Student Login

Home›Chapters›Semiconductor Electronics PYQs›Logic Gates›Question 52

JEE Main 2025 (29 Jan, Shift 1) · Logic Gates

Logic Gates question from JEE Main 2025 (29 Jan, Shift 1)

Q52JEE Main 2025 · 29 JAN · SHIFT 1Medium

Logic Gates

For the circuit shown above, equivalent GATE is:

Show answer and solution

Answer: (d)

NAND gives $\overline{AB}$ ; NOR gives $\overline{A+B}$ .
Final NAND: $Y = \overline{\overline{AB}\cdot\overline{A+B}} = AB + (A+B)$ .
$Y = A + B$ (since $AB$ is already included in $A+B$ ).
So the answer is (d) OR gate.

Report a mistake

More questions on Logic Gates

The Boolean expression Y = ABC + AC can be realised with which of the following gate…JEE Main 2025 (4 Apr, Shift 1)
The output of the circuit is low (zero) for: (A) X = 0, Y = 0 (B) X = 0, Y = 1 (C) X = 1…JEE Main 2025 (24 Jan, Shift 2)
Two different logic gates giving output ' 1 ' for the inputs (1, 0) and then for (0, 1)…MHT CET 2024 (15 May, Shift 1)
The Boolean expression for 'X-OR' gate C = (A B) is equal toMHT CET 2024 (3 May, Shift 2)
When the output of an OR gate is applied as input to a NOT gate, then the combination…NEET 2024
The combination of NAND gates is shown in figure (I) and (II). For the given inputs, the…MHT CET 2024 (3 May, Shift 2)
To get the truth table shown from the following logic circuit, the logic gate G should beMHT CET 2024 (16 May, Shift 2)
The Boolean expression for the given combination of logic gates isMHT CET 2024 (16 May, Shift 2)