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A uniform chain of length L changes partly from a table which is kept in equilibrium by friction. The maximum length that can Withstand without slipping is l, then coefficient of friction between the table and the chain is

  • a)
    1/L

  • b)
    1/L+l

  • c)
    1/L-1

  • d)
    L/L+1

Correct answer is option 'C'. Can you explain this answer?
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Explanation:

Given, a uniform chain of length L changes partly from a table which is kept in equilibrium by friction. The maximum length that can withstand without slipping is l.

Let the length of chain hanging freely from the table be x.

Then, the length of chain on the table = L - x.

For the chain to be in equilibrium, the weight of the chain hanging freely should be balanced by the force of friction between the chain and the table.

Let μ be the coefficient of friction between the chain and the table.

The weight of the chain hanging freely = mg, where m is the mass of the chain and g is the acceleration due to gravity.

The force of friction between the chain and the table = μN, where N is the normal reaction between the chain and the table.

The normal reaction N is equal to the weight of the chain on the table.

Hence, N = mg(L - x)/L.

Therefore, the force of friction = μmg(L - x)/L.

For the chain to be in equilibrium, the force of friction should be equal to the weight of the chain hanging freely.

Hence, μmg(L - x)/L = mgx/L.

Solving for x, we get:

x = μL/(1 + μ)

The maximum length of chain that can withstand without slipping is l.

Hence, μl/(1 + μ) = L - l.

Solving for μ, we get:

μ = l/(L - l) - 1/L

Hence, the coefficient of friction between the table and the chain is l/L - 1/(L - l).

Therefore, the correct option is C.
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