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Question

The length of a solid cylinder is $$4.5$$ times its radius and $$I$$ is the moment of inertia about its natural axis. This solid cylinder is re-casted into a solid sphere, then the moment of inertia of solid sphere about an axis passing through its centre is :


A
5I9
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B
10I9
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C
9I5
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D
9I10
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Solution

The correct option is C $$\dfrac{9I}{5}$$
Let $$m$$ be the mass of the solid cylinder and $$r$$ be its radius.
So MI about its natural axis is $$I=mr^2/2$$
Height of cylinder $$h=4.5r$$
When cylinder is recasted into solid sphere, the radius of the solid sphere $$R$$ is given by $$4\pi R^3/3=\pi r^2\times4.5r\Rightarrow R^3=27r^3/8\Rightarrow R=3r/2$$(as volume and mass remains same.)
So MI of solid sphere is $$2mR^2/5=2m(3r/2)^2/5=9mr^2/10=9I/5$$

Physics

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