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A homogoneous chain of length L lies on ...

A homogoneous chain of length L lies on a table. The coefficient of friction between the chain and the table is `mu`. The maximum length which can hang over the table in equilibrium is

A

`((mu)/(mu+1))L`

B

`(1-mu )/( mu ))L`

C

`(1-mu )/(1+ mu )) L`

D

`((2 mu )/( mu+1))L`

Text Solution

Verified by Experts

The correct Answer is:
A
`((mu)/(mu+1))L`
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