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

\frac{l}{μ}

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\frac{1}{(μ + 1)}

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\frac{μ l}{(μ + l)}

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\frac{μ l}{(μ - 1)}

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Solution

The correct option is C $\frac{μ l}{(μ + l)}$
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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A uniform rope of length l lies on a table. If the coefficient of friction is μ, then the maximum length l1 of the hanging part of the rope which can overhang from the edge of the table without sliding down is

The correct option is C μl(μ+l)In critical case, weight of hanging part = force of friction on the part of rope lying on table. ∴mll1g=μml(l−l1)g Solving this we get, l1=(μ1+μ)l

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

The correct option is C $\frac{μ l}{(μ + l)}$
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,
$l_{1} = (\frac{μ}{1 + μ}) l$