Choose the logic operation for the following circuits:

The correct answer is Option (1) → OR

A NAND gate with both inputs joined acts as a NOT gate: \( \overline{A \cdot A} = \overline{A} \).

Call the two inputs \( A \) and \( B \). The first two NAND gates give \( \overline{A} \) and \( \overline{B} \), which feed the third NAND gate:

    \( \displaystyle y = \overline{\overline{A} \cdot \overline{B}} \)

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

    \( \displaystyle y = A + B \)

This is the OR operation.

    \( A = 0, B = 0 \Rightarrow y = 0 \); \( \ A = 0, B = 1 \Rightarrow y = 1 \)
    \( A = 1, B = 0 \Rightarrow y = 1 \); \( \ A = 1, B = 1 \Rightarrow y = 1 \)

Note: A similar question was also asked in CUET (UG) 2023 on 02-07-2023 (Slot 3).

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):
Logic Gates – “… (iv) NAND Gate This is an AND gate followed by a NOT gate. … NAND gates are also called Universal Gates since by using these gates you can realise other basic gates like OR, AND and NOT (Exercises 14.16 and 14.17).”