A solid sphere of radius R and a hollow sphere of the same mass M and same radius R are released fro

Q: What is NEET?

NEET is a competitive examination conducted in India. This question is from the 2025 NEET Physics paper.

Q: How can I prepare for NEET Physics?

To prepare for NEET Physics, practice previous year questions (PYQs), understand concepts thoroughly, and solve mock tests regularly. This 2025 question covers System of Particles and Rotational Motion topic.

Q: What is the difficulty level of this question?

This NEET 2025 Physics question has a easy difficulty level and is marked as an important question and is one of the most repeated questions.

Q: Where can I find more NEET previous year questions?

You can find more NEET previous year questions on AspirantMitraa. Browse by subject, topic, year, and difficulty level to practice effectively.

Question

A solid sphere of radius R and a hollow sphere of the same mass M and same radius R are released from the same height on a rough inclined plane. Which sphere reaches the bottom first?

Accepted Answer

Correct Answer: B — Solid sphere reaches firstStep-by-step Solution:For rolling without slipping on an inclined plane, the acceleration is:a = g sinθ / (1 + I/(MR²))For solid sphere: I = (2/5)MR² → I/(MR²) = 2/5asolid = g sinθ / (1 + 2/5) = g sinθ / (7/5) = 5g sinθ/7For hollow sphere: I = (2/3)MR² → I/(MR²) = 2/3ahollow = g sinθ / (1 + 2/3) = g sinθ / (5/3) = 3g sinθ/5Comparing: 5/7 > 3/5 (since 25 > 21)asolid > ahollow → Solid sphere reaches the bottom firstWhy Hollow sphere reaches first (A) is wrong: Hollow sphere has higher moment of inertia (more mass at periphery), so it has lower translational acceleration during rolling.Why C is wrong: They have different moments of inertia, so different accelerations during rolling. They only reach simultaneously in sliding (frictionless) conditions.Why D is wrong: As long as rolling without slipping occurs, the coefficient of friction does not affect the acceleration. The result is independent of µ (provided µ is sufficient for rolling).Key Concepts from System of Particles and Rotational Motion (NCERT Class 11, Chapter 7):Acceleration of a rolling body: a = g sinθ/(1 + k²/R²), where k is the radius of gyration (I = Mk²).Solid sphere: k²/R² = 2/5. Hollow sphere: k²/R² = 2/3. Disc: k²/R² = 1/2. Ring: k²/R² = 1.Order from highest to lowest acceleration (rolling): solid sphere > disc > hollow sphere > ring.The object with the smallest moment of inertia (relative to MR²) reaches the bottom first when rolling.

A solid sphere of radius R and a hollow sphere of the same mass M and same radius R are released from the same height on a rough inclined plane. Which sphere reaches the bottom first?

Master NEET with Our Test Series

More Questions from System of Particles and Rotational Motion

Frequently Asked Questions

A solid sphere of radius R and a hollow sphere of the same mass M and same radius R are released from the same height on a rough inclined plane. Which sphere reaches the bottom first?

Master NEET with Our Test Series

More Questions from System of Particles and Rotational Motion

Frequently Asked Questions