UGC NET Questions (Paper – 1)

In a binary number, the rightmost bit represents the smallest positional value, 2^0 or 1. Because it contributes the least weight to the overall number, it is called the least significant bit (LSB). This bit is important in operations like parity checking and incrementing. Therefore, the term least significant correctly describes the rightmost bit.

Option A:
Option A names the "most significant" bit, which actually refers to the leftmost bit that has the highest positional value. It denotes the largest weight in the number and is not located on the right. Hence this cannot describe the rightmost bit.

Option B:
Option B is correct because the rightmost bit carries the smallest positional weight and so is least significant. In many algorithms, changes in the LSB affect the number by one unit, reinforcing its role as the least significant bit. This terminology is standard in digital electronics.

Option C:
Option C refers to a "control" bit, which is not a standard name for the rightmost position. Control bits can appear in various positions depending on the protocol or format; they are not inherently tied to the rightmost place.

Option D:
Option D suggests a "status" bit, which is a general term for bits that signal conditions but not necessarily the rightmost one. Status indicators may reside anywhere in a control word or register.

Statements A, B, D and E are correct binary facts. A describes the positional value system in base 2. B is true because an odd number has last bit 1 and an even number has last bit 0. D correctly states the conversion method from binary to decimal. E is true for multiples of 4, whose binary form ends with two zeros. C is false; adding 1 flips only a block of trailing ones and the bit after that, not necessarily all bits. Therefore, A, B, D and E only are correct.

Option A:
Option A is incomplete since it omits E, thereby missing a useful rule for recognising multiples of 4 from binary patterns, which is sometimes tested in reasoning questions.

Option B:
Option B is incorrect because it includes C, wrongly claiming that adding 1 always flips all bits, which contradicts simple examples like 0010 becoming 0011.

Option C:
Option C is wrong since it leaves out A and includes only B, D and E, losing the fundamental statement about powers of 2 that underlies the entire binary system.

Option D:
Option D is correct because it gathers positional value, parity detection, conversion method and the multiple-of-4 criterion while excluding C, which misstates the effect of incrementing a binary number.

The leftmost bit in a binary number carries the largest positional weight because it corresponds to the highest power of 2 in the representation. This makes it the most significant bit (MSB). Its value greatly influences the magnitude of the entire number. Therefore, it is correctly referred to as the most significant bit.

Option A:
Option A labels it "low order," which is usually associated with bits near the right side that carry smaller weights. Using low-order terminology for the leftmost bit contradicts standard conventions in digital design.

Option B:
Option B suggests a parity bit, which is a special bit added to detect errors. A parity bit may be placed at either end of a word and is not inherently the leftmost data bit. Thus, parity does not define the leftmost bit of any binary number.

Option C:
Option C is correct because the leftmost bit has the greatest influence on the numeric value and is therefore called the most significant. In signed representations, this bit may also indicate the sign of the number, reinforcing its significance.

Option D:
Option D uses the term "overflow" bit, which typically refers to a flag indicating arithmetic overflow in a status register. It is not a positional name for any particular bit within a numeric representation.