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A uniform chain has a mass M and length L.It is placed on a frictionless table with length lo hanging over the edge .The chain begins to slide down .Then the speed v with which the end slide down from the edge is given by1.{g(L+lo)/L}1/22.{g(L-lo)/L}1/23.{g(L2-lo2)/L}1/2

Kaushki , 6 Years ago
Grade 12th pass
anser 1 Answers
Arun

Last Activity: 6 Years ago

Dear student
 
MgL/2 - Mg (lo)² /2L = ½ * M *v²
 
 
v = \large \sqrt[g(L)2 - (lo)² /L ]
 
 
Regards
Arun (askIITians forum expert)

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