let the length of rope on one side be i and on other side be l-x0
mass of string on one side =lσ
mass of string on other side =(l-x0
for string to remain on pulley,the length must be equal on both side i.e one side go up by x0/2
and oter side come down by x0/2.
accln. of string as it moves from l-x0 to l-x0+dl
lσg-T=lσa
T-(l-x0)σg=(l-x0)σa
solving a=x0σg/(2lσ-x0σ)
final vel=0
u2=2as
=2∫x0σg/(2lσ-x0σ)
integrating betn limits 0 and x0/2 ,u can be calculated