A disc has mass 9m. A hole of radius 
is cut from it as shown in the figure. The moment of inertia of the remaining part about an axis passing through the centre ‘O’ of the disc and perpendicular to the plane of the disc is : 
A
8 mR 2
B
4 mR 2
C
mR 2
D
mR 2
Text Solution
Verified by ExpertsThe correct answer is:
B
Ι 0 = Ι 1 – Ι 2
where Ι 1 = (M. Ι . of full disc about O)
Ι 2 (M. Ι . of small removed disc about O)
since mass α area
= 
= 
∴ mass of cut disc = m
∴ Ι 0 = 
– m 
(by theorem of parallel axis.)
= 
– mR 2 = 
– 
= 
= 
4mR 2 .
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