JEE Main 2025 · 29 January · morning · Single correct · +4 / −1
For the circuit shown above, equivalent GATE is :
- (a) NOT gate
- (b) NAND gate
- (c) AND gate
- (d) OR gate
Answer and solution
Answer (d) OR gate
Solution
Physics sir · 2 steps
Write what each gate outputs, feed both into the last NAND, and test all four input pairs.
- Read the circuit. The top gate is a NAND of A and B, giving A̅B̅. The bottom gate is a NOR of the same inputs, giving A̅+̅B̅. Both outputs go into a final NAND gate.
Y = A̅B̅·A̅+̅B̅̅ = AB + (A+B) = A + B
A, B NAND A̅B̅ NOR A̅+̅B̅ Y 0, 0 1 1 0 0, 1 1 0 1 1, 0 1 0 1 1, 1 0 0 1 - So. Y is 0 only when both inputs are 0 and 1 otherwise. That is the truth table of an OR gate, option (d).
The idea to keep. For a combination of gates, write each gate's output, then simplify with De Morgan (X̅·Y̅̅ = X + Y). Confirm with a four-row truth table.
Answer from the NTA final answer key. Solution by PrepWiser.