# 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?

$\begin {array} {1 1} (a)\;\frac{L}{(1+\mu)} & \quad (b)\;\frac{\mu L}{(1+\mu)} \\ (c)\;\frac{L}{(1-\mu)} & \quad (d)\;\frac{\mu L}{(1-\mu)} \end {array}$

$(b)\;\frac{\mu L}{(1+\mu)}$
answered Nov 7, 2013 by 1 flag

Mass per unit length=M/L l =length of hanging mass (L-l)= mass on table surface So, uM/L(L-l)g=M/L.l.g U(L-l)= l On solving l=uL /1+u