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Moment of inertia (M.I.) of four bodies, having same mass and radius, are reported as ; I1 = M.I. of thin circular ring about its diameter. I2 = M.I. of circular disc about an axis perpendicular to the disc and going through the centre. I3 = M.I. of solid cylinder about its axis and I4 = M.I. of solid sphere about its diameter. Then :
- I1 + I3 < I2 + I4
- I1 + I2 = I3 + 5/2 I4
- I1 = I2 = I3 > I4
- I1 = I2 = I3 < I4
Correct answer: I1 = I2 = I3 > I4
Solution
The moment of inertia of a thin circular ring, a circular disc, and a solid cylinder about their respective axes are all greater than that of a solid sphere, which has a more compact mass distribution. This is due to the fact that the sphere's mass is distributed more evenly around its center, resulting in a lower moment of inertia compared to the other shapes.
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