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A chain of mass m and length l is kept on a horizontal frictionless table, such that (1)/(4) th of the length of the chain is hanging out of the table. Find the work done to pull the hanging part of the chain onto the table. |
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Answer» Solution :Mass per UNIT LENGTH of the chain `(m)/(l)` `:.` Mass of hanging part of the chain = `(m)/(l)xx(l)/(4)= (m)/(4)` `:.` Centre of gravity of this portion is at `(l)/(4xx2) =(l)/(8)` below the top of the table. `:. ` Work to be done to LIFT the hanging part of the chain `=(m)/(4)xxgxx(l)/(8)=(mgl)/(32)`. |
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