Question

A uniform chain of length l and mass m is lying on a smooth table and one third of its length hanging find work

Answer 1

mass and length is constant so l/3 portion of length and m/3portion if its mass is hanging therefore w=mgh and due to centre of mass length considered to be used = l/3 ×1/2= l/6
so m/3 ×g× l/6 = mgl/18

Answer 2

by formula

$mgl \div 2 {n}^{2}$
mgl/2
$mgl \div 2 \times {1 \div 3}^{2}$