A uniform chain of 6m length is placed on a table such that a part...

Let, $\lambda =$mass per unit length of the chain and length of mass hanging through the table is $x$.

Normal force applied by the table on the chain will be, $N = \left(l - x\right)\lambdag$.

Tension developed in the chain will balance the gravitational force acting in the downward direction, $T = \lambdaxg$.

Maximum limiting force possible for the part of the table, {f}_\max = \muN = \mu\left(l - x\right)\lambdag

For equilibrium,

{f}_\max = T,

$⇒ \mu\left(l - x\right)\lambdag = \lambdaxg ⇒ \dfrac{1}{2} \times \left(6 - x\right) = x ⇒ x = 2m$