JEE Main · 2026 · 22 Jan 2026 · Shift 1 · Semiconductors

The two upper gates invert their inputs to give $\bar{A}$ and $\bar{B}$, and these feed a NAND gate, so the potential at the anode side of the diode is $P=\overline{\bar{A}\bar{B}}=A+B$. The lower pair of inputs feed a NOR gate, so the potential at the cathode side is $Q=\overline{C+D}$. The LED conducts only when its anode is high and its cathode low, that is when $A+B=1$ together with $C+D=1$. The inputs $A=1,B=1,C=0,D=1$ make $P=1$ and $Q=0$, so the LED glows for the combination $1101$.