From a disc of radius R and mass M, a circular hole of diameter

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From a disc of radius R and mass M, a circular hole of diameter R, whose rim passes through the centre is cut. What is the moment of inertia of the remaining part of the disc about a perpendicular axis, passing
through the centre ?

1. 15 MR²/32
2. 13 MR²/32
3. 11 MR²/32
4. 9 MR²/32

Correct Option: B

Moment of inertia of complete disc about point 'O'.
I_{Total disc} = MR²/2

Mass of removed disc
M_Removed = M/4 (Mass ∝ area)
Moment of inertia of removed disc about point 'O'.
I_Removed (about same perpendicular axis)
= I_cm + mx²

=	M	(R/2)²	+	M		R		²	=	3MR²
=	4	2	+	4		2			=	32

Therefore the moment of inertia of the remaining part of the disc about a perpendicular axis passing through the centre,
I_{Remaing disc} = I_Total – I_Removed

=	MR²	=	3	MR² =	13	MR²
	2		32		32