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A uniform chain of length $L$ is lying on the horizontal table. If the coefficient of friction between the chain and the table top is $\mu$, what is the maximum length of the chain that can hang over the edge of the table without disturbing the rest of the chain on the table?
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The correct answer is:
$\frac{\mu L}{(1+\mu)}$
Let $l^{\prime}$ part of the chain is hanging over the edge of table without sliding.
$\therefore \quad \mu=\frac{\text { Length hanging over the edge }}{\text { Length lying on the table }}$
(As the chain have uniform linear density)
$\begin{array}{ll}
\therefore & \mu=\frac{l^{\prime}}{\left(L-l^{\prime}\right)} \\
\Rightarrow & l^{\prime}=\frac{\mu L}{(1+\mu)}
\end{array}$
$\therefore \quad \mu=\frac{\text { Length hanging over the edge }}{\text { Length lying on the table }}$
(As the chain have uniform linear density)
$\begin{array}{ll}
\therefore & \mu=\frac{l^{\prime}}{\left(L-l^{\prime}\right)} \\
\Rightarrow & l^{\prime}=\frac{\mu L}{(1+\mu)}
\end{array}$
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