Physics-
General
Easy
Question
A uniform chain is just at rest over a rough horizontal table with its (1/n) th part of length hanging vertically. The co-efficient of static friction between the chain and the table is,
The correct answer is:
We see that a portion of the chain is lying on the table top. Let the mass of that portion be m1. Let the mass of the remaining (hanging) portion of the chain be m2. Since the chain is at the point of slipping, the weight of the hanging portion of the chain counterbalances the maximum static frictional force fmax between m1 and the surface.
Þm2g = fmax ; m2g = mN1
where N1 – m1g = 0 for the equilibrium of the portion of chain lying as on the table.
Þm2g – mm1g = 0
Þm = 
= 
Þ m = 
Þm = 
Therefore (C) .
Note: If the direction of frictional force on the part of the chain lying on the table is not properly considered and length of the portions is altered contrary to the given statements, you will lead to wrong answer.
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