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`        Find mi about an axis and posing through center of mass perpendicular to its plane`
one month ago

Sarika Bishnoi
14 Points
```							Let d=mass per unit area         =M/(2a)2-(a)2 = M/3a2Now fill the empty space with a square of +d as well as -d .We will have one square of +d with side 2a and one square of -d with side a.Total moment of inertia= moment of inertia due to                                               square1+ due to square2                                            = M1(L1)2 /6 +M2(L2)2/6                                    =[d(2a)2](2a)2/6+[ -d(a)2](a)2/6                                    =5da4/2                                     5Ma2/6
```
one month 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