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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
7 months 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.

7 months 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