A container filled with liquid up to height h is placed on a smooth horizontal surface. The container is having a small hole at the bottom. As the liquid comes out from the hole, the container moves in a backward direction with acceleration a and finally, when all the liquid is drained out, it acquires a velocity v. Neglect mass of the container.In this case

The velocity with which the liquid comes out is v0=2ghLet cross-sectional area of the hole be A and cross-sectional area of the tank be A' . Then the force exerted by the ejecting fluid due to change in momentum on the container isF=ρAv2=ma,where m is the mass of the liquid inside the container⇒ρAv02=ρA'h×a ⇒ a=2gAA'Acceleration is independent of h.v=u+at ⇒v=2gAA't ⇒here t is the time in which the liquid comes out completely which is dependent on h.