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Grade: 11
        
mass of original disc is 9 m and radius r. A small disc of radius r by 3 is removed as shown in figure ..The moment of inertia of remaining disc about an Axis perpendicular to plane of the disc and passing through the centre of disc is
one year ago

Answers : (1)

Khimraj
3008 Points
							
mass of removed part = (9m/\pi r^{2})*\pi (r/3)^{2} = m
I = Itotal – Iremoved
I = 9mr2/2 – (mr2/18 + 4mr2/9)
I = 4mr2
Hope it clears.
one year ago
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