A uniform rope of length l lies on a table. If the coefficient of friction is $μ$ , then the maximum length $l_{1}$ of the hanging part of the rope which can overhang from the edge of the table without sliding down is

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$\frac{l}{μ}$

$\frac{l}{(μ + 1)}$

$\frac{μ l}{(μ + 1)}$

$\frac{μ l}{(μ - 1)}$

answer is C.

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Detailed Solution

In critical case, weight of hanging part= force of friction on the part of rope lying on table.

$\frac{m}{l} . l_{1} g = μ \frac{m}{l} (l - l_{1}) g$

Solving this we get,

$l_{1} = (\frac{μ}{1 + μ}) l$

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