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A chain of mass $m$ and length $l$ rests on a rough - surfaced table so that one of its ends hangs over the edge. The chain starts sliding off the table all by itself provided the over hanging part equal to $\frac{1}{3}$ of the chain length. then :

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the work done by the frictional forces acting on the table to completely sliding off the chain is $\frac{2}{3} m g l$

the work done by the frictional forces acting on the table to completely sliding off the chain is $[\frac{- m g l}{9}]$

coefficient of friction between table and chain is $\frac{1}{2}$

coefficient of friction between table and chain is $\frac{2}{3}$

answer is A, D.

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Detailed Solution

To slide,

$μ N = T = m g / 3$

or $μ (\frac{2 m g}{3}) = \frac{m g}{3} \Rightarrow μ = \frac{1}{2}$

$W = \int_{2 l / 3}^{0} f d x = \int_{2 l / 3}^{0} μ N d x = \int_{2 l / 3}^{0} (μ \frac{m}{l} x g) d x$

$= \frac{μ m g}{l} {|\frac{x^{2}}{2}|}_{2 l / 3}^{0} = - \frac{μ m g}{2 l} \times \frac{4 l^{2}}{9} = \frac{- \frac{1}{2} m g l \times 4}{2 \times 9} = \frac{- m g l}{9}$

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