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)
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