CBSE · Class 11 · Physics · System of Particles and Rotational MotionThe moment of inertia of a uniform thin rod of mass $M$ and length $L$ about an axis passing through its center and perpendicular to its length is:
Step-by-Step Solution
By integrating $dI = x^2 dm$ from $-L/2$ to $+L/2$, the moment of inertia about the central perpendicular axis is $I = \frac{1}{12}ML^2$.
Detailed Options Breakdown
Option : $\frac{1}{3}ML^2$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of System of Particles and Rotational Motion.
Option 1: $\frac{1}{12}ML^2$ (Correct Answer)
Correct choice. Refer to the step-by-step verified solution guidelines above for details.
Option 2: $\frac{1}{2}ML^2$
Incorrect choice. This distractor represents a common misunderstanding of the core principles of System of Particles and Rotational Motion.
Option 3: $ML^2$
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.