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PhysicsNEET 2025Moment of inertia, theorems of axes

System of Particles and Rotational Motion

A sphere of radius RR is cut from a larger solid sphere of radius 2R2R as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the YY-axis is:

Question 39 as printed in the NEET 2025 booklet

  1. A.78\dfrac{7}{8}
  2. B.740\dfrac{7}{40}
  3. C.757\dfrac{7}{57}✓
  4. D.764\dfrac{7}{64}

Solution

The removed sphere has mass M8\frac{M}{8} (radius halved) and its centre is RR from the YY-axis: Ismall=25⋅M8R2+M8R2=740MR2I_{small} = \frac25\cdot\frac{M}{8}R^2 + \frac{M}{8}R^2 = \frac{7}{40}MR^2. The full sphere: 25M(2R)2=85MR2\frac25 M(2R)^2 = \frac{8}{5}MR^2. The rest: 85−740=5740MR2\frac{8}{5} - \frac{7}{40} = \frac{57}{40}MR^2. Ratio =757= \frac{7}{57}.

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