9. A solid sphere of mass M, radius R and having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness t, whose moment of inertia about an axis passing through its edge and perpendicular to its plane remains I. Then, radius of the disc will be(A) 2R/√15(B) R√(2/15)(C) 4R/√15(D) R/4

Dear Mainak

moment of inertia of solid sphere I =2/5 MR²

mass will remain conserve

 let the new radius is R1

so I = MR₁²/2 + MR₁²

       =3/2 MR₁²

so 3/2 MR₁² =2/5 MR²

    R1 =2R/√15

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