 = 63. Converting 63 to octal by dividing by 8 gives 7 with remainder 7, so the octal value is 77₈. Therefore, 77 is the correct octal equivalent.

Option A:
Option A, 67₈, converts to decimal 6×8 + 7 = 55, which is less than 63. It does not match the decimal value of 3F₁₆.

Option B:
Option B is correct because 3F₁₆ → 63₁₀ → 77₈. The intermediate decimal step verifies that both conversions are consistent.

Option C:
Option C, 73₈, corresponds to 7×8 + 3 = 59 in decimal, which is not equal to 63. Thus it cannot represent 3F₁₆.

Option D:
Option D, 75₈, equals 7×8 + 5 = 61, again not matching the required decimal value.

With n bits, there are 2^n distinct binary patterns ranging from all zeros to all ones. If we interpret these patterns as unsigned integers, the smallest value is 0 and the largest is when all bits are 1. This all-ones pattern corresponds to 2^n - 1. Therefore, the maximum unsigned integer with n bits is 2^n - 1.

Option A:
Option A is correct because subtracting 1 from 2^n removes the zero pattern and leaves the highest possible value. This matches the idea that counting from 0 yields 2^n different values.

Option B:
Option B, 2^n, would be one larger than the highest valid pattern and cannot be represented with n bits. It would require an additional bit.

Option C:
Option C, 2n, assumes only a linear relationship between bit count and value, which contradicts the exponential growth of binary patterns.

Option D:
Option D, 2^(n - 1), is the magnitude associated with the highest single bit, but it is not the maximum value obtainable when all bits are 1.

In normalized binary representation of non-zero values, the convention is that the leading digit before the binary point is 1. This ensures a unique representation for each non-zero number and maximizes precision by using the most significant bit. Therefore, the leading digit is constrained to be 1.

Option A:
Option A, 0, would indicate a denormalized or subnormal number, not a normalized one. In normalized form, a leading 0 would waste the most significant bit.

Option B:
Option B is correct because normalization requires the first non-zero bit to be 1 and positioned just before the binary point. This is standard in IEEE binary floating-point formats.

Option C:
Option C, 2, is not a valid single binary digit since binary digits can only be 0 or 1. It cannot serve as a leading bit.

Option D:
Option D, 3, is also invalid as a binary digit and has no meaning in this context.

Each bit in a binary pattern can independently take one of two values, 0 or 1. With n bits, there are 2 choices for each bit and the total number of combinations is 2×2×…×2 (n times), which equals 2^n. Therefore, n bits can represent 2^n distinct values.

Option A:
Option A, n, would suggest only a linear relationship between bit count and values represented. This contradicts the exponential growth observed in binary combinations.

Option B:
Option B, 2n, also implies a linear relationship and fails to capture how adding a single bit doubles the number of representable values.

Option C:
Option C, n^2, suggests a quadratic relationship, which does not match the binary combinatorial count. The correct growth is exponential, not quadratic.

Option D:
Option D is correct because each additional bit doubles the possible combinations, leading to 2^n distinct patterns. This is a fundamental property used in designing memory and address spaces.

For n-bit 2's complement representation, the range of signed integers is from -2^(n-1) to 2^(n-1) - 1. For n = 4, this becomes -2^3 to 2^3 - 1, which is -8 to +7. Hence, the representable range in 4-bit 2's complement is -8 to +7.

Option A:
Option A, -7 to +8, reverses the endpoints and allows +8, which is not representable in 4-bit 2's complement. The positive limit is +7, not +8.

Option B:
Option B is correct because it aligns with the general formula for 2's complement ranges and matches 4-bit examples used in digital design. Values outside this range would overflow.

Option C:
Option C, -8 to +8, would contain 17 distinct integers, which cannot fit into 4 bits since 4 bits can represent only 16 distinct patterns.

Option D:
Option D, -7 to +7, is symmetric around zero and describes a sign-magnitude or ones-complement-like range, not the asymmetric 2's complement range.