Q.180 Physics ⭐ 360° Option Elimination Active

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NEET 2020 Physics Question 180 Solution

Topic: Semiconductor Electronics: Materials, Devices and Simple Circuits · Official NTA Booklet Code E1

Question

For the logic circuit shown, the truth table is:

✓

Verified Against NTA Final Answer Key

Option (1) : Truth table for AND gate: Output Y is 1 only when A=1 and B=1

📚 Semiconductor Electronics: Materials, Devices and Simple Circuits Easy

📋 Given Parameters & Problem Setup

Input Inverters NOT gates giving A-bar and B-bar

Output Gate NOR gate receiving A-bar and B-bar

⚡

The 30-Second Exam Hall Shortcut: ⚡ "De Morgan Theorem: Y = (A-bar + B-bar)-bar = A · B. The circuit is an AND gate! The truth table outputs 1 only when both inputs are 1. Option (1) / A is the unique match!"

⏱️ 20 seconds

Inputs A and B pass through inverters to yield NOT(A) and NOT(B). These enter a NOR gate, yielding Y = NOT[NOT(A) OR NOT(B)]. By De Morgan's Theorem, NOT(X OR Y) = NOT(X) AND NOT(Y). Here with double negation, Y = A AND B! The circuit is an AND gate, whose truth table has output 1 exclusively when both inputs are 1 (A=1, B=1). Truth table option (1) (Option A) is the exact match!

1 Inputs to NOR gate:

A' = \bar{A}

and

B' = \bar{B}

2 Output of NOR gate:

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

3 Apply De Morgan's Law:

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

4 Circuit is a standard 2-input AND gate.

5 Truth table rows: (0,0)->0; (0,1)->0; (1,0)->0; (1,1)->1.

6 Select Option (1).

🛡️

Why You Can 100% Trust This Shortcut: NCERT Physics · Semiconductor Electronics (Digital Electronics and Logic Gates)

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Trace Logic Signals Through the Circuit

- Input $A$ passes through a NOT gate, producing $\bar{A}$ . - Input $B$ passes through a NOT gate, producing $\bar{B}$ . - Signals $\bar{A}$ and $\bar{B}$ enter a two-input NOR gate.

Step 2

Formulate Boolean Output Expression

The output of a NOR gate is the inverted sum of its inputs: $Y = \overline{\bar{A} + \bar{B}}$ .

Step 3

Apply De Morgan’s First Theorem

According to De Morgan's Theorem: $\overline{X + Y} = \bar{X} \cdot \bar{Y}$ . Letting $X = \bar{A}$ and $Y = \bar{B}$ : $Y = \overline{(\bar{A})} \cdot \overline{(\bar{B})}$ Since the double inversion of a Boolean variable returns the variable itself ( $\overline{\bar{A}} = A$ ): $Y = A \cdot B$ . The composite circuit functions as an AND gate.

Step 4

Construct and Match Truth Table

- When $A = 0, B = 0 \implies Y = 0 \cdot 0 = 0$ - When $A = 0, B = 1 \implies Y = 0 \cdot 1 = 0$ - When $A = 1, B = 0 \implies Y = 1 \cdot 0 = 0$ - When $A = 1, B = 1 \implies Y = 1 \cdot 1 = 1$ This matches Table (1) (Option A).

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan

1 A

✓ 100% Correct

Verdict & Scientific Reason:

Accurate truth table for an AND gate realized by inverting inputs before a NOR gate.

💡

Why this is airtight: None; exact Boolean reduction.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; this is the truth table for an OR gate (Y = A + B).

⚠️

Student Fallacy (Why candidates pick this trap): Confusing AND with OR logic.

3 C

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; this is the truth table for a NAND gate.

⚠️

Student Fallacy (Why candidates pick this trap): Adding an extra unintended inversion.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect; this is the truth table for a basic NOR gate without input inverters.

⚠️

Student Fallacy (Why candidates pick this trap): Ignoring the two input NOT gates.

⚠️

NTA Trap Alert: De Morgan Double Inversion

NOT(NOT A OR NOT B) = A AND B, NOT A OR B. Double negation restores the un-complemented variables.

🛡️ How to avoid the -1 mark: Substitute a test case: A=1, B=1 => inverters give 0, 0 => NOR(0, 0) = 1. Only Table (1) gives 1 for (1,1)!

📖

Core NEET Syllabus Standard

Physics · Semiconductor Electronics: Materials, Devices and Simple Circuits Standard Benchmark

Core Scientific Principle Tested:

Syllabus Focus: De Morgan Realization of AND Gate using Inverters and NOR Gate

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2020 Physics Question 180 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ "De Morgan Theorem: Y = (A-bar + B-bar)-bar = A · B. The circuit is an AND gate! The truth table outputs 1 only when both inputs are 1. Option (1) / A is the unique match!"

Step-by-Step Rigorous Workflow & Mathematical Proof

Trace Logic Signals Through the Circuit

Formulate Boolean Output Expression

Apply De Morgan’s First Theorem

Construct and Match Truth Table

360° Option Elimination Engine Key Feature

NTA Trap Alert: De Morgan Double Inversion

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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