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

\frac{l^{2}}{L - l}

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\frac{l}{L}

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\frac{l}{L - l}

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\frac{L - l}{l}

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Solution

The correct option is C $\frac{l}{L - l}$
Mass per unit length of the chain $= \frac{M}{L}$
Mass of the hanging part $= l (\frac{M}{L})$
Mass of the part on the table $= (L - l) (\frac{M}{L})$
Force equation is:
$μ \frac{M}{L} (L - l) g = \frac{M}{L} (l) g$
$μ = \frac{l}{L - l}$

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Similar questions

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

The correct option is C lL−lMass per unit length of the chain =MLMass of the hanging part =l(ML)Mass of the part on the table =(L−l)(ML)Force equation is:μML(L−l)g=ML(l)gμ=lL−l

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

The correct option is C $\frac{l}{L - l}$
Mass per unit length of the chain $= \frac{M}{L}$
Mass of the hanging part $= l (\frac{M}{L})$
Mass of the part on the table $= (L - l) (\frac{M}{L})$
Force equation is:
$μ \frac{M}{L} (L - l) g = \frac{M}{L} (l) g$
$μ = \frac{l}{L - l}$