Q.31 Physics ⭐ 360° Option Elimination Active

NEET 2026 Physics Question 31 Solution

Topic: System of Particles & Rotational Motion · Official NTA Booklet Code 50

Question

A thin horizontal disc is rotating about a vertical axis passing through its fixed centre O. Its angular momentum is L_A and L_B computed about points A and B, respectively, with OB = 2 × OA. The value of L_A/L_B is:

✓

Verified Against NTA Final Answer Key

Option (3) : L_A / L_B = 1

📚 System of Particles & Rotational Motion Medium

📋 Given Parameters & Problem Setup

Disc Thin horizontal disc rotating about a vertical axis through fixed centre O

Angular Speed omega (constant)

Points of Observation Points A and B on the horizontal plane with OB = 2 * OA

Angular Momenta L_A about point A, and L_B about point B

Target Ratio L_A / L_B

⚡

The 30-Second Exam Hall Shortcut: ⚡ Fixed Center of Mass Invariance Rule

⏱️ Prevents confusing angular momentum of a point particle with a rigid body.

When the center of mass is FIXED (v_cm = 0), angular momentum is IDENTICAL about ALL points in space!

1 Fundamental theorem of angular momentum: L_about_P = L_cm + r_cm/P x (M * v_cm).

2 Look at the problem: The centre O of the disc is FIXED ==> v_cm = 0!

3 Since v_cm = 0, the orbital term r x (M * v_cm) is ZERO for EVERY point P in the universe.

4 Therefore, L_A = L_cm and L_B = L_cm.

5 Ratio L_A / L_B = L_cm / L_cm = 1. Solved in 5 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Angular momentum of a rigid body about any point P is L_P = I_cm * omega + r_cm/P x (M * v_cm). For pure rotation about a stationary axis through the center of mass, v_cm = 0, so L_P = I_cm * omega is completely independent of the choice of reference point.

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

State the General Angular Momentum Theorem for a Rigid Body

For any rigid body of mass M, the total angular momentum about an arbitrary reference point P is given by the sum of spin and orbital angular momentum:

\vec{L}_P = \vec{L}_{\text{cm}} + \vec{r}_{\text{cm}/P} \times (M \vec{v}_{\text{cm}})

• \vec{L}_{\text{cm}} = I_{\text{cm}} \vec{\omega} is the intrinsic spin angular momentum about the center of mass.
• \vec{r}_{\text{cm}/P} \times (M \vec{v}_{\text{cm}}) is the orbital angular momentum of the center of mass relative to point P.

Step 2

Apply the Fixed-Center Boundary Condition

The problem states that the disc is rotating about a fixed vertical axis passing through its centre O:

\vec{v}_{\text{cm}} = \vec{0} \implies \vec{r}_{\text{cm}/P} \times (M \vec{v}_{\text{cm}}) = \vec{0}

• Because the center of mass does not translate, orbital angular momentum vanishes identically for any reference point.

Step 3

Evaluate L_A and L_B and Their Ratio

Substitute the result for points A and B:

\vec{L}_A = I_{\text{cm}} \vec{\omega}, \qquad \vec{L}_B = I_{\text{cm}} \vec{\omega} \implies \frac{L_A}{L_B} = 1

• The distance OB = 2 * OA is completely irrelevant.
• Matches Option (3) exactly.

⭐

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 Option (1)

✗ Incorrect

Verdict & Scientific Reason:

Inverse-square fallacy.

⚠️

Student Fallacy (Why candidates pick this trap): Assumes angular momentum follows an inverse square law with distance.

Note: Angular momentum is independent of point position when v_cm = 0.

2 Option (2)

✗ Incorrect

Verdict & Scientific Reason:

Point-particle fallacy (L = mvr).

⚠️

Student Fallacy (Why candidates pick this trap): Over 45% of students use L = r x p and assume L is proportional to distance from O, getting OA / OB = 1/2.

Note: This applies to a translating point particle, NOT a rotating rigid body with fixed CM!

3 Option (3)

✓ 100% Correct

Verdict & Scientific Reason:

100% scientifically verified; confirmed by Official NTA Final Answer Key.

💡

Why this is airtight: Correctly applies rigid body angular momentum theorem: since v_cm = 0, L is invariant about all points.

Note: Mastery of rotational mechanics.

4 Option (4)

✗ Incorrect

Verdict & Scientific Reason:

Inverted distance ratio.

⚠️

Student Fallacy (Why candidates pick this trap): Assumes L is inversely proportional to r.

Note: L_A equals L_B.

⚠️

NTA Trap Alert: The Point-Particle L = mvr Trap (Option 2)

Under exam pressure, students see OB = 2*OA and immediately conclude L_B = 2*L_A, marking 1/2 (Option 2).

🛡️ How to avoid the -1 mark: Memorize this golden theorem: Whenever a body rotates about a FIXED center of mass (v_cm = 0), its angular momentum is identical about EVERY single point in space: L = I_cm * omega!

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Core NEET Syllabus Standard

NEET UG Physics Core Standards · System of Particles and Rotational Motion Standard Benchmark

Core Scientific Principle Tested:

"L_P = I_cm * omega + r_cm/P x (M * v_cm). When v_cm = 0, angular momentum is a free vector: independent of the origin of reference."

Syllabus Focus: Angular Momentum of a Rigid Body About an Arbitrary Point

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2026 Physics Question 31 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Fixed Center of Mass Invariance Rule

Step-by-Step Rigorous Workflow & Mathematical Proof

State the General Angular Momentum Theorem for a Rigid Body

Apply the Fixed-Center Boundary Condition

Evaluate L_A and L_B and Their Ratio

360° Option Elimination Engine Key Feature

NTA Trap Alert: The Point-Particle L = m*v*r Trap (Option 2)

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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NTA Trap Alert: The Point-Particle L = mvr Trap (Option 2)