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Question

Find the moment of Inertia of a uniform hollow sphere of mass M and radius R, about its diameter


  1. MR2


Solution

The correct option is C


Let us consider a small area element of Rdθ, at an angle θ with x axis as shown in figure.

It forms a ring of radius R Sinθ.

The width of this ring is Rdθ and its periphery is 2πR sinθ. Hence, the area of the ring = (2πRsinθ)(Rdθ)

Mass per unit Area of the given sphere = M4πR2

Mass of ring element = (M4πR2)(2π RSinθ)(Rdθ)

                                                                  = M2Sinθ dθ

Now, the moment of Inertia of this elemental Ring about OX is

dI = (M2Sinθ dθ)(RSinθ)2

        = M2R2Sin3θ dθ

As θ increases from 0 to π, the elemental ring cover the whole spherical Hollow surface.

Therefore, I = π0MR2Sin3θ dθ = 23MR2

 

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