
Initial height of liquid in container's of same cross section are x1 and x2 respectively. Now value is opened find loss in potential energy when water level be become same
loss in PE=Ui−Uf
=[ρ(A)x12x1+ρAx22x2]g−[ρA(2x1+x2)×(4x1+x2)×2]g
=ρAg[2x12+2x22−4(x1+x1)2]=4ρAg(x1−x2)2