MyCompetitiveExams Dynamic 180-question full-length simulator mixed dynamically from all 6 years for 2027 & 2026 aspirants.
The output (Y) of the given logic gate is similar to the output of an/a:
Shorted NAND = NOT(A); Shorted NOR = NOT(B); Output NOR on (Ā, B̄) gives NOT(Ā + B̄) = A · B (AND Gate)!
The top gate is a two-input NAND gate whose inputs are shorted together to receive a common signal A. The Boolean operation of a NAND gate on identical inputs simplifies directly to an inverter (NOT gate):
The bottom gate is a two-input NOR gate whose inputs are shorted together to receive a common signal B. The Boolean operation on identical inputs also simplifies to an inverter (NOT gate):
The signals Y_A = Ā and Y_B = B̄ serve as inputs to the final NOR gate. The output Y is the NOR operation on Ā and B̄:
Applying De Morgan’s law to the complement of a sum yields the product of the complements:
Diagnose every choice: Know exactly why the right option works and why distractors fail.
Represents A + B, whereas the circuit performs A · B.
Boolean algebraic simplification proves Y = A · B identically.
NAND would require an extra inversion at the output (NOT(A · B)).
Only the final physical component is a NOR gate, but its inputs are already inverted.
Hasty students look at the final gate symbol, see a NOR shape, and immediately circle NOR gate without checking the inverted inputs.
"Inverting both inputs to a NOR gate converts its functionality to that of an AND gate (bubbled NOR = AND)."
Test your speed and accuracy in the authentic 200-minute timed NTA simulator.
Spotted a mistake in this solution? Report an error ([email protected])
Cover all 1,140 official NEET questions in 285 days at your own pace. Progress is saved on your device. No login needed.