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        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

Khimraj
3008 Points
							mass of removed part = (9m/$\pi r^{2}$)*$\pi (r/3)^{2}$ = mI = Itotal – IremovedI = 9mr2/2 – (mr2/18 + 4mr2/9)I = 4mr2Hope it clears.

one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions