JEE Main 2026 — Semiconductors Question with Solution

The LED will glow if it is forward biased, which means the potential at the p-side must be higher than the potential at the n-side.
Let $Y_1$ be the output of the upper NAND gate (p-side) and $Y_2$ be the output of the lower NOR gate (n-side).
For the LED to glow, we need $Y_1=1$ and $Y_2=0$.
The upper circuit consists of two NOT gates (NOR gates with shorted inputs) followed by a NAND gate.
Output of first NOT gate is $\bar{A}$, and second is $\bar{B}$.
$Y_1=\overline{\bar{A}\cdot \bar{B}}=\overline{\bar{A}}+\overline{\bar{B}}=A+B$ (using De Morgan's Law).
The lower circuit is a NOR gate with inputs $C$ and $D$.
$Y_2=\overline{C+D}$.
Condition 1: $Y_1=1⟹A+B=1$. This means at least one of $A$ or $B$ must be 1.
Condition 2: $Y_2=0⟹\overline{C+D}=0⟹C+D=1$. This means at least one of $C$ or $D$ must be 1.
Checking the options:
(1) $A=1,B=1,C=0,D=1$: $Y_1=1+1=1$ and $Y_2=\overline{0+1}=0$. (Satisfied)
(2) $A=0,B=0,C=1,D=1$: $Y_1=0+0=0$. (Not satisfied)
(3) $A=1,B=0,C=0,D=0$: $Y_2=\overline{0+0}=1$. (Not satisfied)
(4) $A=0,B=1,C=0,D=0$: $Y_2=\overline{0+0}=1$. (Not satisfied)
Thus, the correct combination is 1101.