The given combination represents the following gate:

The correct answer is Option (1) → OR

In the figure, each input A and B goes to both inputs of a NAND gate, and the two outputs go to a third NAND gate. A NAND gate with its inputs joined acts as a NOT gate:

    \( Y_1 = \overline{A\cdot A} = \overline{A},\qquad Y_2 = \overline{B\cdot B} = \overline{B} \)

The output NAND gate then gives

    \( Y = \overline{Y_1\cdot Y_2} = \overline{\overline{A}\cdot\overline{B}} \)

By De Morgan’s theorem, \( \overline{\overline{A}\cdot\overline{B}} = A + B \), so

    \( Y = A + B \)

which is the Boolean expression of an OR gate. Check: \( Y = 0 \) only for \( A = B = 0 \), and \( Y = 1 \) for the other three input combinations.

Note: This question is from the pre-rationalised NCERT book (2022-23). The topic was removed from the latest NCERT book (2026-27), and it also appears to be out of the CUET 2026 syllabus.

NCERT (Pre-rationalised edition, 2022-23):
Exercises – …