JEE Main 2026 — Physics Modern Physics

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 $Y_1$Y1​ be the output of the upper NAND gate (p-side) and $Y_2$Y2​ be the output of the lower NOR gate (n-side).

For the LED to glow, we need $Y_1 = 1$Y1​=1 and $Y_2 = 0$Y2​=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 $\bar{A}$Aˉ, and second is $\bar{B}$Bˉ.

$Y_1 = \overline{\bar{A} \cdot \bar{B}} = \overline{\bar{A}} + \overline{\bar{B}} = A + B$Y1​=Aˉ⋅Bˉ=Aˉ+Bˉ=A+B (using De Morgan's Law).

The lower circuit is a NOR gate with inputs $C$C and $D$D.

$Y_2 = \overline{C + D}$Y2​=C+D​.

Condition 1: $Y_1 = 1 \implies A + B = 1$Y1​=1⟹A+B=1. This means at least one of $A$A or $B$B must be 1.

Condition 2: $Y_2 = 0 \implies \overline{C + D} = 0 \implies C + D = 1$Y2​=0⟹C+D​=0⟹C+D=1. This means at least one of $C$C or $D$D must be 1.

Checking the options:

(1) $A=1, B=1, C=0, D=1$A=1,B=1,C=0,D=1: $Y_1 = 1+1 = 1$Y1​=1+1=1 and $Y_2 = \overline{0+1} = 0$Y2​=0+1​=0. (Satisfied)

(2) $A=0, B=0, C=1, D=1$A=0,B=0,C=1,D=1: $Y_1 = 0+0 = 0$Y1​=0+0=0. (Not satisfied)

(3) $A=1, B=0, C=0, D=0$A=1,B=0,C=0,D=0: $Y_2 = \overline{0+0} = 1$Y2​=0+0​=1. (Not satisfied)

(4) $A=0, B=1, C=0, D=0$A=0,B=1,C=0,D=0: $Y_2 = \overline{0+0} = 1$Y2​=0+0​=1. (Not satisfied)

Thus, the correct combination is 1101.