Two $4$ bits binary numbers, $A = 1101$ and $B = 1010$ are given in the inputs of a logic circuit shown in figure below. The output $(Y)$ will be: ![JEE Main 2026 Physics, Semiconductors — question figure](https://prepforbharat.s3.ap-south-1.amazonaws.com/exam/615f0e999476412f48314daf/Physics/images/Semiconductors/69d8075381239f6567c4f46d/question_1__q_69d8075381239f6567c4f46d__cdn-question-pool.getmarks.app__04APRS1__q38_d1__175b657270_final_ppt_sync.png)

JEE Main 2026 — Physics Modern Physics

2026mcqeasy

Two $4$ bits binary numbers, $A = 1101$ and $B = 1010$ are given in the inputs of a logic circuit shown in figure below. The output $(Y)$ will be:

JEE Main 2026 Physics, Semiconductors — question figure

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Options

A
$Y = 1101$
B
$Y = 0010$
C
$Y = 0111$
D
$Y = 1000$

Did you get this right?

Solution

From the given logic circuit, the input $A$ passes through a NOT gate, giving $\overline{A}$ .

This output $\overline{A}$ and the input $B$ are then fed into a NAND gate.

The Boolean expression for the final output $Y$ is:

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

Given the $4$ -bit binary inputs:

$A = 1101$

$B = 1010$

First, find the bitwise NOT of $A$ :

$\overline{A} = \overline{1101} = 0010$

Next, find the bitwise AND of $\overline{A}$ and $B$ :

$\overline{A} \cdot B = 0010 \cdot 1010 = 0010$

Finally, apply the NOT operation for the NAND gate:

$Y = \overline{0010} = 1101$

Alternatively, using De Morgan's laws:

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

$Y = 1101 + \overline{1010} = 1101 + 0101 = 1101$

Therefore, the output $Y$ is $1101$ .

Answer: $Y = 1101$

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JEE Main 2026 — Physics Modern Physics

2026mcqeasy

Two $4$ bits binary numbers, $A = 1101$ and $B = 1010$ are given in the inputs of a logic circuit shown in figure below. The output $(Y)$ will be:

JEE Main 2026 Physics, Semiconductors — question figure

Official previous-year question

Held on 4 Apr 2026 · Verified 6 Jul 2026.

Options

A
$Y = 1101$
B
$Y = 0010$
C
$Y = 0111$
D
$Y = 1000$

Did you get this right?

Solution

From the given logic circuit, the input $A$ passes through a NOT gate, giving $\overline{A}$ .

This output $\overline{A}$ and the input $B$ are then fed into a NAND gate.

The Boolean expression for the final output $Y$ is:

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

Given the $4$ -bit binary inputs:

$A = 1101$

$B = 1010$

First, find the bitwise NOT of $A$ :

$\overline{A} = \overline{1101} = 0010$

Next, find the bitwise AND of $\overline{A}$ and $B$ :

$\overline{A} \cdot B = 0010 \cdot 1010 = 0010$

Finally, apply the NOT operation for the NAND gate:

$Y = \overline{0010} = 1101$

Alternatively, using De Morgan's laws:

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

$Y = 1101 + \overline{1010} = 1101 + 0101 = 1101$

Therefore, the output $Y$ is $1101$ .

Answer: $Y = 1101$

Your note

JEE Main Modern Physics in other years

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JEE Main 2026 — Physics Modern Physics

Options

Did you get this right?

Solution

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JEE Main Modern Physics in other years

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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