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⚛️ Physics  ·  System of Particles and Rotational Motion  ·  NEET & JEE

Using the parallel axis theorem, the MI of a rod of mass M, length L about one end is ML²/3. What is the MI about the centre?

Answer: ML²/12.

  • A ML²/12
  • B ML²/6
  • C ML²/3
  • D ML²/4

Correct answer: A. ML²/12

Explanation: I<sub>end</sub> = I<sub>cm</sub> + M(L/2)². So ML²/3 = I<sub>cm</sub> + ML²/4 → I<sub>cm</sub> = ML²/3 - ML²/4 = ML²/12.

axis OPr⊥Fττ = r⊥ × F = r × F × sinθ

Torque about axis O equals the force F multiplied by the perpendicular distance r⊥ from the axis to the line of action of the force, producing rotation.

Concept context

Moment of inertia, torque, angular momentum, rolling motion. Very high JEE weightage.

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