A uniform chain of length $L$ is lying on the horizontal table. If the coefficient of friction between the…
- $\frac{L}{(1+\mu)}$
- $\frac{\mu L}{(1+\mu)}$
- $\frac{L}{(1-\mu)}$
- $\frac{\mu L}{(1-\mu)}$
Solution

$\therefore \quad \mu=\frac{\text { Length hanging over the edge }}{\text { Length lying on the table }}$ (As the chain have uniform linear density) $\begin{array}{ll} \therefore & \mu=\frac{l^{\prime}}{\left(L-l^{\prime}\right)} \\ \Rightarrow & l^{\prime}=\frac{\mu L}{(1+\mu)} \end{array}$
Asked in: AP EAMCET 2011