A uniform rope of length / lies on a table. If the coefficient of friction is μ, then the maximum length x of the part of this rope which can overhang from the edge of the table without sliding down is

Let l be the length of the chain and m be its mass.∴mass per unit length = (m / l)Weight of the hanging part =(m/l)×x×gThis weight provides tension on the portion of the chair lying on the table.weight of the chain on the table =(m/l)(l−x)gThis gives the normal reaction on the table.Hence the friction p .R is given byμR=μ(m/l)(l−x)gFor maximum length of the hanging part, we haveTension = Friction(m/l)(x×g)=μ(m/l)(l−x)gSolving, we get x=μlμ+1.