A solid sphere of mass M and radius R having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness I, 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 solid sphere of mass M and radius R having moment of inertia about a

1.	A solid sphere of mass M and radius R having moment of inertia about an axis passing through the centre of mass as I, is recast into a disc of thickness I, 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
Answer» `(2R)/(sqrt(15))` `R sqrt((2)/(15))` `(4R)/(sqrt(15))` `(R)/(4)` Solution :Moment of INERTIA of solid sphere of mass M and RADIUS R about an axis passing through the centre of mass is `I = (2)/(5) MR^(2)` . Let the radius of the disc be r . Moment of inertia of circular disc of radius r and mass M about an axis passing through the centre of mass and perpendicular to its plane`= (1)/(2) Mr^(2)` Using theorem of parallel axes , moment of inertia of disc about its edge is `I. = (1)/(2) Mr^(2) + Mr^(2) = (3)/(2) Mr^(2)` Given `I = I. therefore (2)/(5) MR^(2) = (3)/(2) Mr^(2) or r = (2R)/(sqrt(15))`