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A uniform chain of length lvand mass m lies on a smooth table .a very small lart of it this chain hangs from table .it begins to fall freely under the weight of hanging chainvfind the velocity of chain when the length of hanging part becomes y


4 years ago

Manas Shukla
102 Points
							Length of the chain = LGiven the chain is of uniform density and shape.Let mass of chain = MHanging part of the chain = ydepth of centre of mass of the chain of hanging part = y/2Since the chain is rigid it will move with the same speed.Mass of the hanging part of the chain = My/LLets balance Energy when hanging part becomes y.$\frac{1}{2} \times M\times v^{2} = \frac{My}{L}\times g\times \frac{y}{2}$$v = y\sqrt{\frac{g}{L}}$

4 years 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