JEE Main 2026 — Physics Modern Physics

2026mcqmedium

Refer to the logic circuit given below. For two inputs $(A=1, B=1)$ and $(A=0, B=1)$ , output $(Y)$ will be __________.

JEE Main 2026 Physics, Semiconductors — question figure

Official previous-year question

Held on 6 Apr 2026 · Verified 6 Jul 2026.

Options

A
$1, 0$ respectively
B
$0, 1$ respectively
C
$0, 0$ respectively
D
$1, 1$ respectively

Did you get this right?

Solution

From the given logic circuit, we can determine the Boolean expression for the output $Y$ .

The output of the NOT gate is $\bar{A}$ .

The inputs to the OR gate are $\bar{A}$ and $B$ , so its output is $\bar{A} + B$ .

The inputs to the AND gate are $\bar{A}$ and $B$ , so its output is $\bar{A} \cdot B$ .

These two outputs are fed into a NOR gate. The final output $Y$ is given by:

$Y = \overline{(\bar{A} + B) + (\bar{A} \cdot B)}$

Simplifying the Boolean expression:

$Y = \overline{\bar{A} + B + \bar{A}B}$

$Y = \overline{\bar{A} + B(1 + \bar{A})}$

$Y = \overline{\bar{A} + B}$

Using De Morgan's laws:

$Y = \overline{\bar{A}} \cdot \bar{B} = A \cdot \bar{B}$

Now, we evaluate $Y$ for the given input combinations:

For $A = 1, B = 1$ :

$Y = 1 \cdot \bar{1} = 1 \cdot 0 = 0$

For $A = 0, B = 1$ :

$Y = 0 \cdot \bar{1} = 0 \cdot 0 = 0$

Therefore, the outputs are $0$ and $0$ respectively.

Answer: $0, 0$ respectively

Your note

JEE Main Modern Physics in other years

Modern Physics 2025 Modern Physics 2024 Modern Physics 2023 Modern Physics 2022 Modern Physics 2021 Modern Physics 2020 Modern Physics 2019 Modern Physics 2018 Modern Physics 2017 Modern Physics 2016 Modern Physics 2015 Modern Physics 2014

Keep practicing Modern Physics

Work through every JEE Main Modern Physics PYQ, year by year.

Browse all Physics topics

Solution

From the given logic circuit, we can determine the Boolean expression for the output $Y$ .

The output of the NOT gate is $\bar{A}$ .

The inputs to the OR gate are $\bar{A}$ and $B$ , so its output is $\bar{A} + B$ .

The inputs to the AND gate are $\bar{A}$ and $B$ , so its output is $\bar{A} \cdot B$ .

These two outputs are fed into a NOR gate. The final output $Y$ is given by:

$Y = \overline{(\bar{A} + B) + (\bar{A} \cdot B)}$

Simplifying the Boolean expression:

$Y = \overline{\bar{A} + B + \bar{A}B}$

$Y = \overline{\bar{A} + B(1 + \bar{A})}$

$Y = \overline{\bar{A} + B}$

Using De Morgan's laws:

$Y = \overline{\bar{A}} \cdot \bar{B} = A \cdot \bar{B}$

Now, we evaluate $Y$ for the given input combinations:

For $A = 1, B = 1$ :

$Y = 1 \cdot \bar{1} = 1 \cdot 0 = 0$

For $A = 0, B = 1$ :

$Y = 0 \cdot \bar{1} = 0 \cdot 0 = 0$

Therefore, the outputs are $0$ and $0$ respectively.

Answer: $0, 0$ respectively

JEE Main 2026 — Physics Modern Physics

Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics

JEE Main 2026 — Physics Modern Physics

Options

Did you get this right?

Solution

Your note

JEE Main Modern Physics in other years

More JEE Main Modern Physics questions

Other Physics topics for JEE Main

Mechanics

Electromagnetism

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Modern Physics