A sphere of radius is cut from a larger solid sphere of radius 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 Y-axis is [2025]

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Solution

Q.
A sphere of radius $R$ is cut from a larger solid sphere of radius $2 R$ 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 Y-axis is [2025]

1
$\frac{7}{57}$

2
$\frac{7}{64}$

3
$\frac{7}{8}$

4
$\frac{7}{40}$

Ans.
(1)

Moment of inertia of a solid sphere of mass M and radius R is

   $I = \frac{2}{5} M R^{2}$

Mass of smaller sphere, $M_{S} = \frac{M}{8}$

Mass of remaining part, $M_{r} = M - \frac{M}{8} = \frac{7 M}{8}$

Using parallel-axes theorem, moment of inertia of smaller sphere about y-axis,

   $I_{S} = \frac{2}{5} M_{S} R^{2} + M_{S} R^{2} = (\frac{2}{5} + 1) M_{S} R^{2} = \frac{7}{5} M_{S} R^{2}$

or   $I_{S} = \frac{7}{5} \times \frac{M}{8} \times R^{2} = \frac{7}{40} M R^{2}$

Moment of inertia of the larger part of the sphere about y-axis,

$I_{L} = \frac{2}{5} M {(2 R)}^{2} = \frac{8}{5} M R^{2}$

Now, moment of inertia of the rest part of the sphere is

$I_{r} = I_{L} - I_{S} = \frac{8}{5} M R^{2} - \frac{7}{40} M R^{2} = \frac{57}{40} M R^{2}$

$∴$    $\frac{I_{S}}{I_{r}} = \frac{\frac{7}{40} M R^{2}}{\frac{57}{40} M R^{2}} = \frac{7}{57}$

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