UGC NET Questions (Paper – 1)

101011110101₂ has 1s at positions 11, 9, 7, 6, 5, 4, 2, and 0. So value = 2¹¹ + 2⁹ + 2⁷ + 2⁶ + 2⁵ + 2⁴ + 2² + 2⁰ = 2048 + 512 + 128 + 64 + 32 + 16 + 4 + 1 = 2805.

Option A:
Option A is correct because summing the powers of 2 for all set bits gives exactly 2805.

Option B:
Option B (3019) would require additional higher-order bits or different set-bit positions.

Option C:
Option C (3442) corresponds to a different binary pattern with larger contributions than present here.

Option D:
Option D (3285) also corresponds to a different set of bit positions and does not match this binary numeral.