Question:

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?

Updated On: Aug 15, 2024
  • $\frac{L}{(1+\mu)}$
  • $\frac{\mu L}{(1+\mu)}$
  • $\frac{L}{(1-\mu)}$
  • $\frac{\mu L}{(1-\mu)}$
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The Correct Option is B

Solution and Explanation

Let $l'$ part of the chain is hanging over the edge of table without sliding.
$\therefore \mu=\frac{\text { Length hanging over the edge }}{\text { Length lying on the table }}$ (As the chain have uniform linear density) $\therefore \mu=\frac{l'}{\left(L-l^{'}\right)}$ $\Rightarrow l'=\frac{\mu L}{(1+\mu)}$
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