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Question

Moment of inertia of solid sphere about its diameter is I. If that sphere is recast into 8 identical small spheres, then the moment of inertia of a small sphere about its diameter is:

A
I8
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B
I16
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C
I24
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D
I32
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Solution

The correct option is C I32
Let Mass and radius of bigger sphere be M and R.
So MI is I=2MR2/5
As this sphere is recast into 8 smaller spheres, the mass of smaller spheres is M/8 and radius is (43πR3/43π8)1/3=R/2
So MI of the smaller sphere is 2(M/8)(R/2)2/5=2MR2/5×32=I/32

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