UGC NET Questions (Paper – 1)

The bits after the binary point correspond to 1/2, 1/4, 1/8, and 1/16. In 0.1011₂, we have 1×1/2 + 0×1/4 + 1×1/8 + 1×1/16 = 0.5 + 0 + 0.125 + 0.0625 = 0.6875. Thus, the decimal value of 0.1011₂ is 0.6875.

Option A:
Option A correctly sums the contributions of the 2⁻¹, 2⁻³, and 2⁻⁴ places and omits the 2⁻² place because its bit is zero. The resulting 0.6875 accurately reflects the binary pattern.

Option B:
Option B, 0.625, equals 0.5 + 0.125 and would correspond to 0.101₂, which has only three fractional bits. It neglects the contribution of the 2⁻⁴ bit.

Option C:
Option C, 0.75, corresponds to 0.11₂ (0.5 + 0.25) and does not match the given bit arrangement.

Option D:
Option D, 0.5625, equals 0.5 + 0.0625, which would correspond to 0.1001₂. This ignores the 2⁻³ bit present in 0.1011₂.

To convert 0.625 to binary, we multiply the fraction repeatedly by 2 and record integer parts. 0.625×2 = 1.25 gives 1, leaving 0.25; 0.25×2 = 0.5 gives 0, leaving 0.5; 0.5×2 = 1.0 gives 1, leaving 0. Thus, the bits after the point are 1,0,1, yielding 0.101₂. Therefore, 0.625₁₀ equals 0.101₂.

Option A:
Option A correctly applies the fractional conversion algorithm and identifies the sequence of integer parts as 1, 0, and 1. This precisely encodes the contributions 1/2 and 1/8, which sum to 0.625, making 0.101₂ the correct binary fraction.

Option B:
Option B, 0.011₂, corresponds to 0/2 + 1/4 + 1/8 = 0.375, which is lower than 0.625 and misses the 1/2 component. Thus, it does not match the given decimal fraction.

Option C:
Option C, 0.110₂, equals 1/2 + 1/4 = 0.75, which overshoots 0.625. This sequence of bits adds too much fractional value, making it incorrect for 0.625.

Option D:
Option D, 0.1001₂, represents 1/2 + 0/4 + 0/8 + 1/16 = 0.5625, which is close but still less than 0.625. It arises from a mistaken final step in the conversion algorithm.

The integer part 111₂ equals 1×4 + 1×2 + 1×1 = 7. The fractional part .001₂ equals 0×1/2 + 0×1/4 + 1×1/8 = 1/8 = 0.125. Adding these components yields 7 + 0.125 = 7.125. Therefore, the decimal equivalent of 111.001₂ is 7.125.

Option A:
Option A correctly computes both the integer and fractional contributions and combines them. It uses the appropriate negative powers of 2 for the bits after the point, giving a precise decimal value of 7.125.

Option B:
Option B, 7.25, would require the fractional part to be .01₂ (1/4) rather than .001₂. This misinterprets which bit is set in the fractional portion and hence is not correct.

Option C:
Option C, 7.5, corresponds to a fractional part of .1₂, where only the 1/2 place is set. This does not align with .001₂, which has only the 1/8 position set.

Option D:
Option D, 7.375, equals 7 + 0.375, which would correspond to fractional bits .011₂. Since the given binary has only the last fractional bit as 1, this value is inconsistent.