UGC NET Questions (Paper – 1)

The numeral 1010 in base b expands as 1×b^3 + 0×b^2 + 1×b + 0. Setting this equal to 10 gives b^3 + b = 10. If we try b = 2, we get 2^3 + 2 = 8 + 2 = 10, which satisfies the equation. Therefore, the base must be 2, so 1010₂ equals 10₁₀.

Option A:
Option A, base 8, would make 1010 equal to 1×8^3 + 1×8 = 512 + 8 = 520. This is much larger than 10, so base 8 is not correct. The equality 1010 = 10 does not hold in base 8.

Option B:
Option B, base 2, gives the expansion 1×2^3 + 1×2 = 8 + 2. This sum equals 10, exactly matching the required decimal value. Hence, base 2 is the correct base for which 1010 represents ten.

Option C:
Option C, base 4, yields 1×4^3 + 1×4 = 64 + 4 = 68. This differs greatly from 10, so base 4 does not give the desired equality. Therefore, this option cannot be correct.

Option D:
Option D, base 5, makes 1010 equal to 1×5^3 + 1×5 = 125 + 5 = 130. Since 130 is not equal to 10, base 5 also fails to satisfy the condition.

Four hexadecimal digits range from 0000₁₆ to FFFF₁₆, and FFFF corresponds to 16⁴ − 1. Evaluating 16⁴ gives 65536, so subtracting 1 yields 65535. Therefore, 65535 is the largest decimal value representable with exactly four hex digits. This follows the general pattern that an n-digit base-b number has a maximum value of bⁿ − 1.

Option A:
16383 is the maximum value for 14 bits in binary, but it does not correspond to four hex digits. It is associated with the range up to 2¹⁴ − 1, not with 16⁴ − 1. Hence this option is not appropriate for four hexadecimal positions.

Option B:
32767 is the maximum value for 15 bits in binary (2¹⁵ − 1) and is often seen for signed 16-bit integers. However, it does not match the range of four hexadecimal digits, which is governed by powers of 16, not 2¹⁵.

Option C:
65534 is one less than 65535, so it would correspond to FFFE₁₆ rather than FFFF₁₆. While it is extremely close, it is not the maximum possible value for four hex digits.

Option D:
65535 equals 16⁴ − 1 and is represented as FFFF₁₆. This is the largest value achievable when all four hexadecimal positions are set to their maximum digit F, satisfying the condition of the question.

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.

To represent all integers from 0 to 10⁶ inclusive, we need at least 10⁶ + 1 distinct bit patterns. In binary, n bits can represent 2ⁿ distinct values, so we require 2ⁿ ≥ 10⁶ + 1. Since 2¹⁹ = 524288 < 10⁶ + 1 and 2²⁰ = 1048576 ≥ 10⁶ + 1, we must take n = 20. Therefore, 20 bits are the minimum needed, making option C correct. Option A: Option A, 19 bits, can represent only 524288 values, fewer than the 1,000,001 needed (0 through 10⁶). So it is insufficient. Option B: Option B, 18 bits, represents an even smaller range, so it cannot work. Option C: Option C is the smallest n with 2ⁿ ≥ 1,000,001, so it is minimal and correct. Option D: Option D, 21 bits, is sufficient but not minimal since 20 bits already cover the required range.

In a positional system, the same digit can represent different values depending on where it appears. Its weight is determined by its position relative to the radix point. Each position corresponds to a specific power of the base. Therefore, the weight of a digit depends primarily on its position in the numeral.

Option A:
Option A, value, is the digit itself, but weight is not determined by digit alone; position controls the power of the base used.

Option B:
Option B is correct because each position has a place value (base^0, base^1, base^2, etc.), which determines the weight.

Option C:
Option C, symbol, is another way to refer to the digit; it does not determine the positional weight by itself.

Option D:
Option D, base, affects the system globally, but the specific weight of a digit in a numeral is determined directly by its position (as a power of the base).