$A$ uniform chain of length $L$ hangs partly from a table and is kept in equilibrium by friction. If the maximum length of the chain that can hang without slipping is $l$,then the coefficient of friction between the table and the chain is:

$A$ small block starts slipping down from a point $B$ on an inclined plane $AB$,which is making an angle $\theta$ with the horizontal. The section $BC$ is smooth and the remaining section $CA$ is rough with a coefficient of friction $\mu$. It is found that the block comes to rest as it reaches the bottom (point $A$) of the inclined plane. If $BC = 2AC$,the coefficient of friction is given by $\mu = k \tan \theta$. The value of $k$ is

$A$ child weighing $25$ kg slides down a rope hanging from the branch of a tall tree. If the force of friction acting against him is $2$ $N$,the acceleration of the child is ........ $m/s^2$. (Take $g = 9.8$ $m/s^2$)

Starting from rest,a body slides down a $45^o$ inclined plane in twice the time it takes to slide down the same distance in the absence of friction. The coefficient of friction between the body and the inclined plane is:

$A$ block starts moving up an inclined plane of inclination $30^{\circ}$ with an initial velocity of $v_{0}$. It comes back to its initial position with velocity $\frac{v_{0}}{2}$. The value of the coefficient of kinetic friction between the block and the inclined plane is close to $\frac{I}{1000}$. The nearest integer to $I$ is......

$A$ chain of length $L$ rests on a rough table. If $\mu$ is the coefficient of friction,the maximum fraction of the chain that can hang over the table will be: