From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed as shown in figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc andpassing through O is

From a circular disc of radius R and mass 9M, a small disc of radius R

1.	From a circular disc of radius R and mass 9M, a small disc of radius R/3 is removed as shown in figure. The moment of inertia of the remaining disc about an axis perpendicular to the plane of the disc andpassing through O is
Answer» `4 MR^(2)` `40/9 MR^(2)` `40 MR^(2)` `37/9 MR^(2)` Solution :MASS per unit area of disc `=(9M)/(piR^(2))` Mass of REMOVED portion of disc `=(9M)/(piR^(2)) xx pi(R/3)^(3) = M` MOMENT of INERTIA of removed portion about an AXIS passing through center of disc and perpendicular to the plane of disc, using theorem of parallel axes is `I_(1)=M/2(R/3)^(2)+1/2(MR^(2))=4MR^(2)`