MP Board · Class 11 · Physics · System of Particles and Rotational MotionThe moment of inertia of a circular ring of mass $M$ and radius $R$ about an axis passing through its center and perpendicular to its plane is:
Step-by-Step Solution
For a ring, all particles are located at distance $R$ from the central perpendicular axis. Thus, $I = \sum m r^2 = M R^2$.
Detailed Options Breakdown
Option : $\frac{1}{2}MR^2$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of System of Particles and Rotational Motion.
Option 1: $MR^2$ (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 2: $\frac{2}{5}MR^2$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of System of Particles and Rotational Motion.
Option 3: $\frac{1}{4}MR^2$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of System of Particles and Rotational Motion.
💡 Study Guide: This question tests core syllabus concepts from System of Particles and Rotational Motion. For formulas, key summaries, and mock exam reference guides, read the full System of Particles and Rotational Motion Revision Notes.