Question

# Find the moment of Inertia of a uniform hollow sphere of mass M and radius R, about its diameter 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 = π∫0MR2Sin3θ dθ = 23MR2

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