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A chain is held on a frictionless talbe ...

A chain is held on a frictionless talbe with `L//4` hanging over. Knowing total mass of the chain is `M` and total length is `L`, the work required to slowly pull hanging part back to the table is `:`

A

`(MgL)/(16)`

B

`(MgL)/(8)`

C

`(MgL)/(32)`

D

`(MgL)/(24)`

Text Solution

Verified by Experts

The correct Answer is:
C
`W_(ext)=-W_(g)`
`=-((M)/(4))g((L)/(8))=(MgL)/(32).`
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