Q.39 Physics ⭐ 360° Option Elimination Active

NEET 2025 Physics Question 39 Solution

Topic: System of Particles and Rotational Motion · Official NTA Booklet Code 45

Question

A sphere of radius R is cut from a larger solid sphere of radius 2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the Y-axis is:

✓

Verified Against NTA Final Answer Key

Option (3) : \frac{7}{57}

📚 System of Particles and Rotational Motion Medium

📋 Given Parameters & Problem Setup

Large Solid Sphere Radius 2R (centered at origin)

Cut Sphere Radius R (diameter 2R, touching origin and outer edge)

Axis of Rotation Y-axis (passing through center of large sphere)

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The 30-Second Exam Hall Shortcut: ⚡ Mass Scaling & Parallel Axis Ratio Trick

⏱️ Save ~45 seconds

Mass: M_large = 8m, m_small = m. I_large = (2/5)(8m)(2R)² = 64/5 mR². I_small = (2/5)mR² + mR² = 7/5 mR². I_rest = 64/5 - 7/5 = 57/5 mR² ⟹ I_small / I_rest = 7 / 57!

1 Mass scales with volume (radius cubed): M_large / m_small = (2R / R)³ = 8 ⟹ M_large = 8m.

2 Center of smaller sphere is at distance d = R from the Y-axis.

3 Total moment of inertia of large sphere about Y-axis: I_total = (2/5) M (2R)² = (2/5)(8m)(4R²) = (64/5) m R².

4 Moment of inertia of smaller sphere about Y-axis using Parallel Axis Theorem: I_small = I_cm + m d² = (2/5) m R² + m R² = (7/5) m R².

5 Moment of inertia of remaining portion: I_rest = I_total - I_small = (64/5) m R² - (7/5) m R² = (57/5) m R².

6 Required ratio: I_small / I_rest = (7/5) / (57/5) = 7 / 57.

7 Match Option (3) in under 25 seconds!

🛡️

Why You Can 100% Trust This Shortcut: Parallel Axis Theorem (NCERT Physics · System of Particles and Rotational Motion: I = I_cm + Md²).

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Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Mass Relationship from Density

For uniform density ρ, mass is proportional to the cube of the radius:

m = \rho \frac{4}{3}\pi R^3, \quad M = \rho \frac{4}{3}\pi (2R)^3 = 8 m

• The smaller sphere has mass m, and the original large sphere has mass 8m.

Step 2

Calculate Moment of Inertia of Large Sphere about Y-Axis

The Y-axis passes through the center of mass of the large sphere:

I_{total} = \frac{2}{5} M (2R)^2 = \frac{2}{5} (8m) (4R^2) = \frac{64}{5} m R^2

• This is the total moment of inertia before cutting.

Step 3

Calculate Moment of Inertia of Small Sphere about Y-Axis

The small sphere has its center of mass at x = R, y = 0. By the parallel-axis theorem:

I_{small} = I_{cm} + m d^2 = \frac{2}{5} m R^2 + m R^2 = \frac{7}{5} m R^2

• Distance d = R between the small sphere’s centroid and the Y-axis.

Step 4

Calculate Moment of Inertia of Remaining Part and Ratio

Subtract the small sphere’s moment of inertia from the total to find the remaining part:

I_{rest} = I_{total} - I_{small} = \frac{64}{5} m R^2 - \frac{7}{5} m R^2 = \frac{57}{5} m R^2

• Ratio of smaller sphere to rest part:
• \frac{I_{small}}{I_{rest}} = \frac{\frac{7}{5} m R^2}{\frac{57}{5} m R^2} = \frac{7}{57}.
• Matches Option (3) with absolute mathematical rigor.

⭐

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

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Corresponds to mass ratio m / (M - m) = 1/7 or similar inverted mass fraction.

⚠️

Student Fallacy (Why candidates pick this trap): Confusing mass ratio with moment of inertia ratio.

2 B

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Obtained if the large sphere radius is taken as R instead of 2R.

⚠️

Student Fallacy (Why candidates pick this trap): Forgetting that large sphere has radius 2R.

3 C

✓ 100% Correct

Verdict & Scientific Reason:

100% correct. I_small = 7/5 mR², I_rest = 57/5 mR² ⟹ ratio = 7/57.

💡

Why this is airtight: None; flawless parallel-axis theorem application.

4 D

✗ Incorrect

Verdict & Scientific Reason:

Incorrect. Represents the ratio of I_small to the TOTAL sphere (7/64), forgetting to subtract I_small from I_total!

⚠️

Student Fallacy (Why candidates pick this trap): Dividing by I_total instead of I_rest.

⚠️

NTA Trap Alert: Dividing by Total Sphere Instead of "Rest Part"

Option 4 (7/64) is the exact ratio of the small sphere to the WHOLE sphere (I_small / I_total = (7/5) / (64/5) = 7/64). The question specifically asked for the ratio to the "REST PART OF THE SPHERE" (64 - 7 = 57)!

🛡️ How to avoid the -1 mark: Highlight the phrase "rest part" in the question: I_rest = I_total - I_cut!

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Core NEET Syllabus Standard

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

Core Scientific Principle Tested:

"I_{total} = I_{remaining} + I_{removed} with I = I_{cm} + Md²."

Syllabus Focus: Theorem of Parallel Axes and Cavity Problems

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Total is 64, Small is 7 ⟹ Rest is 64 - 7 = 57 ⟹ Ratio = 7 / 57!

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NEET 2025 Physics Question 39 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Mass Scaling & Parallel Axis Ratio Trick

Step-by-Step Rigorous Workflow & Mathematical Proof

Mass Relationship from Density

Calculate Moment of Inertia of Large Sphere about Y-Axis

Calculate Moment of Inertia of Small Sphere about Y-Axis

Calculate Moment of Inertia of Remaining Part and Ratio

360° Option Elimination Engine Key Feature

NTA Trap Alert: Dividing by Total Sphere Instead of "Rest Part"

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

🔗 Related Questions — System of Particles and Rotational Motion

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