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A homogeneous chain of length L lies on a table. The coefficient of friction between chain and table is $\mu$. The maximum length which can hang over the table in equilibrium is ( The vertical portion of table is smooth)

$(a)\;\bigg(\frac{\mu}{\mu+1}\bigg)L \quad (b)\;\bigg(\frac{1-\mu}{\mu}\bigg)L \quad (c)\;\bigg(\frac{1-\mu}{1+\mu}\bigg)L \quad (d)\;\bigg(\frac{2\mu}{2\mu+1}\bigg)L$

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Let 'l' be the desired length hanging out so that the chain is in equilibrium chain starts moving when not pulling force is greater than frictional force.
Let m be the mass of length of chain
$g \large\frac{m}{L}$$l=\mu \large\frac{m}{L}$$(L-l)g$
$l=\bigg(\large\frac{\mu}{\mu+1}\bigg)$$L$
Hence a is the correct answer.

answered Jul 10, 2013 by 1 flag
edited Jan 26, 2014 by meena.p
Plz explain the mass part

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