Q.49 Physics ⭐ 360° Option Elimination Active

NEET 2021 Physics Question 49 Solution

Topic: System of Particles and Rotational Motion · Official NTA Booklet Code M1

Question

From a circular ring of mass 'M' and radius 'R' an arc corresponding to a 908 sector is removed. The moment of inertia of the remaining part of the ring about an axis passing through the centre of the ring and perpendicular to the plane of the ring is 'K' times 'MR2'. Then the value of 'K' is:

✓

Verified Against NTA Final Answer Key

Option (1) : 3/4

📚 System of Particles and Rotational Motion Easy

📋 Given Parameters & Problem Setup

Original Ring Mass M

Original Ring Radius R

Removed Arc Sector Angle 90°

Rotation Axis Through center of ring and perpendicular to its plane

⚡

The 30-Second Exam Hall Shortcut: ⚡ Constant Radius: I = M_remaining × R²

⏱️ Save ~20 seconds

All parts of any circular ring or arc lie at the EXACT same perpendicular distance $R$ from the central perpendicular axis! Removing a $90^\circ$ arc removes $\frac{90}{360} = \frac{1}{4}$ of the mass. The remaining arc has mass $M' = \frac{3}{4}M$ . Thus, $I = M' R^2 = \frac{3}{4}MR^2 \implies K = 3/4$ ! Instant 5-second answer matching Option (1)!

1 Total circumference corresponds to

360^\circ

. A

90^\circ

arc is removed, representing a fraction

\frac{90^\circ}{360^\circ} = \frac{1}{4}

2 Mass of the removed arc:

M_{\text{removed}} = \frac{M}{4}

3 Mass of the remaining arc:

M' = M - \frac{M}{4} = \frac{3}{4}M

4 Every mass element

dm

of the circular arc remains at distance

R

from the central perpendicular axis.

5 Moment of inertia:

I = \int r^2 dm = R^2 \int dm = M' R^2 = \frac{3}{4} M R^2

6 Since

I = K M R^2

K = 3/4

. Match Option (1).

🛡️

Why You Can 100% Trust This Shortcut: Canonical definition of moment of inertia for mass distribution at constant radial distance.

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Determining the Mass of the Remaining Portion

For a uniform circular ring of mass $M$ , the linear mass density is $\lambda = \frac{M}{2\pi R}$ . A complete ring spans $360^\circ$ . The fraction of the ring removed by an arc of $90^\circ$ is:

f_{\text{removed}} = \frac{90^\circ}{360^\circ} = \frac{1}{4} \implies M_{\text{remaining}} = M \left(1 - \frac{1}{4}\right) = \frac{3}{4} M

Step 2

Formulating the Moment of Inertia from First Principles

The axis of rotation passes through the geometric center of the original ring and is perpendicular to its plane. For every infinitesimal mass element $dm$ on the circular arc, its perpendicular distance from this axis is identically equal to the radius $R$ :

I = \int r^2 dm = R^2 \int dm = R^2 \cdot M_{\text{remaining}}

Step 3

Evaluating the Proportionality Factor K

Substitute $M_{\text{remaining}} = \frac{3}{4}M$ :

I = \left(\frac{3}{4}M\right) R^2 = \frac{3}{4} M R^2 \implies K = \frac{3}{4}

⭐

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 1

✓ 100% Correct

Verdict & Scientific Reason:

Exact moment of inertia factor: remaining mass is $3/4 M$ , all at distance $R$ , giving $I = 3/4 M R^2$ .

💡

Why this is airtight: None; textbook moment of inertia derivation.

2 2

✗ Incorrect

Verdict & Scientific Reason:

Result of applying an erroneous $45^\circ$ cut or arbitrary arithmetic.

⚠️

Student Fallacy (Why candidates pick this trap): Incorrect angle fraction calculation.

3 3

✗ Incorrect

Verdict & Scientific Reason:

This is the moment of inertia of the REMOVED sector, not the remaining part.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing the removed part with the remaining part.

4 4

✗ Incorrect

Verdict & Scientific Reason:

Arbitrary fractional guess.

⚠️

Student Fallacy (Why candidates pick this trap): Miscalculating sector mass.

⚠️

NTA Trap Alert: Removed Part vs. Remaining Part Trap

NTA explicitly included Option (3) $1/4$ . Candidates who calculate the fraction of the removed sector ( $90/360 = 1/4$ ) often bubble $1/4$ without subtracting from $1$ to find the REMAINING part ( $1 - 1/4 = 3/4$ ).

🛡️ How to avoid the -1 mark: Read the question carefully: "moment of inertia of the REMAINING part of the ring".

📖

Core NEET Syllabus Standard

NCERT Physics · System of Particles and Rotational Motion Standard Benchmark

Core Scientific Principle Tested:

"For any circular arc of mass $m$ and radius $R$ , the moment of inertia about the central perpendicular axis is $I = m R^2$ ."

Syllabus Focus: Moment of Inertia of Circular Bodies

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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NEET 2021 Physics Question 49 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Constant Radius: I = M_remaining × R²

Step-by-Step Rigorous Workflow & Mathematical Proof

Determining the Mass of the Remaining Portion

Formulating the Moment of Inertia from First Principles

Evaluating the Proportionality Factor K

360° Option Elimination Engine Key Feature

NTA Trap Alert: Removed Part vs. Remaining Part Trap

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — System of Particles and Rotational Motion

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