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A Thin Ring Has Mass 0.25 Kg and Radius 0.5 M. Its Moment of Inertia About an Axis Passing Through Its Centre and Perpendicular to Its Plane is _______. - Physics

MCQ

A thin ring has mass 0.25 kg and radius 0.5 m. Its moment of inertia about an axis passing through its centre and perpendicular to its plane is _______.

Options

  • 0. 0625 kg m2

  • 0.625 kg m2

  • 6. 25 kg m2

  • 62. 5 kg m2

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Solution

0.0625 kg m2

 I = mr²
where m is the mass of ring and r is the minimum seperation between axis of rotation and point of observation { e.g., radius of ring }
   here, m = 0.25 Kg 
            r = 0.5 m 
now, I = 0.25 × (0.5)² = 0.25 × 0.25 = 0.0625 Kg.m² 

Concept: Physical Significance of M.I (Moment of Inertia)
  Is there an error in this question or solution?
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