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`        A small object of uniform density Rolls up a curved surface with an initial velocity v.It reaches upto maximum height of 3v^2/4g with respect to initial object.The object is RingDiscSolid sphereHollowsphere `
one year ago

Arun
24739 Points
```							From the law of conservation of energy,mgh = 1/2mv² + 1/2 Iω²I = mk²(where, k= radius of gyration) and, ω = v²/r²by solving the eq.,v²= 2gh/1+(k²/r²)put 'h' = 3v²/4g in this eq. and by solving, you'll get,k²/r² = 1/2which means, given object is a DISC
```
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