M and R be the mass and radius of a disc. A small disc of radius 
is removed from the bigger disc as shown in figure. The moment of inertia of remaining part of bigger disc about an axis 
passing through the center 
and perpendicular to the plane of disc is 
. The value of 
is __________.
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
(9)
Moment of Inertia of the Full Disc:
The moment of inertia (M.I.) of the entire disc without any cavity is given by:
Mass of the Removed Disc:
The mass of the removed disc, which is of radius 
, is calculated as:
Mass of removed disc 
Moment of Inertia of the Removed Disc:
The moment of inertia of the removed disc involves two parts: about its center and due to its position.
About its center:
Due to its position (distance from 0 to the new center of the small disc):
The distance is 
, so:
Total M.I. of removed disc:
Moment of Inertia of the Remaining Part:
Subtract the moment of inertia of the removed disc from the full disc:
Thus, the value of 
is 9.
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