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A chain of uniform mass m and length L i...

A chain of uniform mass `m` and length `L` is held on a frictionless table in such a way that its `(1)/(n)`th part is hanging below the edge of table. The work done to pull the hanging part of chain is `:-`

A

`(mgL^(2))/(2n^(2))`

B

Zero

C

`(mgL)/(2n)`

D

`(mgL)/(2n^(2))`

Text Solution

Verified by Experts

The correct Answer is:
D

Required work done`=` change in potential energy of chain.
Now let Potential energy `(U)=0` at table level so potential energies of chain
`U_(i)=-mg((L)/(2n))=-((M)/(L))(L)/(n)g((L)/(2n))=-(MgL)/(2n^(2))`,
`U_(f)=0`
Work done `=U_(f)-U_(i)=0-(-(MgL)/(2n^(2)))=(MgL)/(2n^(2))`
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Knowledge Check

  • A block A of m ass rests on a horizontal table. A lig h t strin g connected to it passesover a frictionless pulley at the edge of table an d from its other end another block B of mass m_(2) is su sp en d e d . The co efficien t of k in eticfriction between the block and the table is mu_(k) hen the block A is sliding on the table, the ten sio n in the string i s :

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